ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test

More Exercises

Question 1.
Draw a circle of radius 3 cm. Mark its centre as C and mark a point P such that CP = 7 cm. Using ruler and compasses only, Construct two tangents from P to the circle.
Solution:
Steps of Construction :

  1. Draw a circle with centre C and radius 3 cm.
  2. Mark a point P such that CP = 7 cm.
  3. With CP as diameter, draw a circle intersecting the given circle at T and S.
  4. Join PT and PS.
  5. Draw a tangent at Q to the circle given. Which intersects PT at D.
  6. Draw the angle bisector of ∠PDQ intersecting CP at E.
  7. With centre E and radius EQ, draw a circle.
    It will touch the tangent T and PS and the given circle at Q.
    This is the required circle.
    ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test Q1.1

Question 2.
Draw a line AQ = 7 cm. Mark a point P on AQ such that AP = 4 cm. Using ruler and compasses only, construct :
(i) a circle with AP as diameter.
(ii) two tangents to the above circle from the point Q.
Solution:
Steps of construction :

  1. Draw a line segment AQ = 7 cm.
  2. From AQ,cut off AP = 4cm
  3. With AP as diameter draw a circle with centre O.
  4. Draw bisector of OQ which intersect OQ at M.
  5. With centre M and draw a circle with radius MQ
    which intersects the first circle at T and S.
  6. Join QT and QS.
    QT and QS are the tangents to the first circle.
    ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test Q2.1

Question 3.
Using ruler and compasses only, construct a triangle ABC having given c = 6 cm, b = 1 cm and ∠A = 30°. Measure side a. Draw carefully the circumcircle of the triangle.
Solution:
Steps of Construction :

  1. Draw a line segment AC = 7 cm.
  2. At C, draw a ray CX making an angle of 30°
  3. With centre A and radius 6 cm draw an arc
    which intersects the ray CX at B.
  4. Join BA.
  5. Draw perpendicular bisectors of AB and AC intersecting each other at O.
  6. With centre O and radius OA or OB or OC,
    draw a circle which will pass through A, B and C.
    This is the required circumcircle of ∆ABC
    ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test Q3.1

Question 4.
Using ruler and compasses only, construct an equilateral triangle of height 4 cm and draw its circumcircle.
Solution:
Steps of Construction :

  1. Draw a line XY and take a point D on it.
  2. At D, draw perpendicular and cut off DA = 4 cm.
  3. From A, draw rays making an angle of 30°
    on each side of AD meeting the line XY at B and C.
  4. Now draw perpendicular bisector of AC intersecting AD at O.
  5. With centre O and radius OA or OB or OC
    draw a circle which will pass through A, B and C.
    This is the required circumcircle of ∆ABC.
    ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test Q4.1

Question 5.
Using ruler and compasses only :
(i) Construct a triangle ABC with the following data: BC = 7 cm, AB = 5 cm and ∠ABC = 45°.
(ii) Draw the inscribed circle to ∆ABC drawn in part (i).
Solution:
Steps of construction :

  1. Draw a line segment BC = 7 cm.
  2. At B, draw a ray BX making an angle of 45° and cut off BA = 5 cm.
  3. Join AC.
  4. Draw the angle bisectors of ∠B and ∠C intersecting each other at I.
  5. From I, draw a perpendicular ID on BC.
  6. With centre, I and radius ID, draw a circle
    which touches the sides of ∆ABC at D, E and F respectively.
    This is the required inscribed circle.
    ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test Q5.1

Question 6.
Draw a triangle ABC, given that BC = 4cm, ∠C = 75° and that radius of the circumcircle of ∆ABC is 3 cm.
Solution:
Steps of Construction:

  1. Draw a line segment BC = 4 cm
  2. Draw the perpendicular bisector of BC.
  3. From B draw an arc of 3 cm radius which intersects the perpendicular bisector at O.
  4. Draw a ray CX making art angle of 75°
  5. With centre O and radius 3 cm draw a circle which intersects the ray CX at A.
  6. Join AB.
    ∆ABC is the required triangle
    ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test Q6.1

Question 7.
Draw a regular hexagon of side 3.5 cm construct its circumcircle and measure its radius.
Solution:
Steps of construction:

  1. Draw a regular hexagon ABCDEF whose each side is 3.5 cm.
  2. Draw the perpendicular bisector of AB and BC
    which intersect each other at O.
  3. Join OA and OB.
  4. With centre O and radius OA or OB, draw a circle
    which passes through A, B, C, D, E and P.
    Then this is the required circumcircle.
    ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test Q7.1

Question 8.
Construct a triangle ABC with the following data: AB = 5 cm, BC = 6 cm and ∠ABC = 90°.
(i) Find a point P which is equidistant from B and C and is 5 cm from A. How many such points are there ?
(ii) Construct a circle touching the sides AB and BC, and whose centre is equidistant from B and C.
Solution:
Steps of Construction :

  1. Draw a line segment BC = 6 cm.
  2. At B, draw a ray BX making an angle of 90° and cut off BA = 5 cm.
  3. Join AC.
  4. Draw the perpendicular bisector of BC.
  5. From A with 5 cm radius draw arc which intersects the perpendicular bisector of BC at P and P’.
    There are two points.
  6. Draw the angle bisectors of ∠B and ∠C intersecting at 0.
  7. From O, draw OD ⊥ BC.
  8. With centre O and radius OD, draw a circle which will touch the sides AB and BC.
    This is the required circle.
    ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test Q8.1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test are helpful to complete your math homework.

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5

More Exercise

Question 1.
The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 m, find its radius.
Solution:
Diameter of metallic sphere = 6 cm
Radius(r) = \(\\ \frac { 6 }{ 2 } \) = 3cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q1.1

Question 2.
The radius of a sphere is 9 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire in metres.
Solution:
Radius of sphere = 9 cm
Volume = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi \times { \left( 9 \right) }^{ 3 }{ cm }^{ 3 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q2.1
h = 97200 cm = 972 m
Length of wire = 972 m

Question 3.
A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.
Solution:
Radius of a solid hemisphere (r) = 8 cm
Volume = \(\frac { 2 }{ 3 } \pi { r }^{ 3 }=\frac { 2 }{ 3 } \pi \times { \left( 8 \right) }^{ 3 }{ cm }^{ 3 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q3.1

Question 4.
A rectangular water tank of base 11 m x 6 m contains water upto a height of 5 m. if the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.
Solution:
Base of a water tank = 11 m × 6 m
Height of water level in it (h) = 5 m
Volume of water =11 × 6 × 5 = 330 m³
Volume of water in the cylindrical tank
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q4.1

Question 5.
The rain water from a roof of dimensions 22 m x 20 m drains into a cylindrical vessel having diameter of base 2 m and height; 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.
Solution:
Dimensions of roof = 22 m × 20 m
Let rainfall = x m
.’. Volume of water = 22 × 20 × x m³
Volume of water in cylinder = 22 × 20 × x m³
Diameter of its base = 2 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q5.1

Question 6.
The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.
Solution:
Diameter of a cylinder = 40 cm
Radius (r) = \(\\ \frac { 40 }{ 2 } \) = 20 cm
Height(h) = 9 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q6.1

Question 7.
Eight metallic spheres, each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new (single) sphere.
Solution:
Radius of each metallic sphere (r) = 2 cm
Volume of one sphere = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q7.1

Question 8.
A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.
Solution:
Height of disc cylindrical shaped = 2.5 mm
and base radius = 12 cm
Volume of the disc = πr²h
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q8.1

Question 9.
Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.
Solution:
Weight of first sphere = 1 kg
and weight of second sphere = 7 kg
Radius of smaller sphere = 3 cm
Let r be the radius of a larger sphere
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q9.1
R = 3 x 2 = 6 cm
Diameter of big sphere = 2 x 6 = 12 cm

Question 10.
A hollow copper pipe of inner diameter 6 cm and outer diameter 10 cm is melted and changed into a solid circular cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.
Solution:
Inner diameter of a hollow pipe = 6 cm
and outer diameter = 10 cm
Inner radius (r) = \(\\ \frac { 6 }{ 2 } \) = 3cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q10.1

Question 11.
A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm and height is 72 cm, find the uniform thickness of the cylinder.
Solution:
Radius of a solid sphere (r) = 6 cm
Volume = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi \times { \left( 6 \right) }^{ 3 }{ cm }^{ 3 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q11.1

Question 12.
A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal of the tube is \(\\ \frac { 1 }{ 2 } \) cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone correct to one decimal place.
Solution:
Internal radius of a hollow metallic cylindrical tube (r) = 3 cm
and height (h) = 21 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q12.1

Question 13.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm. Find the height of the cone. (2002)
Solution:
Internal diameter of a hollow sphere = 4 cm
and external diameter = 8 cm
Internal radius (r) = 2 cm
and external radius (R) = 4 cm
Volume of hollow sphere
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q13.1

Question 14.
A well with inner diameter 6 m is dug 22 m deep. Soil taken out of it has been spread evenly all round it to a width of 5 m to form an embankment. Find the height of the embankment.
Solution:
Inner diameter of a well = 6 m
Depth (h) = 22 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q14.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q14.2

Question 15.
A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.
Solution:
Internal diameter of cylindrical can = 21 cm
Radius (R) = \(\\ \frac { 21 }{ 2 } \) cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q15.2

Question 16.
There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water go down when the sphere is removed?
Solution:
Radius of the cylindrical jar (R) = 8 cm
Height of water level (h) = 14 cm
Volume of water = πR²h
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q16.1

Question 17.
A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid cone submerged? (2002)
Solution:
Height of conical vessel (h) = 20 cm
and diameter = 16.8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q17.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q17.2

Question 18.
A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder?
Solution:
Radius of a solid metallic cylindrical (r) = 14 cm
and height (h) = 12 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q18.1

Question 19.
How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm x 11 cm x 12 cm?
Solution:
Diameter of a shot = 3 cm
Radius (r) = \(\\ \frac { 3 }{ 2 } \) cm
Volume of one shot = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q19.1

Question 20.
How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm?
Solution:
Diameter of lead shot = 4 cm
Radius (r) = \(\\ \frac { 4 }{ 2 } \) = 2 cm and
volume =\(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q20.1

Question 21.
Find the number of metallic circular discs with 1.5 cm base diameter and height 0.2 cm to be melted to form a circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
Radius of the circular disc (r) = 0.75 cm
Height of circular disc (h) = 0.2 cm
Radius of cylinder (R) = 2.25 cm
Height of cylinder (H) = 10 cm
Now,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q21.1

Question 22.
A solid metal cylinder of radius 14 cm and height 21 cm is melted down and recast into spheres of radius 3.5 cm. Calculate the number of spheres that can be made.
Solution:
Radius of a solid metallic cylinder (r) = 14 cm
and height (h) = 21 cm
Volume of cylinder = πr²h
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q22.1

Question 23.
A metallic sphere of radius 10.5 cm is melted and then recast into small cenes, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained. (2005)
Solution:
Radius of a metallic sphere (r) = 10.5 cm
Volume = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
= \(\\ \frac { 4 }{ 3 } \) × π × 10.5 × 10.5 × 10.5 = 1543.5π cm³
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q23.1

Question 24.
A certain number of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the number of cones. (2016)
Solution:
Radius of each cone (r) = 2 cm
and height (h) = 3 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q24.1

Question 25.
A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, \(\\ \frac { 2 }{ 5 } \) of the water flows out. Find the number of lead shots dropped into the vessel. (2003)
Solution:
Radius of the top of the inverted conical vessel (R) = 2.5 cm
and height (h)= 11 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q25.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q25.2

Question 26.
The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained? (2007)
Solution:
Surface area of a metallic sphere = 616 cm²
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q26.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q26.2

Question 27.
The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate
(i) the radius of the solid sphere.
(ii) the number of cones recast. (Use π = 3.14).
Solution:
Surface area of a solid metallic sphere = 1256 cm²
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q27.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q27.2

Question 28.
Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboid pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?
Solution:
Speed of water flow = 15 km/h
Diameter of pipe = 14 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q28.1

Question 29.
A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate :
(i) the total surface area of the can in contact with water when the sphere is in it.
(ii) the depth of the water in the can before the sphere was put into the can. Given your answer as proper fractions.
Solution:
Radius of a cylindrical can = 3.5 cm
Radius of the sphere = 3.5 cm
and height of water level in the can = 3.5 × 2 = 7 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q29.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 Q29.2

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2

More Exercises

Question 1.
If O is the centre of the circle, find the value of x in each of the following figures (using the given information):
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q1.1
Solution:
From the figure
(i) ABCD is a cyclic quadrilateral
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q1.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q1.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q1.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q1.5

Question 2.
(a) In the figure (i) given below, O is the centre of the circle. If ∠AOC = 150°, find (i) ∠ABC (ii) ∠ADC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q2.1
(b) In the figure (i) given below, AC is a diameter of the given circle and ∠BCD = 75°. Calculate the size of (i) ∠ABC (ii) ∠EAF.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q2.2
Solution:
(a) Given, ∠AOC = 150° and AD = CD
We know that an angle subtends by an arc of a circle
at the centre is twice the angle subtended by the same arc
at any point on the remaining part of the circle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q2.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q2.4

Question 3.
(a) In the figure, (i) given below, if ∠DBC = 58° and BD is a diameter of the circle, calculate:
(i) ∠BDC (ii) ∠BEC (iii) ∠BAC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q3.1
(b) In the figure (if) given below, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find:
(i) ∠CAD (ii) ∠CBD (iii) ∠ADC (2008)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q3.2
Solution:
(a) ∠DBC = 58°
BD is diameter
∠DCB = 90° (Angle in semi circle)
(i) In ∆BDC
∠BDC + ∠DCB + ∠CBD = 180°
∠BDC = 180°- 90° – 58° = 32°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q3.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q3.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q3.5

Question 4.
(a) In the figure given below, ABCD is a cyclic quadrilateral. If ∠ADC = 80° and ∠ACD = 52°, find the values of ∠ABC and ∠CBD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q4.1
(b) In the figure given below, O is the centre of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q4.2
Solution:
(a) In the given figure, ABCD is a cyclic quadrilateral
∠ADC = 80° and ∠ACD = 52°
To find the measure of ∠ABC and ∠CBD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q4.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q4.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q4.5

Question 5.
(a) In the figure (i) given below, ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. Given that ∠BEF = 80°, find ∠ABC.
(b) In the figure (ii) given below, ABCD is a cyclic trapezium in which AD is parallel to BC and ∠B = 70°, find:
(i)∠BAD (ii) DBCD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q5.1
Solution:
(a) ADFE is a cyclic quadrilateral
Ext. ∠FEB = ∠ADF
⇒ ∠ADF = 80°
ABCD is a parallelogram
∠B = ∠D = ∠ADF = 80°
or ∠ABC = 80°
(b)In trapezium ABCD, AD || BC
(i) ∠B + ∠A = 180°
⇒ 70° + ∠A = 180°
⇒ ∠A = 180° – 70° = 110°
∠BAD = 110°
(ii) ABCD is a cyclic quadrilateral
∠A + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180° – 110° = 70°
∠BCD = 70°

Question 6.
(a) In the figure given below, O is the centre of the circle. If ∠BAD = 30°, find the values of p, q and r.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q6.1
(b) In the figure given below, two circles intersect at points P and Q. If ∠A = 80° and ∠D = 84°, calculate
(i) ∠QBC (ii) ∠BCP
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q6.2
Solution:
(a) (i) ABCD is a cyclic quadrilateral
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q6.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q6.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q6.5

Question 7.
(a) In the figure given below, PQ is a diameter. Chord SR is parallel to PQ.Given ∠PQR = 58°, calculate (i) ∠RPQ (ii) ∠STP
(T is a point on the minor arc SP)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q7.1
(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q7.2
Solution:
(a) In ∆PQR,
∠PRQ = 90° (Angle in a semi circle) and ∠PQR = 58°
∠RPQ = 90° – ∠PQR = 90° – 58° = 32°
SR || PQ (given)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q7.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q7.4

Question 8.
(a) In the figure (i) given below, AB is a diameter of the circle. If ∠ADC = 120°, find ∠CAB.
(b) In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, the sides AD and BC are produced to meet at F. If x : y : z = 3 : 4 : 5, find the values of x, y and z.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q8.1
Solution:
(a) Construction: Join BC, and AC then
ABCD is a cyclic quadrilateral.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q8.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q8.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q8.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q8.5

Question 9.
(a) In the figure (i) given below, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E. If ∠ADE = 70° and ∠OBA = 45°, calculate
(i) ∠OCA (ii) ∠BAC
(b) In figure (ii) given below, ABF is a straight line and BE || DC. If ∠DAB = 92° and ∠EBF = 20°, find :
(i) ∠BCD (ii) ∠ADC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q9.1
Solution:
(a) ABCD is a cyclic quadrilateral
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q9.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q9.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q9.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q9.5

Question 10.
(a) In the figure (ii) given below, PQRS is a cyclic quadrilateral in which PQ = QR and RS is produced to T. If ∠QPR = 52°, calculate ∠PST.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q10.1
(b) In the figure (ii) given below, O is the centre of the circle. If ∠OAD = 50°, find the values of x and y.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q10.2
Solution:
(a) PQRS is a cyclic quadrilateral in which
PQ = QR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q10.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q10.4

Question 11.
(a) In the figure (i) given below, O is the centre of the circle. If ∠COD = 40° and ∠CBE = 100°, then find :
(i) ∠ADC
(ii) ∠DAC
(iii) ∠ODA
(iv) ∠OCA.
(b) In the figure (ii) given below, O is the centre of the circle. If ∠BAD = 75° and BC = CD, find :
(i) ∠BOD
(ii) ∠BCD
(iii) ∠BOC
(iv) ∠OBD (2009)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q11.1
Solution:
(a) (i) ∴ ABCD is a cyclic quadrilateral.
∴ Ext. ∠CBE = ∠ADC
⇒ ∠ADC = 100°
(ii) Arc CD subtends ∠COD at the centre
and ∠CAD at the remaining part of the circle
∴ ∠COD = 2 ∠CAD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q11.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q11.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q11.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q11.5

Question 12.
In the given figure, O is the centre and AOE is the diameter of the semicircle ABCDE. If AB = BC and ∠AEC = 50°, find :
(i) ∠CBE
(ii) ∠CDE
(iii) ∠AOB.
Prove that OB is parallel to EC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q12.1
Solution:
In the given figure,
O is the centre of the semi-circle ABCDE
and AOE is the diameter. AB = BC, ∠AEC = 50°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q12.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q12.3

Question 13.
(a) In the figure (i) given below, ED and BC are two parallel chords of the circle and ABE, ACD are two st. lines. Prove that AED is an isosceles triangle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q13.1
(b) In the figure (ii) given below, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = RS.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q13.2
Solution:
(a) Given: Chord BC || ED,
ABE and ACD are straight lines.
To Prove: ∆AED is an isosceles triangle.
Proof: BCDE is a cyclic quadrilateral.
Ext. ∠ABC = ∠D …(i)
But BC || ED (given)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q13.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q13.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q13.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q13.6

Question 14.
In the given figure, ABC is an isosceles triangle in which AB = AC and circle passing through B and C intersects sides AB and AC at points D and E. Prove that DE || BC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q14.1
Solution:
In the given figure,
∆ABC is an isosceles triangle in which AB = AC.
A circle passing through B and C intersects
sides AB and AC at D and E.
To prove: DE || BC
Construction : Join DE.
∵ AB = AC
∠B = ∠C (angles opposite to equal sides)
But BCED is a cyclic quadrilateral
Ext. ∠ADE = ∠C
= ∠B (∵ ∠C = ∠B)
But these are corresponding angles
DE || BC
Hence proved.

Question 15.
(a) Prove that a cyclic parallelogram is a rectangle.
(b) Prove that a cyclic rhombus is a square.
Solution:
(a) ABCD is a cyclic parallelogram.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q15.1
To prove: ABCD is a rectangle
Proof: ABCD is a parallelogram
∠A = ∠C and ∠B = ∠D
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q15.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q15.3

Question 16.
In the given figure, chords AB and CD of the circle are produced to meet at O. Prove that triangles ODB and OAC are similar. Given that CD = 2 cm, DO = 6 cm and BO = 3 cm, area of quad. CABD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q16.1
Solution:
In the given figure, AB and CD are chords of a circle.
They are produced to meet at O.
To prove : (i) ∆ODB ~ ∆OAC
If CD = 2 cm, DO = 6 cm, and BO = 3 cm
To find : AB and also area of the
\(\frac { Quad.ABCD }{ area\quad of\quad \Delta OAC } \)
Construction : Join AC and BD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q16.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q16.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 Q16.4

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1

More Exercises

Question 1.
State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm):
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q1.2
Solution:
Given
(i) In ∆ABC and PQR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q1.3

Question 2.
It is given that ∆DEF ~ ∆RPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P ? Why?
Solution:
∆DEF ~ ∆RPQ
∠D = ∠R and ∠F = ∠Q not ∠P
No, ∠F ≠ ∠P

Question 3.
If in two right triangles, one of the acute angle of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles are similar? Why?
Solution:
In two right triangles,
one of the acute angles of the one triangle is
equal to an acute angle of the other triangle.
The triangles are similar. (AAA axiom)

Question 4.
In the given figure, BD and CE intersect each other at the point P. Is ∆PBC ~ ∆PDE? Give reasons for your answer.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q4.1
Solution:
In the given figure, two line segments intersect each other at P.
In ∆BCP and ∆DEP
∠BPC = ∠DPE
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q4.2

Question 5.
It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm.
Find the lengths of the remaining sides of the triangles.
Solution:
∆ABC ~ ∆EDF
AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q5.1

Question 6.
(a) If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, then find the perimeter of ∆ABC.
(b) If ∆ABC ~ ∆PQR, Perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then find the length of AC.
Solution:
(a) ∆ABC ~ ∆DEF
AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q6.2

Question 7.
Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4 cm, 7 cm and 8 cm.
Solution:
Let ∆ABG ~ ∆DEF in which shortest side of
∆ABC is BC = 6 cm.
In ∆DEF, DE = 8 cm, EF = 4 cm and DF = 7 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q7.1

Question 8.
(a) In the figure given below, AB || DE, AC = , 3 cm, CE = 7.5 cm and BD = 14 cm. Calculate CB and DC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q8.1
(b) In the figure (2) given below, CA || BD, the lines AB and CD meet at G.
(i) Prove that ∆ACO ~ ∆BDO.
(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q8.2
Solution:
(a) In the given figure,
AB||DE, AC = 3 cm, CE = 7.5 cm, BD = 14 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q8.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q8.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q8.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q8.6

Question 9.
(a) In the figure
(i) given below, ∠P = ∠RTS.
Prove that ∆RPQ ~ ∆RTS.
(b) In the figure (ii) given below,
∠ADC = ∠BAC. Prove that CA² = DC x BC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q9.1
Solution:
(a) In the given figure, ∠P = ∠RTS
To prove : ∆RPQ ~ ∆RTS
Proof : In ∆RPQ and ∆RTS
∠R = ∠R (common)
∠P = ∠RTS (given)
∆RPQ ~ ∆RTS (AA axiom)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q9.2

Question 10.
(a) In the figure (1) given below, AP = 2PB and CP = 2PD.
(i) Prove that ∆ACP is similar to ∆BDP and AC || BD.
(ii) If AC = 4.5 cm, calculate the length of BD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q10.1
(b) In the figure (2) given below,
∠ADE = ∠ACB.
(i) Prove that ∆s ABC and AED are similar.
(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q10.2
(c) In the figure (3) given below, ∠PQR = ∠PRS. Prove that triangles PQR and PRS are similar. If PR = 8 cm, PS = 4 cm, calculate PQ.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q10.3
Solution:
In the given figure,
AP = 2PB, CP = 2PD
To prove:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q10.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q10.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q10.6
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q10.7

Question 11.
In the given figure, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC. Prpve that BM x NP = CN x MP.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q11.1
Solution:
In the given figure, ABC in which AB = AC.
P is a point on BC such that PM ⊥ AB and PN ⊥ AC
To prove : BM x NP = CN x MP
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q11.2

Question 12.
Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
Solution:
Given : ∆ABC ~ ∆PQR .
To prove : Ratio in their perimeters k
the same as the ratio in their corresponding sides.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q12.1

Question 13.
In the given figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that \(\frac { AO }{ OC } =\frac { BO }{ OD } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q13.1
Using the above result, find the value(s) of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.
Solution:
ABCD is a trapezium in which AB || DC
Diagonals AC and BD intersect each other at O.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q13.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q13.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q13.4

Question 14.
In ∆ABC, ∠A is acute. BD and CE are perpendicular on AC and AB respectively. Prove that AB x AE = AC x AD.
Solution:
In ∆ABC, ∠A is acute
BD and CE are perpendiculars on AC and AB respectively
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q14.1

Question 15.
In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC. Prove that \(\frac { BE }{ DE } =\frac { AC }{ BC } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q15.1
Solution:
In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC
To prove : \(\frac { BE }{ DE } =\frac { AC }{ BC } \)
Proof: In ∆ABC and ∆DEB
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q15.2

Question 16.
(a) In the figure (1) given below, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that ∆ABE ~ ∆CFB.
(b) In the figure (2) given below, PQRS is a parallelogram; PQ = 16 cm, QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.
(i) Prove that triangle RLQ is similar to triangle PLN. Hence, find PN.
(ii) Name a triangle similar to triangle RLM. Evaluate RM.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q16.1
Solution:
(a) In the given figure, ABCD is a ||gm
E is a point on AD and produced
and BE intersects CD at F.
To prove : ∆ABE ~ ∆CFB
Proof : In ∆ABE and ∆CFB
∠A = ∠C (opposite angles of a ||gm)
∠ABE = ∠BFC (alternate angles)
∆ABE ~ ∆CFB (AA axiom)
(b) In the given figure, PQRS is a ||gm PQ = 16 cm,
QR = 10 cm
L is a point on PR such that
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q16.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q16.3

Question 17.
The altitude BN and CM of ∆ABC meet at H. Prove that
(i) CN . HM = BM . HN .
(ii) \(\frac { HC }{ HB } =\sqrt { \frac { CN.HN }{ BM.HM } } \)
(iii) ∆MHN ~ ∆BHC.
Solution:
In the given figure, BN ⊥ AC and CM ⊥ AB of ∆ABC
which intersect each other at H.
To prove:
(i) CN.HM = BM.HN
(ii) \(\frac { HC }{ HB } =\sqrt { \frac { CN.HN }{ BM.HM } } \)
(iii) ∆MHN ~ ∆BHC.
Construction: Join MN
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q17.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q17.2

Question 18.
In the given figure, CM and RN are respectively the medians of ∆ABC and ∆PQR. If ∆ABC ~ ∆PQR, prove that:
(i) ∆AMC ~ ∆PNR
(ii) \(\frac { CM }{ RN } =\frac { AB }{ PQ } \)
(iii) ∆CMB ~ ∆RNQ.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q18.1
Solution:
In the given figure, CM and RN are medians of ∆ABC and ∆PQR
respectively and ∆ABC ~ ∆PQR
To prove:
(i) ∆AMC ~ ∆PNR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q18.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q18.3>

Question 19.
In the given figure, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. Prove that
(i) EF = FC
(ii) AG : GD = 2 : 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q19.1
Solution:
In the given figure,
AD and BE are the medians of ∆ABC
intersecting each other at G
DF || BE is drawn
To prove :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q19.2

Question 20.
(a) In the figure given below, AB, EF and CD are parallel lines. Given that AB =15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate
(i) EF
(ii) AC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q20.1
(b) In the figure given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm, BE = x and AE = y. Find the values of x and y.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q20.2
Solution:
In the given figure,
AB || EF || CD
AB = 15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm
Calculate :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q20.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q20.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q20.5

Question 21.
In the given figure, ∠A = 90° and AD ⊥ BC If BD = 2 cm and CD = 8 cm, find AD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q21.1
Solution:
In ∆ABC, we have ∠A = 90°
Also, AD ⊥ BC
Now,
In ∆ABC, we have,
∠BAC = 90°
⇒ ∠BAD + ∠DAC = 90°…..(i)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q21.2

Question 22.
A 15 metres high tower casts a shadow of 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.
Solution:
Height of a tower AB = 15 m
and its shadow BC = 24 m
At the same time and position
Let the height of a telephone pole DE = x m
and its shadow EF = 16 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q22.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q22.2

Question 23.
A street light bulb is fixed on a pole 6 m above the level of street. If a woman of height casts a shadow of 3 m, find how far she is away from the base of the pole?
Solution:
Height of height pole(AB) = 6m
and height of a woman (DE) = 1.5 m
Here shadow EF = 3 m .
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 Q23.1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test

More Exercises

Question 1.
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, – 3). If origin is the mid-point of the base BC, find the coordinates of the points A and B
Solution:
Base BC of an equilateral ∆ABC lies on y-axis
co-ordinates of point C are (0, – 3),
origin (0, 0) is the mid-point of BC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q1.2

Question 2.
A and B have co-ordinates (4, 3) and (0, 1), Find
(i) the image A’ of A under reflection in the y – axis.
(ii) the image of B’ of B under reflection in the lineAA’.
(iii) the length of A’B’.
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q2.1
(i) Co-ordinates of A’, the image of A (4, 3)
reflected in y-axis will be ( – 4, 3).
(ii) Co-ordinates of B’ the image of B (0, 1)
reflected in the line AA’ will be (0, 5).
(iii) Length A’B’
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q2.2

Question 3.
Find the co-ordinates of the point that divides the line segment joining the points P (5, – 2) and Q (9, 6) internally in the ratio of 3 : 1.
Solution:
Let R be the point whose co-ordinates are (x, y)
which divides PQ in the ratio of 3:1.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q3.1

Question 4.
Find the coordinates of the point P which is three-fourth of the way from A (3, 1) to B ( – 2, 5).
Solution:
Co-ordinates of A (3, 1) and B ( – 2, 5)
P lies on AB such that
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q4.1

Question 5.
P and Q are the points on the line segment joining the points A (3, – 1) and B ( – 6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q.
Solution:
Given
AP = PQ = QB
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q5.2

Question 6.
The centre of a circle is (α + 2, α – 5). Find the value of a given that the circle passes through the points (2, – 2) and (8, – 2).
Solution:
Let A (2, -2), B (8, -2) and centre of the circle be
O (α + 2, α – 5)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q6.2

Question 7.
The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.
Solution:
Given
(3, 5) is the mid-point of A (2, p) and B (q, 4)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q7.1

Question 8.
The ends of a diameter of a circle have the co-ordinates (3, 0) and ( – 5, 6). PQ is another diameter where Q has the coordinates ( – 1, – 2). Find the co-ordinates of P and the radius of the circle.
Solution:
Let AB be the diameter where co-ordinates of
A are (3, 0) and of B are (-5, 6).
Co-ordinates of its origin O will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q8.2

Question 9.
In what ratio does the point ( – 4, 6) divide the line segment joining the points A( – 6, 10) and B (3, – 8) ?
Solution:
Let the point (-4, 6) divides the line segment joining the points
A (-6, 10) and B (3, -8), in the ratio m : n
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q9.1

Question 10.
Find the ratio in which the point P ( – 3, p) divides the line segment joining the points ( – 5, – 4) and ( – 2, 3). Hence find the value of p.
Solution:
Let P (-3, p) divides AB in the ratio of m1 : m2 coordinates of
A (-5, -4) and B (-2, 3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q10.2

Question 11.
In what ratio is the line joining the points (4, 2) and (3, – 5) divided by the x-axis? Also find the co-ordinates of the point of division.
Solution:
Let the point P which is on the x-axis, divides the line segment
joining the points A (4, 2) and B (3, -5) in the ratio of m1 : m2.
and let co-ordinates of P be (x, 0)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q11.1

Question 12.
If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points ( – 4 – 3) and (6, 3). Hence, find the co-ordinates of P.
Solution:
Let co-ordinates of A be (-4, 3) and of B (6, 3) and of P be (2, y)
Let the ratio in which the P divides AB be m1 : m2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q12.1

Question 13.
Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, – 2) and B (3, 7). Also find the co-ordinates of the point of division.
Solution:
Points are given A (2, -2), B (3, 7)
and let the line 2x + y – 4 = 0 divides AB in the ratio m1 : m2
at P and let co-ordinates of
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q13.1

Question 14.
The point A(2, – 3) is reflected in the v-axis onto the point A’. Then the point A’ is reflected in the line x = 4 onto the:point A”.
(i) Write the coordinates of A’ and A”.
(ii) Find the ratio in which the line segment AA” is divided by the x-axis. Also find the coordinates of the point of division.
Solution:
A’ is the reflection of A(2, -3) in the x-axis
(i) ∴ Co-ordinates of A’ will be (2, 3)
Draw a line x = 4 which is parallel to y-axis
A” is the reflection of A’ (2, 3)
∴Co-ordinates OA” will be (6, 3)
(ii) Join AA” which intersects x-axis at P whose
co-ordinate are (4, 0)
Let P divide AA” in the ratio in m1 : m2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q14.1
Hence P(4, 0) divides AA” in the ratio 1 : 1

Question 15.
ABCD is a parallelogram. If the coordinates of A, B and D are (10, – 6), (2, – 6) and (4, – 2) respectively, find the co-ordinates of C.
Solution:
Let the co-ordinates of C be (x, y) and other three vertices
of the given parallelogram are A (10, – 6), B, (2, – 6) and D (4, – 2)
∴ ABCD is a parallelogram
Its diagonals bisect each other.
Let AC and BD intersect each other at O.
∴O is mid-points of BD
∴ Co-ordinates of O will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q15.1

Question 16.
ABCD is a parallelogram whose vertices A and B have co-ordinates (2, – 3) and ( – 1, – 1) respectively. If the diagonals of the parallelogram meet at the point M(1, – 4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. find the perimeter of the parallelogram.
Solution:
ABCD is a || gm , m which co-ordinates of A are (2, -3) and B (-1, -1)
Its diagonals AC and BD bisect each other at M (1, -4)
∴ M is the midpoint of AC and BD
Let co-ordinates of C be (x1, y1) and of D be (x2, y2)
when M is the midpoint of AC then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q16.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q16.2

Question 17.
In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q17.1
Solution:
A lies on x-axis and
B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y)
and P (3, 1) divides it in the ratio of 2 : 3.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q17.2

Question 18.
Given, O, (0, 0), P(1, 2), S( – 3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram.
Find : (i) the co-ordinates of Q.
(ii)the co-ordinates of R.
(iii) the ratio in which RQ is divided by y-axis.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q18.1
Solution:
(i) Let co-ordinates of Q be (x’, y’) and of R (x”, y”)
Point P (1, 2) divides OQ in the ratio of 2 : 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q18.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q18.3

Question 19.
If A (5, – 1), B ( – 3, – 2) and C ( – 1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.
Solution:
A (5, -1), B (-3, -2) and C (-1, 8) are the vertices of ∆ABC
D, E and F are the midpoints of sides BC, CA and AB respectively
and G is the centroid of the ∆ABC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q19.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test Q19.2

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test

More Exercises

Question 1.
Find the equation of a line whose inclination is 60° and y-intercept is – 4.
Solution:
Angle of inclination = 60°
Slope = tan θ = tan 60° = √3
Equation of the line will be,
y = mx + c = √3x + ( – 4)
⇒ y – √3x – 4

Question 2.
Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.
Solution:
Slope of the line 3y + 2x = 12
⇒ 3y = 12 – 2x
⇒ 3y = -2x + 12
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q2.1

Question 3.
If the equation of a line is y – √3x + 1, find its inclination.
Solution:
In the line
y = √3 x + 1
Slope = √3
⇒ tan θ = √3
⇒ θ = 60° (∵ tan 60° = √3)

Question 4.
If the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.
Solution:
The equation of line y = mx + c
∵ it passes through (2, – 4) and ( – 3, 1)
Now substituting the value of these points -4 = 2m + c …(i)
and 1 = -3m + c …(ii)
Subtracting we get,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q4.1

Question 5.
If the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.
Solution:
Let the points to be A (1, 4), B (3, -2) and C (p, -5) are collinear and let B (3, -2)
divides AC in the ratio of m1 : m2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q5.1

Question 6.
Find the inclination of the line joining the points P (4, 0) and Q (7, 3).
Solution:
Slope of the line joining the points P (4, 0) and Q (7, 3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q6.1

Question 7.
Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to \(– \frac { 3 }{ 7 } \)
Solution:
Equation of lines are
2x + y = 5 …(i)
x – 2y = 5 …(ii)
Multiply (i) by 2 and (ii) by 1, we get
4x + 2y = 10
x – 2y = 5
Adding we get,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q7.1

Question 8.
If point A is reflected in the y-axis, the co-ordinates of its image A1, are (4, – 3),
(i) Find the co-ordinates of A
(ii) Find the co-ordinates of A2, A3 the images of the points A, A1, Respectively under reflection in the line x = – 2
Solution:
(i) ∵ A is reflected in the y-axis and its image is A1 (4, -3)
Co-ordinates of A will be (-4, -3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q8.2

Question 9.
If the lines \(\frac { x }{ 3 } +\frac { y }{ 4 } =7 \) and 3x + ky = 11 are perpendicular to each other, find the value of k.
Solution:
Given Equation of lines are
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q9.1

Question 10.
Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).
Solution:
The equation of the line is x – 2y + 8 = 0
⇒ 2y = x + 8
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q10.1

Question 11.
Write down the equation of the line passing through ( – 3, 2) and perpendicular to the line 3y = 5 – x.
Solution:
Equations of the line is
3y = 5 – x ⇒ 3y = -x + 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q11.1

Question 12.
Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.
Solution:
Let slope of the line joining the points A (1, 2) and B (6, 7) be m1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q12.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q12.2

Question 13.
The points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.
Solution:
Slope of line AC (m1)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q13.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q13.2

Question 14.
A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q14.1
(i) the co-ordinates of A and B
(ii) the equation of the line AB
Solution:
A lies on x-axis and B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y)
and P (2, 1) divides BA in the ratio 3 : 1.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q14.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q14.3

Question 15.
A straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the line passes through the point ( – 3, 8), find its equation.
Solution:
Let the line make intercept a and b with the
x-axis and y-axis respectively then the line passes through
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q15.2

Question 16.
If the coordinates of the vertex A of a square ABCD are (3, – 2) and the quation of diagonal BD is 3 x – 7 y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the centre of the square.
Solution:
Co-ordinates of A are (3, -2).
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q16.1
Diagonals AC and BD of the square ABCD
bisect each other at right angle at O.
∴ O is the mid-point of AC and BD Equation
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q16.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q16.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q16.4

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS

More Exercises

Choose the correct answer from the given four options (1 to 13) :

Question 1.
The slope of a line parallel to y-axis is
(a) 0
(b) 1
(c) – 1
(d) not defined
Solution:
Slope of a line parallel to y-axis is not defined. (b)

Question 2.
The slope of a line which makes an angle of 30° with the positive direction of x-axis is
(a) 1
(b) \(\frac { 1 }{ \sqrt { 3 } } \)
(c) √3
(d) \(– \frac { 1 }{ \sqrt { 3 } } \)
Solution:
Slope of a line which makes an angle of 30°
with positive direction of x-axis = tan 30°
= \(\frac { 1 }{ \sqrt { 3 } } \) (b)

Question 3.
The slope of the line passing through the points (0, – 4) and ( – 6, 2) is
(a) 0
(b) 1
(c) – 1
(d) 6
Solution:
Slope of the line passing through the points (0, -4) and (-6, 2)
\(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 2+4 }{ -6-0 } =\frac { 6 }{ -6 } =-1 \) (c)

Question 4.
The slope of the line passing through the points (3, – 2) and ( – 7, – 2) is
(a) 0
(b) 1
(c) \(– \frac { 1 }{ 10 } \)
(d) not defined
Solution:
Slope of the line passing through the points (3, -2) and (-7, -2)
\(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -2+2 }{ -7-3 } =\frac { 0 }{ -10 } =0 \) (a)

Question 5.
The slope of the fine passing through the points (3, – 2) and (3, – 4) is
(a) – 2
(b) 0
(c) 1
(d) not defined
Solution:
The slope of the line passing through (3, -2) and (3, -4)
\(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -4+2 }{ 3-3 } =\frac { -2 }{ 0 } \) (d)

Question 6.
The inclination of the line y = √3x – 5 is
(a) 30°
(b) 60°
(c) 45°
(d) 0°
Solution:
The inclination of the line y = √3x – 5 is
√3 = tan 60° = 60° (b)

Question 7.
If the slope of the line passing through the points (2, 5) and (k, 3) is 2, then the value of k is
(a) -2
(b) -1
(c) 1
(d) 2
Solution:
Slope of the line passing through the points (2, 5) and (k, 3) is 2, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS Q7.1

Question 8.
The slope of a line parallel to the line passing through the points (0, 6) and (7, 3) is
(a) \(– \frac { 1 }{ 5 } \)
(b) \(\\ \frac { 1 }{ 5 } \)
(c) -5
(d) 5
Solution:
Slope of the line parallel to the line passing through (0, 6) and (7, 3)
Slope of the line = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 3-6 }{ 7-0 } =\frac { -3 }{ 7 } \) (b)

Question 9.
The slope of a line perpendicular to the line passing through the points (2, 5) and ( – 3, 6) is
(a) \(– \frac { 1 }{ 5 } \)
(b) \(\\ \frac { 1 }{ 5 } \)
(c) -5
(d) 5
Solution:
Slope of the line joining the points (2, 5), (-3, 6)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS Q9.1

Question 10.
The slope of a line parallel to the line 2x + 3y – 7 = 0 is
(a) \(– \frac { 2 }{ 3 } \)
(b) \(\\ \frac { 2 }{ 3 } \)
(c) \(– \frac { 3 }{ 2 } \)
(d) \(\\ \frac { 3 }{ 2 } \)
Solution:
The slope of a line parallel to the line 2x + 3y – 7 = 0
slope of the line
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS Q10.1

Question 11.
The slope of a line perpendicular to the line 3x = 4y + 11 is
(a) \(\\ \frac { 3 }{ 4 } \)
(b) \(– \frac { 3 }{ 4 } \)
(c) \(\\ \frac { 4 }{ 3 } \)
(d) \(– \frac { 4 }{ 3 } \)
Solution:
slope of a line perpendicular to the line 3x = 4y + 11 is
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS Q11.1

Question 12.
If the lines 2x + 3y = 5 and kx – 6y = 7 are parallel, then the value of k is
(a) 4
(b) – 4
(c) \(\\ \frac { 1 }{ 4 } \)
(d) \(– \frac { 1 }{ 4 } \)
Solution:
lines 2x + 3y = 5 and kx – 6y = 7 are parallel
Slope of 2x + 3y = 5 = Slope of kx – 6y = 7
⇒ 3y – 2x + 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS Q12.1

Question 13.
If the line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0 are perpendicular to each other, then the value of k is
(a) \(\\ \frac { 3 }{ 2 } \)
(b) \(– \frac { 3 }{ 2 } \)
(c) \(\\ \frac { 2 }{ 3 } \)
(d) \(– \frac { 2 }{ 3 } \)
Solution:
line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0
are perpendicular to each other
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS Q13.1

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If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

More Exercises

Question 1.
In the given figure, ∠1 = ∠2 and ∠3 = ∠4. Show that PT x QR = PR x ST.
Solution:
Given: In the given figure,
∠1 = ∠1 and ∠3 = ∠4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q1.1
To prove : PT × QR = PR × ST
Proof: ∠1 = ∠2
Adding ∠6 to both sides
∠1 + ∠6 = ∠2 + ∠6
∠SPT = ∠QPR
In ∆PQR and ∆PST
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q1.2

Question 2.
In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q2.1
Solution:
Given : In the given figure,
AB = AC, PM ⊥ AB and PN ⊥ AC
To prove : PM × PC = PN × PB
Proof: In ∆ABC, AB = AC
∠B = ∠C
Now in ∆CPN and ∆BPM,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q2.2

Question 3.
(a) In the figure (1) given below. ∠AED = ∠ABC. Find the values of x and y.
(b) In the fig. (2) given below, CD = \(\\ \frac { 1 }{ 2 } \) AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :
(i) CE || AG
(ii) 3 ED = GD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q3.1
Solution:
(a) Given : In following figure, ∠AED = ∠ABC
Required: The values of x and y.
Now, in ∆ABC and ∆ADE
∠AED = ∠ABC (given)
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE
(By A.A. axiom of similarity)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q3.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q3.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q3.4

Question 4.
In the given figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:
(i) DF || BH
(ii) AH = 3 AF.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q4.1
Solution:
Given: E is the mid-point of BD
and F is mid-point of AC
also 2 AD = BD and EC || BH
To Prove : (i) DF || BH
(ii) AH = 3 AF
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q4.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q4.3

Question 5.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.
Solution:
Given : In ∆ABC, D and E are the points
on the sides AB and AC respectively
DE || BC
AD = 2.4 cm, AE = 3.2 cm,
DE = 2 cm, BC = 5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q5.2

Question 6.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.
Solution:
In ∆ABC, D and E are points on the sides AB and AC respectively
AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q6.2

Question 7.
In a ∆ABC, DE is parallel to the base BC, with D on AB and E on AC. If \(\frac { AD }{ DB } =\frac { 2 }{ 3 } ,\frac { BC }{ DE } \)
Solution:
In ∆ABC, DE || BC
D is on AB and E is on AC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q7.2

Question 8.
If the area of two similar triangles are 360 cm² and 250 cm² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.
Solution:
Let ∆ABC and ∆DEF are similar and area of
∆ABC = 360 cm²
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q8.2

Question 9.
In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find
(i) BC
(ii) DC
(iii) area of ∆ACD : area of ∆BCA.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q9.1
Solution:
In ∆ABC and ∆ACD
∠C = ∠C (Common)
∠ABC = ∠CAD (given)
∴ ∆ABC ~ ∆ACD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q9.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q9.3

Question 10.
In the adjoining figure the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of DAOE : area of ||gm ABCD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q10.1
Solution:
(a) In the figure
Diagonals of parallelogram ABCD are
AC and BD which intersect each other at O.
OE is drawn parallel to CB to meet AB in E.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q10.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q10.3

Question 11.
In the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of ∆AOB and ∆COD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q11.1
Solution:
In the given figure, ABCD is trapezium in
which AB || DC, 2AB = 3DC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q11.2

Question 12.
In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that .
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q12.1
(i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4.
Solution:
Given : In || gm ABCD,
E is the midpoint of BC and DE meets the diagonal AC at O
and meet AB produced at F.
To prove : (i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4
Proof: In ∆AOD and ∆EDC
∠AOD = ∠EOC (vertically opposite angle)
∠OAD = ∠OCB (alt. angles)
∆AOD ~ ∆EOC (AA postulate)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q12.2

Question 13.
A model of a ship is made to a scale of 1 : 250. Calculate :
(i) the length of the ship, if the length of model is 1.6 m.
(ii) the area of the deck of the ship, if the area of the deck of model is 2.4 m².
(iii) the volume of the model, if the volume of the ship is 1 km³.
Solution:
Scale factor (k) of the model of the ship = \(\\ \frac { 1 }{ 250 } \)
(i) Length of model = 1.6 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test Q13.1

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS

More Exercises

Choose the correct answer from the given four options (1 to 22):

Question 1.
In the given figure, ∆ABC ~ ∆QPR. Then ∠R is
(a) 60°
(b) 50°
(c) 70°
(d) 80°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q1.1
Solution:
In the given figure,
∆ABC ~ ∆QPR
∴ ∠A = ∠Q, ∠B = ∠P and ∠C = ∠R
But ∠A = 70°, ∠B = 50°
∴ ∠C = 180° – (70° + 500) = 180° – 120° = 60°
∠C = ∠R
∴ ∠R = ∠C = 60°

Question 2.
In the given figure, ∆ABC ~ ∆QPR. The value of x is
(a) 2.25 cm
(b) 4 cm
(c) 4.5 cm
(d) 5.2 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q2.1
Solution:
In the given figure
∆ABC ~ ∆QPR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q2.2

Question 3.
In the given figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to
(a) 50°
(b) 30°
(c) 60°
(d) 100°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q3.1
Solution:
In the given figure two line segments AC and BD intersect each other at P
and PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°
In ∆APB and ∆CPD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q3.2

Question 4.
If in two triangles ABC and PQR,
\(\frac { AB }{ QR } =\frac { BC }{ PR } =\frac { CA }{ PQ } \)
then
(a) ∆PQR ~ ∆CAB
(b) ∆PQR ~ ∆ABC
(c) ∆CBA ~ ∆PQR
(d) ∆BCA ~ ∆PQR
Solution:
In two ∆ABC and ∆PQR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q4.1

Question 5.
In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Solution:
In ∆ABC and ∆DEF
∠B = ∠E, ∠F = ∠C
AB = 3DE
∵ Two angles of the one triangles are equal
to corresponding two angles of the other
But sides are not equal
∵ Triangles are similar but not congruent, (b)

Question 6.
In the given figure, if D, E and F are midpoints of the sides BC, CA and AB respectively, then the two triangles ABC and DEF are
(a) similar
(b) congruent
(c) both similar and congruent
(d) neither similar nor congruent
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q6.1
Solution:
D, E and F are the mid-points of the sides BC, CA and AB of ∆ABC
then two triangles ABC and DEF are similar. (a)

Question 7.
The given figure, AB || DE. The length of CD is
(a) 2.5 cm
(b) 2.7 cm
(c) \(\\ \frac { 10 }{ 3 } \) cm
(d) 3.5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q7.1
Solution:
In the given figure, AB || DE
∆ABC ~ ∆DCE
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q7.2

Question 8.
If in triangles ABC and DEF,\(\frac { AB }{ DE } =\frac { BC }{ FD } \) , then they will be similar when
(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠A = ∠F
Solution:
In two triangles ABC and DEF
\(\frac { AB }{ DE } =\frac { BC }{ FD } \)
They will be similar if their included angles are equal
∠B = ∠D (c)

Question 9.
If ∆PQR ~ ∆ABC, PQ = 6 cm, AB = 8 cm and perimeter of ∆ABC is 36 cm, then perimeter of ∆PQR is
(a) 20.25 cm
(b) 27 cm
(c) 48 cm
(d) 64 cm
Solution:
∆PQR ~ ∆ABC
PQ = 6 cm, AB = 8 cm
Perimeter of ∆ABC = 36 cm
Let perimeter of ∆PQR be x cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q9.1

Question 10.
In the given figure, DE || BC and all measurements are in centimetres. The length of AE is
(a) 2 cm
(b) 2.25 cm
(c) 3.5 cm
(d) 4 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q10.1
Solution:
In the given figure,
DE || BC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q10.2

Question 11.
In the given figure, PQ || CA and all lengths are given in centimetres. The length of BC is
(a) 6.4 cm
(b) 7.5 cm
(c) 8 cm
(d) 9 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q11.1
Solution:
In the given figure,
PQ || CA
Let BC = x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q11.2

Question 12.
In the given figure, MN || QR. If PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm, then PM is
(a) 4 cm
(b) 3.6 cm
(c) 2 cm
(d) 3 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q12.1
Solution:
In the given figure, MN || QR
PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm
Let PM = x cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q12.2

Question 13.
D and E are respectively the points on the sides AB and AC of a ∆ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then the length of DE is
(a) 2.5 cm
(b) 3 cm
(c) 5 cm
(d) 6 cm
Solution:
D and E are the points on sides AB and AC of ∆ABC,
AD = 2 cm, BD = 3 cm,
BC = 7.5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q13.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q13.2

Question 14.
It is given that ∆ABC ~ ∆PQR with \(\frac { BC }{ QR } =\frac { 1 }{ 3 } \) then \(\frac { area\quad of\quad \Delta PQR }{ area\quad of\quad \Delta ABC } \) equal to
(a) 9
(b) 3
(c) \(\\ \frac { 1 }{ 3 } \)
(d) \(\\ \frac { 1 }{ 9 } \)
Solution:
∆ABC ~ ∆PQR
\(\frac { BC }{ QR } =\frac { 1 }{ 3 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q14.1

Question 15.
If the areas of two similar triangles are in the ratio 4 : 9, then their corresponding sides are in the ratio
(a) 9 : 4
(b) 3 : 2
(c) 2 : 3
(d) 16 : 81
Solution:
Ratio in the areas of two similar triangles = 4 : 9
Ratio in their corresponding sides
= √4 : √9
= 2 : 3 (c)

Question 16.
If ∆ABC ~ ∆PQR, BC = 8 cm and QR = 6 cm, then the ratio of the areas of ∆ABC and ∆PQR is
(a) 8 : 6
(b) 3 : 4
(c) 9 : 16
(d) 16 : 9
Solution:
∆ABC ~ ∆PQR, BC = 8 cm, QR = 6 cm
\(\frac { area\quad of\quad \Delta ABC }{ area\quad of\quad \Delta PQR } =\frac { { BC }^{ 2 } }{ { QR }^{ 2 } } \)
= \(\frac { { 8 }^{ 2 } }{ { 6 }^{ 2 } } =\frac { 64 }{ 36 } =\frac { 16 }{ 9 } \) (d)

Question 17.
If ∆ABC ~ ∆QRP \(\frac { area\quad of\quad \Delta ABC }{ area\quad of\quad \Delta PQR } \) AB = 18 cm and BC = 15 cm, then the length of PR is equal to
(a) 10 cm
(b) 12 cm
(c) \(\\ \frac { 20 }{ 3 } \)
(d) 8 cm
Solution:
∆ABC ~ ∆QRP
\(\frac { area\quad of\quad \Delta ABC }{ area\quad of\quad \Delta PQR } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q17.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q17.2

Question 18.
If ∆ABC ~ ∆PQR, area of ∆ABC = 81 cm², area of ∆PQR = 144 cm² and QR = 6 cm, then length of BC is
(a) 4 cm
(b) 4.5 cm
(c) 9 cm
(d) 12 cm
Solution:
∆ABC ~ ∆PQR,
area of ∆ABC = 81 cm² and area of ∆PQR = 144 cm²,
QR = 6 cm, BC = ?
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q18.1

Question 19.
In the given figure, DE || CA and D is a point on BD such that BD : DC = 2 : 1. The ratio of area of ∆ABC to area of ∆BDE is
(a) 4 : 1
(b) 9 : 1
(c) 9 : 4
(d) 3 : 2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q19.1
Solution:
In the given figure, DE || CA ,
D is a point on BC and BD : DC = 2 : 1
BD : BC = 2 : (2 + 1) = 2 : 3
In ∆ABC, DE || CA
∆ABC ~ ∆BDE
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q19.2

Question 20.
If ABC and BDE are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC and BDE is
(a) 2 : 1
(b) 1 : 2
(c) 1 : 4
(d) 4 : 1
Solution:
∆ABC and ∆BDE are an equilateral triangles
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q20.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q20.2

Question 21.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. If an altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle is
(a) 9 cm
(b) 7 cm
(c) 6 cm
(d) 4.5 cm
Solution:
Areas of two similar triangles are 81 cm² and 49 cm²
The altitude of the smaller triangle is 3.5 cm
Let the altitude of the bigger triangle is x, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q21.1

Question 22.
Given ∆ABC ~ ∆PQR, area of ∆ABC = 54 cm² and area of ∆PQR = 24 cm². If AD and PM are medians of ∆’s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is
(a) \(\\ \frac { 49 }{ 3 } \) cm
(b) \(\\ \frac { 20 }{ 3 } \) cm
(c) 15 cm
(d) 22.5 cm
Solution:
∆ABC ~ ∆PQR
area of ∆ABC = 54 cm²
and of ∆PQR = 24 cm²
AD and PM are their median respectively
PM = 10 cm
Let AD = x cm, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS Q22.1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2

More Exercises

Question 1.
State which one of the following is true : The straight lines y = 3x – 5 and 2y = 4x + 7 are
(i) parallel
(ii) perpendicular
(iii) neither parallel nor perpendicular.
Solution:
Slope of line y = 3x – 5 = 3
and slope of line 2y = 4x + 7
⇒ y = 2x + \(\\ \frac { 7 }{ 2 } \) = 2.
∴ Slope of both the lines are neither equal nor their product is – 1.
∴ These line are neither parallel nor perpendicular.

Worksheets for Class 10 Maths

Question 2.
If 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.
Solution:
In equation
6x + 5 y – 7 = 0
⇒ 5y = -6x + 7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q2.1

Question 3.
Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b. (1991)
Solution:
In equation 2x – by + 5 = 0
⇒ – by = – 2x – 5
⇒ y = \(\\ \frac { 2 }{ b } \) + \(\\ \frac { 5 }{ b } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q3.1

Question 4.
Given that the line \(\\ \frac { y }{ 2 } \) = x – p and the line ax + 5 = 3y are parallel, find the value of a. (1992)
Solution:
In equation y = x – p
⇒ y = 2x – 2p
Slope (m1) = 2
In equation ax + 5 = 3y
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q4.1

Question 5.
If the lines y = 3x + 7 and 2y + px = 3 perpendicular to each other, find the value of p. (2006)
Solution:
Gradient m1 of the line y = 3x + 7 is 3
2y + px = 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q5.1

Question 6.
Find the value of k for which the lines kx – 5y + 4 = 0 and 4x – 2y + 5 = 0 are perpendicular to each other. (2003)
Solution:
Given
In equation, kx – 5y + 4 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q6.1

Question 7.
If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.
Solution:
Given
In the equation 3x + by + 5 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q7.1

Question 8.
Is the line through ( – 2, 3) and (4, 1) perpendicular to the line 3x = y + 1 ?
Does the line 3x = y + 1 bisect the join of ( – 2, 3) and (4, 1). (1993)
Solution:
Slope of the line passing through the points
(-2, 3) and (4, 1) = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q8.1
Hence the line 3x = y + 1 bisects the line joining the points (-2, 3), (4, 1).

Question 9.
The line through A ( – 2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b.
Solution:
Gradient (m1) of the line passing through the
points A (-2, 3) and B (4, b)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q9.1

Question 10.
If the lines 3x + y = 4, x – ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the value of a and b.
Solution:
In the line 3x + y = 4 …(i)
⇒ y = – 3x + 4
Slope (m1) = – 3
In the line x – ay + 7 = 0…..(ii)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q10.1
⇒ -b = -6 ⇒ b = 6
Hence a = 3, b = 6

Question 11.
Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x – 3y – 7 = 0. Find the coordinates of the point where it cuts the x-axis. (1998)
Solution:
In the given line 2x – 3y – 7 = 0
⇒ 3y = 2x – 7
⇒ \(y=\frac { 2 }{ 3 } x-\frac { 7 }{ 3 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q11.1

Question 12.
Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept – 3 units.
Solution:
In the line 2x + 5y + 7 = 0
⇒ 5y = – 2x – 7
⇒ \(y=\frac { -2 }{ 5 } x-\frac { 7 }{ 5 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q12.1

Question 13.
Find the equation of a st. line perpendicular to the line 3x – 4y + 12 = 0 and having same y-intercept as 2x – y + 5 = 0.
Solution:
In the given line 3x – 4y + 12 = 0
⇒ 4y = 3x + 12
⇒ y = \(\\ \frac { 3 }{ 4 } x+3\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q13.1

Question 14.
Find the equation of the line which is parallel to 3x – 2y = – 4 and passes through the point (0, 3). (1990)
Solution:
In the given line 3x – 2y = – 4
⇒ 2y = 3x + 4
⇒ y = \(\\ \frac { 3 }{ 2 } x+2\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q14.1

Question 15.
Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0. (1999)
Solution:
In the given equation 3x + 5y + 15 = 0
⇒ 5y = – 3x – 15
⇒ y = \(\\ \frac { -3 }{ 5 } x-3\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q15.1

Question 16.
The equation of a line is y = 3x – 5. Write down the slope of this line and the intercept made by it on the y-axis. Hence or otherwise, write down the equation of a line which is parallel to the line and which passes through the point (0, 5).
Solution:
In the given line y = 3x – 5
Here slope (m1) = 3
Substituting x = 0, then y = – 5
y-intercept = – 5
The slope of the line parallel to the given line
will be 3 and passes through the point (0, 5).
Equation of the line will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q16.1

Question 17.
Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point ( – 1, – 2).
Solution:
In the given line 3x + 8y = 12
⇒ 8y = -3x + 12
⇒ \(y=\frac { -3 }{ 8 } x+\frac { 12 }{ 8 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q17.1

Question 18.
(i) The line 4x – 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A.
(ii) Determine the equation of the line passing through A and perpendicular to 4x – 3y + 12 = 0. (1993)
Solution:
(i) In the line 4x – 3y + 12 = 0 …(i)
3y = 4x + 12
⇒ y = \(\\ \frac { 4 }{ 3 } x+4\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q18.1

Question 19.
Find the equation of the line that is parallel to 2x + 5y – 7 = 0 and passes through the mid-point of the line segment joining the points (2, 7) and ( – 4, 1).
Solution:
The given line 2x + 5y – 7 = 0
5y = -2x + 7
⇒ \(y=\frac { -2 }{ 5 } x+\frac { 7 }{ 5 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q19.1

Question 20.
Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes through the mid-point of the line segment joining the points (5, – 2), (2, 2).
Solution:
In the given line 3x + 2y – 8=0
⇒ 2y = – 3x + 8
⇒ y = \(\\ \frac { -3 }{ 2 } x+4\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q20.1

Question 21.
Find the equation of a straight line passing through the intersection of 2x + 5y – 4 = 0 with x-axis and parallel to the line 3x – 7y + 8 = 0.
Solution:
Let the point of intersection of the line 2x + 5y – 4 = 0 and x-axis be (x, 0)
Substituting the value of y in the equation
2x + 5 × 0 – 4 = 0
⇒ 2x – 4 = 0
⇒ 2x = 4
⇒ x = \(\\ \frac { 4 }{ 2 } \) = 2
Coordinates of the points of intersection will be (2, 0)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q21.1

Question 22.
The equation of a line is 3x + 4y – 7 = 0. Find
(i) the slope of the line. .
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0. (2010)
Solution:
(i) Equation of the line is 3x + 4y – 1 = 0
⇒ 4y = 7 – 3x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q22.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q22.2

Question 23.
Find the equation of the line perpendicular from the point (1, – 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.
Solution:
In the equation 4x – 3y – 5 = 0,
⇒ 3y = 4x – 5
⇒ \(y=\frac { 4 }{ 3 } x-\frac { 5 }{ 3 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q23.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q23.2

Question 24.
Prove that the line through (0, 0) and (2, 3) is parallel to the line through (2, – 2) and (6, 4).
Solution:
Given that
Slope of the line through (0, 0) and (2, 3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q24.1

Question 25.
Prove that the line through,( – 2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (4, 24).
Solution:
Given that
Slope of the line through (-2, 6) and (4, 8)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q25.1

Question 26.
Show that the triangle formed by the points A (1, 3), B (3, – 1) and C ( – 5, – 5) is a right angled triangle by using slopes.
Solution:
Slope (m1) of line by joining the points
A(1, 3), B (3, -1) = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ x_{ 2 }-{ x }_{ 1 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q26.1

Question 27.
Find the equation of the line through the point ( – 1, 3) and parallel to the line joining the points (0, – 2) and (4, 5).
Solution:
Slope of the line joining the points (0, -2) and (4, 5) = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ x_{ 2 }-{ x }_{ 1 } } \)
= \(\frac { 5+2 }{ 4-0 } =\frac { 7 }{ 4 } \)
Slope of the line parallel to it passing through (-1, 3) = \(\\ \frac { 7 }{ 4 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q27.1

Question 28.
A ( – 1, 3), B (4, 2), C (3, – 2) are the vertices of a triangle.
(i) Find the coordinates of the centroid G of the triangle.
(ii) Find the equation of the line through G and parallel to AC
Solution:
Given, A (-1, 3), B (4, 2), C (3, -2)
(i) Coordinates of centroid G
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q28.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q28.2

Question 29.
The line through P (5, 3) intersects y-axis at Q.
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the coordinates of Q.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q29.1
Solution:
(i) Here θ = 45°
So, slope of the line = tan θ = tan 45° = 1
(ii) Equation of the line through P and Q is
y – 3 = 1(x – 5) ⇒ y – x + 2 = 0
(iii) Let the coordinates of Q be (0, y)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q29.2

Question 30.
In the adjoining diagram, write down
(i) the co-ordinates of the points A, B and C.
(ii) the equation of the line through A parallel to BC. (2005)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q30.1
Solution:
From the given figure, it is clear that
co-ordinates of A are (2, 3) of B are ( -1, 2)
and of C are (3, 0).
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q30.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q30.3

Question 31.
Find the equation of the line through (0, – 3) and perpendicular to the line joining the points ( – 3, 2) and (9, 1).
Solution:
The slope (m1) of the line joining the points (-3, 2) and (9, 1)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q31.1

Question 32.
The vertices of a triangle are A (10, 4), B (4, – 9) and C ( – 2, – 1). Find the equation of the altitude through A. The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.
Solution:
Vertices of ∆ABC are A (10, 4), B (4, -9) and C( – 2, – 1)
Slope of the line BC (m1) = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q32.1

Question 33.
A (2, – 4), B (3, 3) and C ( – 1, 5) are the vertices of triangle ABC. Find the equation of :
(i) the median of the triangle through A
(ii) the altitude of the triangle through B
Solution:
(i) D is the mid-point of BC
Co-ordinates of D will
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q33.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q33.2

Question 34.
Find the equation of the right bisector of the line segment joining the points (1, 2) and (5, – 6).
Solution:
Slope of the line joining the points (1, 2) and (5, -6)
\({ m }_{ 1 }=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -6-2 }{ 5-1 } =\frac { -8 }{ 4 } =-2\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q34.1

Question 35.
Points A and B have coordinates (7, – 3) and (1, 9) respectively. Find
(i) the slope of AB.
(ii) the equation of the perpendicular bisector of the line segment AB.
(iii) the value of ‘p’ if ( – 2, p) lies on it.
Solution:
Coordinates of A are (7, -3), of B = (1, 9)
(i) ∴ slope (m)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q35.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q35.2

Question 36.
The points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.
Solution:
Slope of BD (m1) = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
= \(\frac { 8-3 }{ 6-1 } =\frac { 5 }{ 5 } =1\)
Diagonal AC is perpendicular bisector of diagonal BD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q36.1

Question 37.
ABCD is a rhombus. The co-ordinates of A and C are (3, 6) and ( – 1, 2) respectively. Write down the equation of BD. (2000)
Solution:
Co-ordinates of A (3, 6), C (-1, 2)
Slope of AC (m1) = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
= \(\frac { 2-6 }{ -1-3 } =\frac { -4 }{ -4 } =1\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q37.1

Question 38.
Find the equation of the line passing through the intersection of the lines 4x + 3y = 1 and 5x + 4y = 2 and
(i) parallel to the line x + 2y – 5 = 0
(ii) perpendicular to the x-axis.
Solution:
4x + 3y = 1 …(i)
5x + 4y = 2 …(ii)
Multiplying (i) by 4 and (ii) by 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q38.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q38.2

Question 39.
(i) Write down the co-ordinates of the point P that divides the line joining A ( – 4, 1) and B (17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP where 0 is the origin
(iii) In what ratio does the y-axis divide the line AB?
Solution:
(i) Co-ordinate A (-4, 1) and B (17, 10) P divides it in the ratio of 1 : 2
Let the co-ordinates of P will be (x, y)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q39.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q39.2

Question 40.
Find the image of the point (1, 2) in the line x – 2y – 7 = 0
Solution:
Draw a perpendicular from the point P(1, 2) on the line, x – 2y – 7 = 0
Let P’ is the image of P and let its
co-ordinates sue (α, β) slope of line x – 2y – 7 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q40.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q40.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q40.3

Question 41.
If the line x – 4y – 6 = 0 is the perpendicular bisector of the line segment PQ and the co-ordinates of P are (1, 3), find the co-ordinates of Q.
Solution:
Let the co-ordinates of Q be (α, β) and let the line x – 4y – 6 = 0 is the
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q41.1
perpendicular bisector of PQ and it intersects the line at M.
M is the mid point of PQ Now slope of line x – 4y – 6 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q41.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q41.3

Question 42.
OABC is a square, O is the origin and the points A and B are (3, 0) and (p, q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.
Solution:
OA = \(\sqrt { { (3-0) }^{ 2 }+{ (0-0) }^{ 2 } } \)
\(\sqrt { { (3-0) }^{ 2 }+{ (0) }^{ 2 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q42.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 Q42.2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3

More Exercises

Question 1.
Given that ∆s ABC and PQR are similar.
Find:
(i) The ratio of the area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3.
(ii) the ratio of their corresponding sides if area of ∆ABC : area of ∆PQR = 25 : 36.
Solution:
(i) ∴ ∆ABC ~ ∆PQR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q1.1
(By theorem 15.1)
But BC = QR =1 : 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q1.2

Question 2.
∆ABC ~ DEF. If area of ∆ABC = 9 sq. cm., area of ∆DEF =16 sq. cm and BC = 2.1 cm., find the length of EF.
Solution:
Let EF = x
Given that
∆ABC ~ ∆DEF,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q2.1

Question 3.
∆ABC ~ ∆DEF. If BC = 3 cm, EF = 4 cm and area of ∆ABC = 54 sq. cm. Determine the area of ∆DEF.
Solution:
∆ABC ~ ∆DEF
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q3.1

Question 4.
The area of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.
Solution:
Let ABC ~ ∆DEF, AL and DM are their altitudes
then area of ∆ABC = 36 cm²
area of ∆DEF = 25 cm² and AL = 2.4 cm.
Let DM = x
Now ∆ABC ~ ∆DEF
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q4.1

Question 5.
(a) In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm², find the area of ∆QOA. (2006)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q5.1
(b) In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm.
(i) Prove that ∆OAB ~ ∆OCD.
(ii) Find CD and OB.
(iii) Find the ratio of areas of ∆OAB and ∆OCD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q5.2
Solution:
In ∆AOQ and ∆BOP, we have
∠ OAQ = ∠ OBP [Each = 90°]
∠ AOQ= ∠BOP
[Vertically opposite angles]
∆AOQ ~ ∆BOP [A.A. similarity]
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q5.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q5.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q5.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q5.6
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q5.7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q5.8

Question 6.
(a) In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm, find the area of ∆ABC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q6.1
(b) In the figure (iii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of ∆ADE and trapezium DBCE.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q6.2
Solution:
(a) In the figure,
DE || BC
∠D = ∠B and ∠E = ∠C
(Corresponding angles)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q6.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q6.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q6.5

Question 7.
In the given figure, DE || BC.
(i) Prove that ∆ADE and ∆ABC are similar.
(ii) Given that AD = \(\\ \frac { 1 }{ 2 } \) BD, calculate DE if BC = 4.5 cm.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q7.1
(iii) If area of ∆ABC = 18cm², find the area of trapezium DBCE
Solution:
(i) Given : In ∆ABC, DE || BC.
To prove : ∆ADE ~ ∆ABC
Proof: In ∆ADE and ∆ABC,
∠A = ∠A (common)
∠ADE = ∠ABC (corresponding angles)
∴ ∆ADE ~ ∆ABC. (AA axiom)
(ii) ∴ ∆ADE ~ ∆ABC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q7.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q7.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q7.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q7.5

Question 8.
In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm, calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q8.1
Solution:
(i) To prove : ∆ABC ~ ∆DEC
In ∆ABC and ∆DEC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q8.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q8.3

Question 9.
In the adjoining figure, ABC is a triangle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q9.1
(ii) Prove that ∆DEF is similar to ∆CBF.
Hence, find \(\\ \frac { EF }{ FB } \).
(iii) What is the ratio of the areas of ∆DEF and ∆CBF ? (2007)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q9.2

Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q9.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q9.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q9.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q9.6

Question 10.
In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:
(i) area ∆APO : area ∆ABC.
(ii) area ∆APO : area ∆CQO. (2008)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q10.1
Solution:
In the figure,
PQ || BC and PO is produced to Q such that CQ || BA
and AP : PB = 2 : 3.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q10.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q10.3

Question 11.
(a) In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of ∆AOB and ∆COD.
(b) In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find
(i) AB
(ii) BC
(iii) area of ∆ADM : area of ∆ANB.
(c) In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find
(i) EF : AD
(ii) area of ∆BEF : area of ∆ABD
(iii) area of ∆ABD : area of trap. AFED
(iv) area of ∆FEO : area of ∆OBC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q11.1
Solution:
(a) In trapezium ABCD, AB || DC.
∠OAB = ∠OCD [alternate angles]
∠OBA = ∠ODC
∆AOB ~ ∆COD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q11.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q11.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q11.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q11.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q11.6

Question 12.
In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q. If area of ∆CPQ = 20 cm², find
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q12.1
(i) area of ∆BPQ.
(ii) area ∆CDP.
(iii) area of || gm ABCD.
Solution:
In the figure, ABCD is a parallelogram.
P is a point on BC such that BP : PC = 1 : 2
and DP is produced to meet ABC produced at Q.
Area ∆CPQ = 20 cm²
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q12.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q12.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q12.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q12.5

Question 13.
(a) In the figure (i) given below, DE || BC and the ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q13.1
(b) In the figure (ii) given below, AB || DC and AB = 2 DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find (i) ED (ii) BE (iii) area of ∆EDC : area of trapezium ABCD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q13.2
Solution:
(a) In ∆ABC, DE || BC
Now in ∆ABC and ∆ADE
∠A = ∠A (common)
∠D = ∠B and ∠E = ∠C
(Corresponding angles)
∆ADE ~ ∆ABC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q13.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q13.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q13.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q13.6

Question 14.
(a) In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BB = 12 cm., find
(i) BP
(ii) the ratio of areas of ∆APB and ∆DPC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q14.1
(b) In the figure given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA
(ii) Find BC and CD
(iii) Find the area of ∆ACD : area of ∆ABC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q14.2
Solution:
(a) In trapezium ABCD, DC || AB
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q14.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q14.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q14.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q14.6

Question 15.
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:
(i) ∆ADE ~ ∆ACB.
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of ∆ADE : area of quadrilateral BCED.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q15.1
Solution:
(i) Consider DADE and DACB
∠A = ∠A (Common)
m∠B = m∠E = 90°
Thus by angle-angle similarity, triangles,
∆ACB ~ ∆ADE
(ii) Consider ∆ADE and ∆ACB
Since they are similar triangles,
the sides are proportional Thus, we have,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q15.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q15.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q15.4

Question 16.
Two isosceles triangles have equal vertical angles and their areas are in the ratio 7 : 16. Find the ratio of their corresponding height.
Solution:
In two isosceles ∆s ABC and DEF
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q16.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q16.2

Question 17.
On a map drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements :
AB = 3 cm, BC = 4 cm and ∠ABC = 90°. Calculate
(i) the actual length of AB in km.
(ii) the area of the plot in sq. km:
Solution:
Scale factor k = 1 : 250000 = \(\\ \frac { 1 }{ 250000 } \)
Length on map,
AB = 3 cm, BC = 4 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q17.1

Question 18.
On a map drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD has the following measurements AB = 12 cm and BG = 16 cm.
Calculate:
(i) the distance of a diagonal of the plot in km.
(ii) the area of the plot in sq. km.
Solution:
Scale factor (k) = \(\\ \frac { 1 }{ 25000 } \)
Measurements of plot ABCD on the map are
AB = 12 cm and BC = 16 cm.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q18.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q18.2

Question 19.
The model of a building is constructed with the scale factor 1 : 30.
(i) If the height of the model is 80 cm, find the actual height of the building in metres.
(ii) If the actual volume of a tank at the top of the building is 27 m³, find the volume of the tank on the top of the model. (2009)
Solution:
(i)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q19.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q19.2

Question 20.
A model of a ship is made to a scale of 1 : 200.
(i) If the length of the model is 4 m, find the length of the ship.
(ii) If the area of the deck of the ship is 160000 m², find the area of the deck of the model.
(iii) If the volume of the model is 200 litres, find the volume of the ship in m³.
(100 litres = m³)
Solution:
Scale = 1 : 200
(i) Length of a model of ship = 4 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q20.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 Q20.2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.