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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1

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 15 Circles Ex 15.1

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1
• 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 15.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

Question 1.
Using the given information, find the value of x in each of the following figures :


Solution:
(i) ∠ADB and ∠ACB are in the same segment.
∠ADB = ∠ACB = 50°
Now in ∆ADB,
∠DAB + X + ∠ADB = 180°

Question 2.
If O is the centre of the circle, find the value of x in each of the following figures (using the given information):


Solution:
(i) ∠ACB = ∠ADB
(Angles in the same segment of a circle)
But ∠ADB = x°

Question 3.
(a) In the figure (i) given below, AD || BC. If ∠ACB = 35°. Find the measurement of ∠DBC.
(b) In the figure (ii) given below, it is given that O is the centre of the circle and ∠AOC = 130°. Find∠ABC

Solution:
(a) Construction: Join AB
∠A = ∠C = 35° [∵ Alt angles]

Question 4.
(a) In the figure (i) given below, calculate the values of x and y.
(b) In the figure (ii) given below, O is the centre of the circle. Calculate the values of x and y.

Solution:
(a) ABCD is a cyclic quadrilateral

Question 5.
(a) In the figure (i) given below, M, A, B, N are points on a circle having centre O. AN and MB cut at Y. If ∠NYB = 50° and ∠YNB = 20°, find ∠MAN and the reflex angle MON.
(b) In the figrue (ii) given below, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°, find
(i) ∠ACB
(ii) ∠OBC
(iii) ∠OAB
(iv) ∠CBA

Solution:
(a) ∠NYB = 50°, ∠YNB = 20°.

Question 6.
(a) In the figure (i) given below, O is the centre of the circle and ∠PBA = 42°. Calculate the value of ∠PQB
(b) In the figure (ii) given below, AB is a diameter of the circle whose centre is O. Given that ∠ECD = ∠EDC = 32°, calculate
(i) ∠CEF
(ii) ∠COF.

Solution:
(a) In ∆APB,
∠APB = 90° (Angle in a semi-circle)
But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle)
∠A + 90° + 42°= 180°
∠A + 132° = 180°
⇒ ∠A = 180° – 132° = 48°
But ∠A = ∠PQB

Question 7.
(a) In the figure (i) given below, AB is a diameter of the circle APBR. APQ and RBQ are straight lines, ∠A = 35°, ∠Q = 25°. Find (i) ∠PRB (ii) ∠PBR (iii) ∠BPR.
(b) In the figure (ii) given below, it is given that ∠ABC = 40° and AD is a diameter of the circle. Calculate ∠DAC.

Solution:
(a) (i) ∠PRB = ∠BAP
(Angles in the same segment of the circle)
∴ ∠PRB = 35° (∵ ∠BAP = 35° given)
(ii) In ∆PRQ,

Question 8.
(a) In the figure given below, P and Q are centres of two circles intersecting at B and C. ACD is a st. line. Calculate the numerical value of x.

(b) In the figure given below, O is the circumcentre of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate
(i)∠CAB
(ii)∠OAC

Solution:
Given that
(a) Arc AB subtends ∠APB at the centre
and ∠ACB at the remaining part of the circle

Question 9.
(a) In the figure (i) given below, chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

(b) In the figure (ii) given below, C is a point on the minor arc AB of the circle with centre O. Given ∠ACB = p°, ∠AOB = q°, express q in terms of p. Calculate p if OACB is a parallelogram.
Solution:
(a) ∠CBE = ∠CAE
(Angle in the same segment of a circle)
⇒ ∠CAE = 65°
∠AEC = 90° (Angle in a semi circle)
Now in ∆AEC
∠AEC + ∠CAE + ∠ACE = 180° (Angle of a triangle)
⇒ 90°+ 65° +∠ACE = 180°
⇒ 155° + ∠ACE = 180°
⇒ ∠ACE = 180° – 155° – 25°
∵AC || ED (given)
∴∠ACE = ∠DEC (alternate angles)
∴∠DEC = 25°

Question 10.
(a) In the figure (i) given below, straight lines AB and CD pass through the centre O of a circle. If ∠OCE = 40° and ∠AOD = 75°, find the number of degrees in :
(i) ∠CDE
(ii) ∠OBE.
(b) In the figure (ii) given below, I is the incentre of ∆ABC. AI produced meets the circumcircle of ∆ABC at D. Given that ∠ABC = 55° and ∠ACB = 65°, calculate
(i) ∠BCD
(ii) ∠CBD
(iii) ∠DCI
(iv) ∠BIC.

Solution:
(a) (i) ∠CED = 90° (Angle in semi-circle)
In ∆CED
∠CED + ∠CDE + ∠DCE = 180°
⇒ 90° +∠CDE + 40° = 180°
⇒ 130° + ∠CDE = 180°
⇒ ∠CDE = 180° – 130° = 50°

NCVT MIS 2019

Question 11.
O is the circumcentre of the triangle ABC and D is mid-point of the base BC. Prove that ∠BOD = ∠A.
Solution:
In the given figure, O is the centre of circumcentre of ∆ABC.
D is mid-point of BC. BO, CO and OD are joined.

Question 12.
In the given figure, AB and CD are equal chords. AD and BC intersect at E. Prove that AE = CE and BE = DE.

Solution:
In the given figure, AB and CD are two equal chords
AD and BC intersect each other at E.
To prove : AE = CE and BE = DE
Proof:
In ∆AEB and ∆CED
AB = CD (given)
∠A = ∠C (angles in the same segment)
∠B = ∠D (angles in the same segment)
∴ ∆AEB ≅ ∆CED (ASA axiom)
∴ AE = CE and BE = DE (c.p.c.t.)

Question 13.
(a) In the figure (i) given below, AB is a diameter of a circle with centre O. AC and BD are perpendiculars on a line PQ. BD meets the circle at E. Prove that AC = ED.
(b) In the figure (ii) given below, O is the centre of a circle. Chord CD is parallel to the diameter AB. If ∠ABC = 25°, calculate ∠CED.

Solution:
(a) Given: AB is the diameter of a circle with centre O.
AC and BD are perpendiculars on a line PQ,
such that BD meets the circle at E.

Question 14.
In the adjoining figure, O is the centre of the given circle and OABC is a parallelogram. BC is produced to meet the circle at D.
Prove that ∠ABC = 2 ∠OAD.

Solution:
Given: In the figure,
OABC is a || gm and O is the centre of the circle.
BC is produced to meet the circle at D.
To Prove : ∠ABC = 2∠OAD.
Construction: Join AD.

Question 15.
(a) In the figure (i) given below, P is the point of intersection of the chords BC and AQ such that AB = AP. Prove that CP = CQ.

(b) In the figure (i) given below, AB = AC = CD, ∠ADC = 38°. Calculate :
(i) ∠ABC (ii) ∠BEC.

Solution:
(a) Given: Two chords AQ and BC intersect each other at P
inside the circle. AB and CQ are joined and AB = AP.
To Prove : CP = CQ
Construction : Join AC.
Proof: In ∆ABP and ∆CQP
∴ ∠B = ∠Q

Question 16.
(a) In the figure (i) given below, CP bisects ∠ACB. Prove that DP bisects ∠ADB.
(b) In the figure (ii) given below, BDbisects ∠ABC. Prove that \(\frac { AB }{ BD } =\frac { BE }{ BC } \)

Solution:
(a)Given: In the figure, CP is the bisector of
∠ACB meeting the circle at P.
PD is joined

Question 17.
(a) In the figure (ii) given below, chords AB and CD of a circle intersect at E.
(i) Prove that triangles ADE and CBE are similar.
(ii) Given DC =12 cm, DE = 4 cm and AE = 16 cm, calculate the length of BE.

(b) In the figure (ii) given below, AB and CD are two intersecting chords of a circle. Name two triangles which are similar. Hence, calculate CP given that AP = 6cm, PB = 4 cm, and CD = 14 cm (PC > PD).

Solution:
(a) Given: Two chords AB and CD intersect each other
at E inside the circle.

Question 18.
In the adjoining figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE. (2008)

Solution:
In the figure, AE and BC intersect each other at D.
AB is joined.

Question 19.
(a) In the figure (i) given below, PR is a diameter of the circle, PQ = 7 cm, QR = 6 cm and RS = 2 cm. Calculate the perimeter of the cyclic quadrilateral PQRS.
(b) In the figure (ii) given below, the diagonals of a cyclic quadrilateral ABCD intersect in P and the area of the triangle APB is 24 cm². If AB = 8 cm and CD = 5 cm, calculate the area of ∆DPC.

Solution:
(a) PR is the diameter of the circle
PQ = 7 cm, QR = 6 cm, RS = 2 cm.

Question 20.
(a) In the figure (i) given below, QPX is the bisector of ∠YXZ of the triangle XYZ. Prove that XY : XQ = XP : XZ,
(b) In the figure (ii) given below, chords BA and DC of a circle meet at P. Prove that:
(i) ∠PAD = ∠PCB
(ii) PA. PB = PC . PD.

Solution:
(a) Given: ∆XYZ is inscribed in a circle.
Bisector of ∠YXZ meets the circle at Q.
QY is joined.
To Prove : XY : XQ = XP : XZ

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 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 14 Locus Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Ex 14
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Chapter Test

Question 1.
Draw a straight line AB of length 8 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
Solution:
(i) Draw a line segment AB = 8 cm.

(ii) Draw the perpendicular bisector of AB intersecting AB at D.
∴ Every point P on it will be equidistant from A and B.
(iii) Take a point P on the perpendicular bisector.
(iv) Join PA and PB.
Proof: In ∆PAD and ∆PBD
PD = PD (common)
AD = BD (D is mid-point of AB)
∠PDA = ∠PDB (each 90°)
∴ ∆ PAD ≅ ∆ PBD (SAS axiom of congruency)
∴PA = PB (c.p.c.t.)
Similarly, we can prove any other point on the
perpendicular bisector of AB is equidistant from A and B.
Hence Proved.

Question 2.
A point P is allowed to travel in space. State the locus of P so that it always remains at a constant distance from a fixed point C.
Solution:
The point P is moving in the space and
it is at a constant distance from a fixed point C.
∴ Its locus is a sphere.

Question 3.
Draw a line segment AB of length 7 cm. Construct the locus of a point P such that area of triangle PAB is 14 cm².
Solution:
Base of ∆PAB = 7 cm
and its area = 14 cm²

Now draw a line XY parallel to AB at a distance of 4 cm.
Now take any point P on XY
Join PA and PB
area of ∆PAB = 14 cm.
Hence locus of P is the line XY
which is parallel to AB at a distance of 4 cm.

Question 4.
Draw a line segment AB of length 12 cm. Mark M, the mid-point of AB. Draw and describe the locus of a point which is
(i) at a distance of 3 cm from AB.
(ii) at a distance of 5 cm from the point M. Mark the points P, Q, R, S which satisfy both the above conditions. What kind of quadrilateral is PQRS? Compute the area of the quadrilateral PQRS.
Solution:
Steps of Construction :

(i) Take a line AB = 12 cm
(ii) Take M, the midpoint of AB.
(iii) Draw straight lines CD and EF parallel to AB at a distance of 3 cm.
(iv) With centre M and radius 5 cm,
draw areas which intersect CD at P and Q and EF at R and S.
(v) Join QR and PS.
PQRS is a rectangle where the length PQ = 8 cm.
Area of rectangle PQRS = PQ x RS = 8 x 6 = 48 cm²

Question 5.
AB and CD are two intersecting lines. Find the position of a point which is at a distance of 2 cm from AB and 1.6 cm from CD.
Solution:
(i) AB and CD are the intersecting lines which intersect each other at O.

(ii) Draw a line EF parallel to AB and GH parallel to CD intersecting each other at P
P is the required point.

Question 6.
Two straight lines PQ and PK cross each other at P at an angle of 75°. S is a stone on the road PQ, 800 m from P towards Q. By drawing a figure to scale 1 cm = 100 m, locate the position of a flagstaff X, which is equidistant from P and S, and is also equidistant from the road.
Solution:
1 cm = 100 cm
800 m = 8 cm.
Steps of Construction :
(i) Draw the lines PQ and PK intersecting each other
at P making an angle of 75°.

(ii) Take a point S on PQ such that PS = 8 cm.
(iii) Draw the perpendicular bisector of PS.
(iv) Draw the angle bisector of ∠KPS intersecting
the perpendicular bisector at X.
X is the required point which is equidistant from P and S
and also from PQ and PK.

Question 7.
Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively. Find by construction a point X equidistant from PQ, PS and equidistant from R, S. Measure XR.
Solution:
Steps of Construction :

(i) Take PR = 8 cm and draw the perpendicular bisector
of PR intersecting it at O.
(ii) From O, out. off OS = OQ = 3 cm
(iii) Join PQ, QR, RS and SP.
PQRS is a rhombus. Whose diagonal are PR and QS.
(iv) PR is the bisector of ∠SPQ.
(v) Draw the perpendicular bisector of SR intersecting PR at X
∴ X is equidistant from PQ and PS and also from S and R.
On measuring length of XR = 3.2 cm (approx)

Question 8.
Without using set square or protractor, construct the parallelogram ABCD in which AB = 5.1 cm. the diagonal AC = 5.6 cm and the diagonal BD = 7 cm. Locate the point P on DC, which is equidistant from AB and BC.
Solution:
Steps of Construction :

(i) Take AB = 5.1 cm
(ii) At A, with readius \(\\ \frac { 5.6 }{ 2 } \) = 2.8 cm
and at B with radius \(\\ \frac { 7.0 }{ 2 } \) = 3.5 cm,
draw two arcs intersecting each other at O.
(iii) Join AO and produce it to C such that
OC = AD = 2.8 cm and
join BO and produce it to D such that
BO = OD = 3.5 cm
(iv) Join BC, CD, DA
ABCD is a parallelogram.
(v) Draw the angle bisector of ∠ABC intersecting CD at P.
P is the required point which is equidistant from AB and BC.

Question 9.
By using ruler and compass only, construct a quadrilateral ABCD in which AB = 6.5 cm, AD = 4cm and ∠DAB = 75°. C is equidistant from the sides AB and AD, also C is equidistant from the points A and B.
Solution:
Steps of Construction :

(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a ray making an angle of 75° and cut off AD = 4 cm.
(iii) Draw the bisector of ∠DAB.
(iv) Draw perpendicular bisector of AB intersecting the angle bisector at C.
(v) Join CB and CD.
ABCD is the required quadrilateral.

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus 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 17.3

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 17 Mensuration Ex 17.3

More Exercise

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4
• 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 MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

Question 1.
Find the surface area of a sphere of radius :
(i) 14 cm
(ii) 10.5 cm
Solution:
(i) Radius (r) = 14 cm
Surface area = \(4\pi { r }^{ 2 }=4\times \frac { 22 }{ 7 } \times 14\times 14 \) cm²
= 2964 cm²

Question 2.
Find the volume of a sphere of radius :
(i) 0.63 m
(ii) 11.2 cm
Solution:
(i) Radius (r) = 0.63 m
Volume = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

Question 3.
Find the surface area of a sphere of diameter: (i) 21 cm (ii) 3.5 cm
Solution:
(i) Diameter = 21 cm
Radius (r) = \(\\ \frac { 21 }{ 2 } \) cm

Question 4.
A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-put.
Solution:
Radius of the metallic shot-put = 4.9 cm
Volume = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

Question 5.
Find the diameter of a sphere whose surface area is 154 cm².
Solution:
Surface area of a sphere = 154 cm²

Question 6.
Find:
(i) the curved surface area.
(ii) the total surface area of a hemisphere of radius 21 cm.
Solution:
Radius of a hemisphere = 21 cm
(i) Curved surface area = 2πr²

Question 7.
A hemispherical brass bowl has inner- diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm².
Solution:
The inner diameter of hemispherical bowl = 10.5 cm

Question 8.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface areas of the balloon in two cases.
Solution:
Original radius of balloon = 7 cm
Radius after filling the air in it = 14 cm
The surface area of balloon, the original position

Question 9.
A sphere and a cube have the same surface. Show that the ratio of the volume of the sphere to that of the cube is √6 : √π
Solution:
Let the edge of a cube = a
Surface area = 6a²
and surface area of sphere = 6a²

Question 10.
(a) If the ratio of the radii of two sphere is 3 : 7, find :
(i) the ratio of their volumes.
(ii) the ratio of their surface areas.
(b) If the ratio of the volumes of the two sphere is 125 : 64, find the ratio of their surface areas.
Solution:
(a) Ratio in radii of two spheres = 3 : 7
Let radius of the first sphere = 3x
and radius of the second sphere = 7x

Question 11.
A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between.
Solution:
Side of a cube = 4 cm
Volume (side)³ = 4 × 4 × 4 = 64 cm³
Diameter of sphere contained by this cube is d = 4 cm

Question 12.
Find the volume of a sphere whose surface area is 154 cm².
Solution:
Given that
Surface area of a sphere = 154 cm²

Question 13.
If the volume of a sphere is \(179 \frac { 2 }{ 3 } \) cm³, find its radius and the surface area.
Solution:
Given that
Volume of a sphere = \(179 \frac { 2 }{ 3 } \) cm³
= \(\\ \frac { 539 }{ 3 } \) cm³

Question 14.
A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Solution:
Radius of a hemispherical bowl (r) = 3.5 cm
= \(\\ \frac { 7 }{ 2 } \) cm

Question 15.
The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill to its capacity. Find the volume of water pumped into the tank.
Solution:
Internal diameter of a hemispherical tank (r) = 14 m
Radius of the tank = \(\\ \frac { 14 }{ 2 } \) = 7 m
Water stored in it = 50 kilolitres of water

Question 16.
The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14.
Solution:
Surface area of a solid sphere = 1256 cm²
By cutting it into two hemisphere,
Curved surface area of each hemisphere

Question 17.
Write whether the following statements are true or false. Justify your answer :
(i) The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.
(ii) The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals the volume of a hemisphere of radius r.
(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1 : 2 : 3.
Solution:
(i) The volume of a sphere is equal to the two third of the volume of a cylinder
whose height and diameter are equal to the diameter of the sphere.

(ii) The volume of the longest right circular cone that can be filled in a cube
whose edge is 2r equal to the volume of a hemisphere of radius r

(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height.
The ratio of their volumes is 1 : 2 : 3

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Ex 14

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 14 Locus Ex 14

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Ex 14
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Chapter Test

Question 1.
A point moves such that its distance from a fixed line AB is always the same. What is the relation between AB and the path traveled by P ?
Solution:
Let point P moves in such a way that
it is at a fixed distance from the fixed line AB.

∴ It is a set of two lines l and m parallel to AB
drawn on either side of it at equal distance from it.

Question 2.
A point P moves so that its perpendicular distance from two given lines AB and CD are equal. State the locus of the point P.
Solution:
(i) When two lines AB and CD are parallel,
then the locus of the point P which is equidistant
from AB and CD is a line (l)
in the midway of AB and CD and parallel to them

(ii) If AB and CD are intersecting lines,
then the locus of the point P will be a
pair of the straight lines l and m which bisect
the angles between the given lines AB and CD.

Question 3.
P is a fixed point and a point Q moves such that the distance PQ is constant, what is the locus of the path traced out by the point Q ?
Solution:
∴ P is a fixed point and Q is a moving point
such that it is always at an equidistant from P.

∴ P is the centre of the path of Q which is a circle.
The distance between P and Q is the radius of the circle.
Hence locus of point Q is a circle with P as centre.

Question 4.
(i) AB is a fixed line. State the locus of the point P so that ∠APB = 90°.
(ii) A, B are fixed points. State the locus of the point P so that ∠APB = 60°.
Solution:
(i) AB is a fixed line and P is a point
such that ∠APB = 90°.

The locus of P will be the circle whose diameter is AB.
We know that the angle in a semi-circle is always equal to 90°.
∠APB = 90°
(ii) AB is a fixed line and P is a point such that ∠APB = 60°.
The locus of P will be a major segment of a circle whose AB is a chord.

Question 5.
Draw and describe the locus in each of the following cases :
(i) The locus of points at a distance 2.5 cm from a fixed line.
(ii) The locus of vertices of all isosceles triangles having a common base.
(iii) The locus of points inside a circle and equidistant from two fixed points on the circle.
(iv) The locus of centres of all circles passing through two fixed points.
(v) The locus of a point in rhombus ABCD which is equidistant from AB and AD. (1998)
(vi) The locus of a point in the rhombus ABCD which is equidistant from points A and C.
Solution:
1. Draw a given line AB.
2. Draw lines of l and m parallel to AB at a distance of 2.5 cm.
Lines l and m are the locus of point P which is at a distance of 2.5 cm.

(ii) ∆ABC is an isosceles triangle in which AB = AC.
From A, draw AD perpendicular to BC.
AD is the locus of the point A the vertices of ∆ABC.
In rt. ∆ABD and ∆ACD
Side AD = AD (Common)
Hyp. AB = AC (given)
∴ ∆ABD = ∆ACD (R.H.S. Axiom)
∴ BD = DC (c.p.c.t.)
Hence locus of vertices of isosceles triangles
having common base is the perpendicular bisector of BC.
(iii) (i) Draw a circle with centre O.

(ii) Take points A and B on it and join them.
(iii) Draw a perpendicular bisector of AB
which passes from O and meets the circle at C.
CE the diameter, which is the locus of a point inside the circle
and equidistant from two points A and B at the circle.
(iv) Let C1, C2, C3 be the centres of the circle
which pass through the two fixed points A and B.

Draw a line XY passing through these centres C1, C2, C3.
Hence locus of centres of circles passing through two points A and B
is the perpendicular bisector of the line segment joining the two fixed points.
(v) In rhombus ABCD, Join AC.

AC is the diagonal of rhombus ABCD
∴ AC bisect ∠A.
∴ Any point on AC, is the locus which is equidistant from AB and AD.
(vi) ABCD is a rhombus. Join BD

BD is the locus of a point in the rhombus which is equidistant from A and C.
Diagonal BD bisects  ∠B and ∠D.
Any point on BD will be equidistant from A and C.

Question 6.
Describe completely the locus of points in each of the following cases :
(i) mid-point of radii of a circle.
(ii) centre of a ball, rolling along a straight line on a level floor.
(iii) point in a plane equidistant from a given line.
(iv) point in a plane, at a constant distance of 5 cm from a fixed point (in the plane).
(v) centre of a circle of varying radius and touching two arms of ∠ADC.
(vi) centre of a circle of varying radius and touching a fixed circle, centre O, at a fixed point A on it.
(vii) centre of a circle of radius 2 cm and touching a fixed circle of radius 3 cm with centre 0.
Solution:
(i) The locus of midpoints of the radii of a circle
is another concentric circle with radius is
half of the radius of the given circle.

(ii) AB is the straight line on the ground and the ball is rolling on it
∴ locus of the centre of the ball is a line parallel A lo the given line AB.

(iii) AB is the given line and P is a point in the plane.

From P, draw a line CD and another line EF from P’ parallel to AB.
Thus CD and EF are the lines which are the locus of the point equidistant from AB
(iv) Take a point O and another point P such that OP = 5 cm.
with centre O and radius equal to OP, draw a circle.
Thus this circle is the locus of point P
which is at a distance of 5 cm from O, the given point.

(v) Draw the bisector BX of ∠ABC.
This bisector of an angle is the locus of the centre of a circle with different radii.
Any point on BX, is equidistant from the arms BA and BC of the ∠ABC.

(vi) A circle with centre O is given and one point A on it.
The locus of the centre of a circle which touches the circle
at the fixed point A on it, is a line joining the points O and A.

(vii) (a) If the circle with 2 cm as radius touches the given circle
externally then the locus of the centre of the circle
will be a concentric circle with radius 3+2 = 5 cm.

If the circle with 2 cm as radius touches the given circle with 3 cm as radius internally,
then the locus of the centre of the circle will be a concentric circle with radius 3-2 = 1 cm.

Question 7.
Using ruler and compasses construct :
(i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm.
(ii) the locus of points equidistant from A and C.
Solution:
(i) Draw BC = 3.4 and mark the arcs of 5.5 add 4.9 cm from B and C.
Join A, B and C.
ABC is the required triangle.
(ii) Draw ⊥ bisector of AC.
(iii) Draw an angle of 90° at AB at A which intersects ⊥ bisector at O.
Draw circle taking 0 as centre and OA as the radius.

Question 8.
Construct triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C and
(ii) P is equidistant from AB and BC
(iii) Measure and record the length of PB. (2000)
Solution:

(i) Take BC = 8 cm a long line segment. At B,
draw a ray BX making an angle of 60° with BC.
Cut off BA = 7 cm. and join AC.
(i) Draw the perpendicular bisector of BC.
(ii) Draw the angle bisector of ∠B which intersect
the perpendicular bisector of BC at P. P is the required point.
(iii) On measuring the length of BP = 4.6 cm (approx.)

Question 9.
A straight line AB is 8 cm long. Locate by construction the locus of a point which is :
(i) Equidistant from A and B.
(ii) Always 4 cm from the line AB.
(iii) Mark two points X and Y, which are 4 cm from AB and equidistant from A and B. Name the figure AXBY.(2008)
Solution:
Steps of construction:

(i) Draw a line segment AB = 8 cm.
(ii) With the help of compasses and ruler,
draw the perpendicular bisector l of AB which intersects AB at O.
(iii) Then any point on l, is equidistant from A and B.
(iv) Cut off OX = OY = 4 cm. The X and Y are the required loci,
which is equidistant from AB and also from A and B.
(v) Join AX, XB, BY and YA.
The figures so formed AXBY is the shape of a square because
its diagonals are equal and bisect each other at right angles.

Question 10.
Use ruler and compasses only for this question.
(i) Construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure arid record the length of PB. (2010)
Solution:
In ∆ABC, AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°

Steps of construction :
(a) (i) Draw a line segment BC = 6 cm
(ii) At, B draw a ray BX making an angle of 60° and cut off BA = 3.5 cm.
(iii) Join AC.
The ∆ABC is the required triangle.
(b) Draw the bisector BY of ∠ABC.
(c) Draw the perpendicular bisector of BC which intersects BY at P.
P is the required point P which is equidistant from BC and BA
and also equidistant from B and C.
On measuring PB it is 3.4 cm (approx)

Question 11.
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Solution:
Steps of construction :

Construct the triangle ABC with AB = 5.5 cm
∠BAC = 105° and AC = 6 cm
(i) Points which are equidistant from BA and BC lies on the bisector of ∠ABC.
(ii) Points equidistant from B and C lies on the perpendicular bisector of BC.
Draw perpendicular bisector of BC.
The required point P is the point of intersection of the bisector of
∠ABC and the perpendicular bisector of BC.
(iii) Required length of PC = 4.8 cm.

Question 12.
In the given diagram, A, B and C are fixed collinear points; D is a fixed point outside the line: Locate

(i) the point P on AB such that CP = DP.
(ii) the points Q such that CQ = DQ = 3 cm. How many such points are possible?
(iii) the points R on AB such that DR = 4 cm. How many such points are possible?
(iv) the points S such that CS = DS and S is 4 cm away from the line CD. How many such points are possible?
(v) Are the points P, Q, R collinear?
(vi) Are the points P, Q, S collinear?
Solution:
Points A, B and C are collinear and D is any point outside AB.

(i) Join CD.
(ii) Draw the perpendicular bisector of CD which meets AB in P.
(iii) P is the required point such that CP = DP
(iv) With centres C and D, draw two arcs with 3 cm radius
which intersect each other at Q and Q’.
Hence there are two points Q and Q’ which are equidistant from C and D.
(v) With centre D, and radius 4 cm draw an arc which intersects AB at R and R’
∴ R and R’ are the two point on AB.
(vi) With centre C and D, draw arcs with a radius equal to
4 cm which intersects each other in S and S’.
∴ There can be two such points which are equidistant from C and D.
(vii) No P, Q, R are not collinear.
(viii) Yes, P, Q, S are collinear.

Question 13.
Points A, B and C represent position of three towers such that AB = 60 m, BC = 73m and CA = 52 m. Taking a scale of 10 m to 1 cm, make an accurate drawing of ∆ABC. Find by drawing, the location of a point which is equidistant from A, B and C, and its actual distance from any of the towers.
Solution:
AB = 60 mm = 6.0 cm, BC = 73 mm = 7.3 cm
and CA = 52 mm = 5.2 cm.

(i) Draw a line segment BC = 7.3cm
(ii) With Centre B and radius 6cm and with centre C
and radius 5.2 cm, draw two arcs intersecting each other at A
(iii) Joining AB and AC.
(iv) Draw the perpendicular bisector of AB, BC and CA respectively,
which intersect each other at point P. Join PB.
P is equidistant from A, B and C on measuring PB = 3.7 cm.
Actual distance = 37 m.

Question 14.
Draw two intersecting lines to include an angle of 30°. Use ruler and compasses to locate points which are equidistant from these lines and also 2 cm away from their point of intersection. How many such points exist ? (1990)
Solution:
(i) Two lines AB and CD intersect each other at O.

(ii) Draw the bisector of ∠BOD and ∠AOD
(iii) With centre O and radius equal to 2 cm.
marks points on the bisector of angles at P, Q, R and S respectively.
Hence there are four points which are equidistant
from AB and CD and 2 cm from 0, the point of intersection of AB and CD.

Question 15.
Without using set square or protractor, construct the quadrilateral ABCD in which ∠BAD = 45°, AD = AB = 6 cm, BC = 3.6 cm and CD = 5 cm.
(i) Measure ∠BCD.
(ii) Locate the point P on BD which is equidistant from BC and CD. (1992)
Solution:
(i) Take AB = 6 cm long
(ii) AT A, draw the angle of 45° and cut off AD = 6 cm

(iii) With centre D and radius 5 cm and with centre B,
and radius 3.5 cm draw two arcs intersecting each other at C.
(iv) Join CD and CB and join BD
ABCD is the required quadrilateral.
(v) On measuring ∠BCD = 65°.
(vi) Draw the bisector of ∠BCD which intersects BD at P.
P is the required point which is equidistant from CD and CB.

Question 16.
Without using set square or protractor, construct rhombus ABCD with sides of length 4 cm and diagonal AC of length 5 cm. Measure ∠ABC. Find the point R on AD such that RB = RC. Measure the length of AR. (1990)

Solution:
(i) Take AB = 4 cm
(ii) With centre A, draw an arc of 5 cm radius
and with B draw another arc of radius 4 cm intersecting each other at C.
(iii) Join AC and BD.
(iv) Again with centre A and C,
draw two arcs of radius 4 cm intersecting each other on D.
(v) Join AD and CD.
ABCD is the required rhombus and on measuring the ∠ABC, it is 78°.
(vi) Draw perpendicular bisector of BC intersecting AD at R.
On measuring the length of AR, it is equal to 1.2 cm.

Question 17.
Without using set-squares or protractor construct :
(i) Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
(ii) Draw the locus of a point which moves so that it is always 2.5 cm from B.
(iii) Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
(iv) Mark the point of intersection of the loci with the letter P and measure PC. (1994)
Solution:
Steps of Construction :
(i) Draw BC = 3.2 cm long.
(ii) With centre B and radius 5.5 cm
and with centre C and radius 4.8 cm
draw arcs intersecting each other at A.

(iiii) Join AB and AC.
(iv) Draw the bisector of ∠BCA.
(v) With centre B and radius 2.5 cm,
draw an arc intersecting the angle bisector of ∠BCA at P and P’.
P and P’ are two loci which satisfy the given condition.
On measuring CP and CP’
CP = 3.6 cm and CP’ =1.1 cm.

Question 18.
By using ruler and compasses only, construct an isosceles triangle ABC in which BC = 5 cm, AB = AC and ∠BAC = 90°. Locate the point P such that :
(i) P is equidistant from the sides BC and AC.
(ii) P is equidistant from the points B and C.
Solution:
Steps of Construction :
(i) Take BC = 5.0 cm and bisect it at D.

(ii) Taking BC as diameter, draw a semicircle.
(iii) At D, draw a perpendicular intersecting the circle at A
(iv) Join AB and AC.
(v) Draw the angle bisector of C intersecting
the perpendicular at P. P is the required point.

Question 19.
Using ruler and compasses only, construct a quadrilateral ABCD in which AB = 6 cm, BC = 5 cm, ∠B = 60°, AD = 5 cm and D is equidistant from AB and BC. Measure CD. (1983)
Solution:
Steps of Construction :
(i) Draw AB = 6 cm

(ii) At B, draw angle of 60° and cut off BC = 5 cm.
(iii) Draw the angle bisector of ∠B.
(iv) With centre A and radius 5 cm. draw an arc
which intersects the angle bisector of ∠B at D
(v) Join AD and DC.
ABCD in the required quadrilateral.
On measuring CD, it is 5.3 cm (approx).

Question 20.
Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB (2001)
Solution:
Steps of Construction :
(i) Draw a line AB = 6 cm

(ii) With centre A and B and radius 4 cm,
draw two arcs intersecting each other at C.
(iii) Join CA and CB
(iv) Draw the bisector of ∠C and cut off CP = 5 cm
(v) Draw a line XY parallel to AB at a distance of 5 cm.
(vi) From P, draw arcs of radius 5 cm each intersecting
the line XY at Q and R. Hence Q and R are the required points.

Question 21.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of length 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC, (1995)
Solution:
Steps of Construction :
(i) With centre 0 and radius 4 cm draw a circle.
(ii) Take point A on this circle.

(iii) With centre A and radius 6 cm draw an arc cutting the circle at B.
(iv) Again with radius 5 cm, draw another arc cutting the circle at C.
(v) Join AB and AC.
(vi) Draw the perpendicular bisector of AC.
Any point on it, will be equidistant from A and C.
(vii) Draw the angle bisector of ∠A intersecting
the perpendicular bisector of AC at P. P is the required locus.

Question 22.
Ruler and compasses only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm. and ∠ABC = 60°.
(ii) Construct the locus of all points, inside ∆ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to ∆ABC.
(iv) Mark the point Q, in your construction, which would make ∆QBC equal in area to ∆ABC, and isosceles.
(v) Measure and record the length of CQ. (1998)
Solution:
Steps of Construction :
(i) Draw AB = 9 cm
(ii) At B draw an angle of 60° and cut off BC = 6 cm.
(iii) Join AC

(iv) Draw perpendicular bisector of BC.
All points on it will be equidistant from B and C.
(v) From A, draw a line XY parallel to BC.
(vi) Produce the perpendicular bisector of BC to meet XY in Q.
(vii) Join QC and QB.
∆QBC will be the triangle equal in area to ∆ABC
because these are on the same base BC and between the same parallel lines.
On measuring, the length of CQ is 8.2 cm (approx.).

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Ex 14 are helpful to complete your math homework.

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2

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 17 Mensuration Ex 17.2

More Exercise

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4
• 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 MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

Take π = \(\\ \frac { 22 }{ 7 } \) unless stated otherwise.

Question 1.
Write whether the following statements are true or false. Justify your answer.
(i) If the radius of a right circular cone is halved and its height is doubled, the volume will remain unchanged.
(ii) A cylinder and a right circular cone are having the same base radius and same height. The volume of the cylinder is three times the volume of the cone.
(iii) In a right circular cone, height, radius and slant height are always the sides of a right triangle.
Solution:
(i) If the radius of a right circular cone is halved and its height is doubled,
then the volume will remain unchanged
It is wrong as

(ii) A cylinder and a right circular cone are having the same base radius
and same height the volume of the cylinder is three times the volume of cone – It is true as
Volume of cylinder = \(\pi { r }^{ 2 }h=3\times \frac { 1 }{ 3 } \pi { r }^{ 2 }h\) = 3(volume of cone)
(iii) In a right circular cone, height, radius and slant height are always the sides of a right triangle
It is true as in a cone and in a right-angled triangle.
Hypotenuse (slant x height) = r² + h²
and cone is formed by revolving the right triangle about the perpendicular.

Question 2.
Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Solution:
10 Slant height of a cone (l) = 10 cm
and radius of the base = 7 cm
Curved surface area = πrl
= \(\\ \frac { 22 }{ 7 } \) × 7 × 10 = 220 cm²

Question 3.
Diameter of the base of a cone is 10.5 cm and slant height is 10 cm. Find its curved surface area.
Solution:
The diameter of the base of a cone = 10.5cm
Its radius (r) = \(\\ \frac { 10.5 }{ 7 } \) = 5.25 cm
and slant height (l) = 10 cm
Curved surface area = πrl
= \(\\ \frac { 22 }{ 7 } \) × 5.25 × 10 cm²
= 165.0 cm²

Question 4.
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find ,
(i)radius of the base
(ii)total surface area of the cone.
Solution:
Curved surface area of a cone = 308 cm²
Slant height = 14 cm

Question 5.
Find the volume of the right circular cone with
(i) radius 6 cm and height 7 cm
(ii) radius 3.5 cm and height 12 cm.
Solution:
(i) Radius of cone (r) = 6 cm
and height (h) = 7 cm
Volume = \(\frac { 1 }{ 3 } \pi { r }^{ 2 }h \)

Question 6.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
Solution:
(i) Radius = 7 cm
and slant height (l) = 25 cm

Question 7.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?
Solution:
Diameter of top of conical pit = 3.5 m
Radius (r) = \(\\ \frac { 3.5 }{ 2 } \) = 1.75 m
and depth (h) = 12m

Question 8.
If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Solution:
Volume of a right circular cone = 48π cm³
Height (h) = 9 cm

Question 9.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)
Solution:
Height of cone (h) = 15 cm
Volume = 1570 cm³

Question 10.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100 m².
Solution:
Slant height of conical tomb (l) = 25 m
and base diameter = 14 m
Radius(r) = \(\\ \frac { 14 }{ 2 } \) = 7m

Question 11.
A conical tent is 10 m high and the radius of its base is 24 m. Find :
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70.
Solution:
Height of a conical tent (h) = 10 m
and radius (r) = 24 m

Question 12.
A Jocker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the cloth required to make 10 such caps.
Solution:
Base radius of a conical cap = 7 cm
and height (h) = 24 cm

= 550 x 10
= 5500 cm²

Question 13.
(a) The ratio of the base radii of two right circular cones of the same height is 3 : 4. Find the ratio of their volumes.
(b) The ratio of the heights of two right circular cones is 5 : 2 and that of their base radii is 2 : 5. Find the ratio of their volumes.
(c) The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find:
(i) the ratio of their volumes.
(ii) the ratio of their lateral surface areas.
Solution:
(i) The ratio in base radii of two right circular cones of the same height = 3 : 4
Let h be the height and radius of first cone = 3x and
Radius of second cone = 4x

Question 14.
Find what length of canvas 2 m in width is required to make a conical tent 20 m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of Rs 80 per metre.
Solution:
Diameter of the base of the conical tent = 20 m
Radius (r) = \(\\ \frac { 20 }{ 2 } \) = 10 m
and slant height (h) = 42 m

Question 15.
The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.
Solution:
Perimeter of the base of a cone = 44 cm

Question 16.
The volume of a right circular cone is 9856 cm³ and the area of its base is 616 cm². Find
(i) the slant height of the cone.
(ii) total surface area of the cone.
Solution:
Volume of a circular cone = 9856 cm³
Area of the base = 616 cm²

Question 17.
A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed. (Take π = 3.14)
Solution:
Sides of a right triangle are 6 cm and 8 cm
It is revolved around 8 cm side
Radius (r) = 6 cm
Height (h) = 8 cm
Slant height (l) = 10 cm

Question 18.
The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be \(\\ \frac { 1 }{ 27 } \) of the volume of the given cone, at what height above the base is the section cut?
Solution:
Height of a cone (H) = 30 cm
A small cone is cut off from the top of the cone given


h = 10 cm
∴ A line parallel to base at a distance of 30 – 10 = 20 cm is drawn.

Question 19.
A semi-circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find
(i) the radius of the cone.
(ii) the (lateral) surface area of the cone.
Solution:
Radius of a semi-circular lamina = 35 cm
By folding it a cone is formed whose slant height (l) = r = 35
and half circumference = circumference of the top of the cone

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1

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 17 Mensuration Ex 17.1

More Exercise

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4
• 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 MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

Take π = \(\\ \frac { 22 }{ 7 } \) unless stated otherwise.

Question 1.
Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of π.
Solution:
Radius of the cylinder (r) = 5 cm
Height (h) = 10 cm
Total surface area = 2πr (h + r)
= 2π x 5(10 + 5) cm²
= 10 x 15π
= 150π cm²

Question 2.
An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1.2m. Neglecting the thickness of its walls, calculate
(i) its outer lateral surface area,
(ii) its capacity in litres.
Solution:
Diameter of cylindrical geyser = 35 cm
Radius (r) = \(\\ \frac { 35 }{ 2 } \) cm
Height = 1.2 m = 120 cm

Question 3.
A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto a height of 12 cm, find how many litres of milk is needed to serve 1600 students.
Solution:
Number of students = 1600
Diameter of cylindrical glasses = 7 cm
Radius (r) = \(\\ \frac { 7 }{ 2 } \) cm

Question 4.
In the given figure, a rectangular tin foil of size 22 cm by 16 cm is wrapped around to form a cylinder of height 16 cm. Find the volume of the cylinder.

Solution:
Length of rectangular tin foil (l) = 22 cm
and breadth (b) = 16 cm
By folding lengthwise, the radius of the cylinder

Question 5.
(i) How many cubic metres of soil must be dug out to make a well 20 metres deep and 2 metres in diameter?
(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of Rs 50 per m², find the cost of plastering.
Solution:
(i) Depth of well (h) = 20 m
and diameter = 2 m

Question 6.
A roadroller (in the shape of a cylinder) has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120 m by 44 m.
Solution:
Diameter of a road roller = 0.7 m
Radius (r) = \(\\ \frac { 0.7 }{ 2 } \) = 0.35 m
and width (h) = 1.2 m

Question 7.
If the volume of a cylinder of height 7 cm is 448 π cm³, find its lateral surface area and total surface area.
Solution:
Volume of a cylinder = 448 π cm³
Height (h) = 7 cm

Question 8.
A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m³.
Solution:
Height of a wooden pole (h) = 7 m
Diameter = 20 cm

Question 9.
The area of the curved surface of a cylinder is 4400 cm², and the circumference of its base is 110 cm. Find
(i) the height of the cylinder.
(ii) the volume of the cylinder.
Solution:
Area of the curved surface of a cylinder = 4400 cm²
Circumference of base = 110 cm

Question 10.
A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm². Find
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place. (Take π = 3.14)
Solution:
Diameter of a cylinder = 20 cm
Radius (r) = \(\\ \frac { 20 }{ 2 } \) = 10 cm
Curved surface area = 1000 cm²

Question 11.
The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?
Answer correct to the nearest. 100 words.
Solution:
Height of cylindrical barrel of a pen (h) = 7 cm
Diameter = 5 mm

Question 12.
Find the ratio between the total surface area of a cylinder to its curved surface area given that its height and radius are 7.5 cm and 3.5 cm.
Solution:
Radius of a cylinder (r) = 3.5 cm
and height (h) = 7.5 cm
Total surface area = 2πr(r + h)

Question 13.
The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder?
Solution:
Let the radius of the base of a right circular cylinder = r
and height (h) = h
Volume = πr²h

Question 14.
(i) The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm². Find the height and the volume of the cylinder.
(ii) The total surface area of a cylinder is 352 cm². If its height is 10 cm, then find the diameter of the base.
Solution:
Sum of radius and height of a cylinder = 37 cm
Total surface area = 1628 cm²
Let r be radius and h be height, then r × h = 37
and 2πr(r + h) = 1628

Question 15.
The ratio between the curved surface and the total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm².
Solution:
Ratio between curved surface area and total surface area = 1 : 2
Total surface area = 616 cm²

Question 16.
Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.
Solution:
Volume of two cylinders is the same
Diameter of both cylinder are in the ratio = 3 : 4

Question 17.
A rectangular sheet of tin foil of size 30 cm x 18 cm can be rolled to form a cylinder in two ways along length and along breadth. Find the ratio of volumes of the two cylinders thus formed.
Solution:
Size of the sheet = 30 cm × 18 cm
(i) By rolling lengthwise,
The circumference of the cylinder = 2πr = 30

Question 18.
A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the volume of the metal.
Solution:
Internal diameter of a metal tube = 11.2 cm
and radius (r) = \(\\ \frac { 11.2 }{ 2 } \) = 5.6 cm
Length (h) = 21 cm
Thickness of metal = 0.4 cm
External radius (R) = 5.6 + 0.4 = 6.0 cm

Question 19.
The given figure shows a metal pipe 77 cm long. The inner diameter of a cross-section is 4 cm and the outer one is 4.4 cm. Find its
(i) inner curved surface area
(ii) outer curved surface area
(iii) total surface area.

Solution:
In the given figure,
Length of metal pipe (h) = 77 cm
Inner diameter = 4 cm

Question 20.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution:
Diameter of the pencil = 7 mm
Radius (R) = \(\\ \frac { 7 }{ 2 } \) mm = \(\\ \frac { 7 }{ 20 } \) cm
Diameter of graphite (lead) = 1 mm

Question 21.
A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Solution:
(i) Base of the tin of rectangular base = 5 cm × 4 cm
Height = 15 cm
Volume = lbh = 5 × 4 × 15 = 300 cm³
(ii) Base diameter of cylindrical plastic cylinder = 7 cm

Question 22.
A cylindrical roller made of iron is 2 m long. Its inner diameter is 35 cm and the thickness is 7 cm all round. Find the weight of the roller in kg, if 1 cm³ of iron weighs 8 g.
Solution:
Length of cylindrical roller (h) = 2 m = 200 cm
Diameter = 35 cm
Inner radius = \(\\ \frac { 35 }{ 2 } \) cm
Thickness = 7 cm

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.3

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 15 Circles Ex 15.3

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1
• 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 15.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

Question 1.
Find the length of the tangent drawn to a circle of radius 3 cm, from a point distant 5 cm from the centre.
Solution:
In a circle with centre O and radius 3 cm
and P is at a distance of 5 cm.

Question 2.
A point P is at a distance 13 cm from the centre C of a circle and PT is a tangent to the given circle. If PT = 12 cm, find the radius of the circle.
Solution:
CT is the radius
CP = 13 cm and tangent PT = 12 cm

Question 3.
The tangent to a circle of radius 6 cm from an external point P, is of length 8 cm. Calculate the distance of P from the nearest point of the circle.

Solution:
Radius of the circle = 6 cm
and length of tangent = 8 cm
Let OP be the distance
i.e. OA = 6 cm, AP = 8 cm .
OA is the radius

Question 4.
Two concentric circles are of the radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.
Solution:
Two concentric circles with centre O
OP and OB are the radii of the circles respectively, then
OP = 5 cm, OB = 13 cm.

Question 5.
Two circles of radii 5 cm and 2-8 cm touch each other. Find the distance between their centres if they touch :
(i) externally
(ii) internally.
Solution:
Radii of the circles are 5 cm and 2.8 cm.
i.e. OP = 5 cm and CP = 2.8 cm.

(i) When the circles touch externally,
then the distance between their centres = OC = 5 + 2.8 = 7.8 cm.
(ii) When the circles touch internally,
then the distance between their centres = OC = 5.0 – 2.8 = 2.2 cm

Question 6.
(a) In figure (i) given below, triangle ABC is circumscribed, find x.
(b) In figure (ii) given below, quadrilateral ABCD is circumscribed, find x.

Solution:
(a) From A, AP and AQ are the tangents to the circle
∴ AQ = AP = 4cm

Question 7.
(a) In figure (i) given below, quadrilateral ABCD is circumscribed; find the perimeter of quadrilateral ABCD.
(b) In figure (ii) given below, quadrilateral ABCD is circumscribed and AD ⊥ DC ; find x if radius of incircle is 10 cm.

Solution:
(a) From A, AP and AS are the tangents to the circle
∴AS = AP = 6
From B, BP and BQ are the tangents
∴BQ = BP = 5
From C, CQ and CR are the tangents

Question 8.
(a) In the figure (i) given below, O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, find the radius of the circle.
(b) In the figure (ii) given below, from an external point P, tangents PA and PB are drawn to a circle. CE is a tangent to the circle at D. If AP = 15 cm, find the perimeter of the triangle PEC.

Solution:
(i) Join OB
∠OBA = 90°
(Radius through the point of contact is
perpendicular to the tangent)

Question 9.
(a) If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by
\(r= \frac { a+b-c }{ 2 } \)
(b) In the given figure, PB is a tangent to a circle with centre O at B. AB is a chord of length 24 cm at a distance of 5 cm from the centre. If the length of the tangent is 20 cm, find the length of OP.

Solution:
(a) Let the circle touch the sides BC, CA and AB
of the right triangle ABC at points D, E and F respectively,
where BC = a, CA = b
and AB = c (as showing in the given figure).

Question 10.
Three circles of radii 2 cm, 3 cm and 4 cm touch each other externally. Find the perimeter of the triangle obtained on joining the centres of these circles.
Solution:
Three circles with centres A, B and C touch each other externally
at P, Q and R respectively and the radii of these circles are
2 cm, 3 cm and 4 cm.

Question 11.
(a) In the figure (i) given below, the sides of the quadrilateral touch the circle. Prove that AB + CD = BC + DA.
(b) In the figure (ii) given below, ABC is triangle with AB = 10cm, BC = 8cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with vertices A, B and C as their centres. Find the radii of the three circles

Solution:
(a) Given: Sides of quadrilateral ABCD touch the circle at
P, Q, R and S respectively.

Question 12.
(a) ln the figure (i) PQ = 24 cm, QR = 7 cm and ∠PQR = 90°. Find the radius of the inscribed circle ∆PQR

(b) In the figure (ii) given below, two concentric circles with centre O are of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP = 12cm, find BP.

Solution:
(a) In the figure, a circle is inscribed in the triangle PQR
which touches the sides. O is centre of the circle.
PQ = 24cm, QR = 7 cm ∠PQR = 90°
OM is joined.

Question 13.
(a) In the figure (i) given below, AB = 8 cm and M is mid-point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all three semi-circles as shown, find its radius.

(b) In the figure (ii) given below, equal circles with centres O and O’ touch each other at X. OO’ is produced to meet a circle O’ at A. AC is tangent to the circle whose centre is O. O’D is perpendicular to AC. Find the value of :
(i) \(\\ \frac { AO’ }{ AO } \)
(ii) \(\frac { area\quad of\quad \Delta ADO’ }{ area\quad of\quad \Delta ACO } \)

Solution:
(a) Let x be the radius of the circle
with centre C and radii of each equal

Question 14.
The length of the direct common tangent to two circles of radii 12 cm and 4 cm is 15 cm. Calculate the distance between their centres.
Solution:
Let R and r be the radii of the circles
with centre A and B respectively
Let TT’ be their common tangent.

Hence distance between their centres = 17 cm Ans.

Question 15.
Calculate the length of a direct common tangent to two circles of radii 3 cm and 8 cm with their centres 13 cm apart.
Solution:
Let A and B be the centres of the circles
whose radii are 8 cm and 3 cm and
let TT’ length of their common tangent and AB = 13 cm.

Question 16.
In the given figure, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively. Given that AB = 8 cm, calculate PQ.

Solution:
AC is a transverse common tangent to the two circles
with centre P and Q and of radii 6 cm and 3 cm respectively
AB = 8 cm. Join AP and CQ.

Question 17.
Two circles with centres A, B are of radii 6 cm and 3 cm respectively. If AB = 15 cm, find the length of a transverse common tangent to these circles.
Solution:
AB = 15 cm.
Radius of the circle with centre A = 6 cm
and radius of second circle with radius B = 3 cm.

Question 18.
(a) In the figure (i) given below, PA and PB are tangents at a points A and B respectively of a circle with centre O. Q and R are points on the circle. If ∠APB = 70°, find (i) ∠AOB (ii) ∠AQB (iii) ∠ARB
(b) In the figure (ii) given below, two circles touch internally at P from an external point Q on the common tangent at P, two tangents QA and QB are drawn to the two circles. Prove that QA = QB.

Solution:
(a) To find : (i) ∠AOB, (ii) ∠AQB, (iii) ∠ARB
Given: PA and PB are tangents at the points A and B respectively
of a circle with centre O and OA and OB are radii on it.
∠APB = 70°
Construction: Join AB

Question 19.
In the given figure, AD is a diameter of a circle with centre O and AB is tangent at A. C is a point on the circle such that DC produced intersects the tangent of B. If ∠ABC = 50°, find ∠AOC.

Solution:
Given AB is tangent to the circle at A and OA is radius, OA ⊥ AB
In ∆ABD

Question 20.
In the given figure, tangents PQ and PR are drawn from an external point P to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ, Find ∠RQS

Solution:
In the given figure,
PQ and PR are tangents to the circle with centre O drawn from P
∠RPQ = 30°
Chord RS || PQ is drawn
To find ∠RQS
∴ PQ = PR (tangents to the circle)

Question 21.
(a) In the figure (i) given below, PQ is a tangent to the circle at A, DB is a diameter, ∠ADB = 30° and ∠CBD = 60°, calculate (i) ∠QAB (ii) ∠PAD (iii) ∠CDB.

(b) In the figure (ii) given below, ABCD is a cyclic quadrilateral. The tangent to the circle at B meets DC produced at F. If ∠EAB = 85° and ∠BFC = 50°, find ∠CAB.

Solution:
(a) PQ is tangent and AD is chord
(i) ∴ ∠QAB = ∠BDA = 30°
(Angles in the alternate segment)
(ii) In ∆ADB,
∠DAB = 90° (Angle in a semi-circle)



⇒ ∠CAB = 35°

Question 22.
(a) In the figure (i) given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the value of x, y and z. (2015)
(b) In the figure (ii) given below, O is the centre of the circle. PS and PT are tangents and ∠SPT = 84°. Calculate the sizes of the angles TOS and TQS.

Solution:
Consider the following figure:
TS ⊥ SP,
∠TSR = ∠OSP = 90°
In ∆TSR,
∠TSR + ∠TRS + ∠RTS = 180°

Question 23.
In the given figure, O is the centre of the circle. Tangents to the circle at A and B meet at C. If ∠ACO = 30°, find
(i) ∠BCO (ii) ∠AOR (iii) ∠APB

Solution:
(i) ∠BCO = ∠ACO = 30°
(∵ C is the intersecting point of tangent AC and BC)
(ii) ∠OAC = ∠OBC = 90°
∵∠AOC = ∠BOC = 180° – (90° + 30°) = 60°
(∵ sum of the three angles a ∆ is 180°)

Question 24.
(a) In the figure (i) given below, O is the centre of the circle. The tangent at B and D meet at P. If AB is parallel to CD and ∠ ABC = 55°. find: (i)∠BOD (ii) ∠BPD
(b) In the figure (ii) given below. O is the centre of the circle. AB is a diameter, TPT’ is a tangent to the circle at P. If ∠BPT’ = 30°, calculate : (i)∠APT (ii) ∠B OP.

Solution:
(a) AB || CD
(i) ∠ABC = ∠BCD (Alternate angles)
⇒ ∠BCD = 55°

Question 25.
In the adjoining figure, ABCD is a cyclic quadrilateral.

The line PQ is the tangent to the circle at A. If ∠CAQ : ∠CAP = 1 : 2, AB bisects ∠CAQ and AD bisects ∠CAP, then find the measure of the angles of the cyclic quadrilateral. Also prove that BD is a diameter of the circle.
Solution:
ABCD is a cyclic quadrilateral.
PAQ is the tangent to the circle at A.
∠CAD : ∠CAP = 1 : 2.
AB and AD are the bisectors of ∠CAQ and ∠CAP respectively

Question 26.
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at P, Q and R respectively. Calculate (i) ∠QOR (ii) ∠QPR given that ∠A = 60°.
Solution:
OQ and OR are the radii and AC and AB are tangents.
OQ ⊥ AC and OR ⊥ AB
Now in the quad. AROQ

Question 27.
(a) In the figure (0 given below, AB is a diameter. The tangent at C meets AB produced at Q, ∠CAB = 34°. Find
(i) ∠CBA (ii) ∠CQA (2006)

(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. Given ∠APB = 60°, calculate.
(i) ∠AOB (ii) ∠OAB (iii) ∠ACB.

Solution:
(a) AB is the diameter.

Question 28.
(a) In the figure (i) given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY. (1994)
(b) In the figure (ii) given below, O is the centre of the circle and PT is the tangent to the circle at P. Given ∠QPT = 30°, calculate (i) ∠PRQ (ii) ∠POQ.

Solution:
(a) Join OY, OX and OY are the radii of the circle
and XT and YT are the tangents to the circle.

Question 29.
Two chords AB, CD of a circle intersect internally at a point P. If
(i) AP = cm, PB = 4 cm and PD = 3 cm, find PC.
(ii) AB = 12 cm, AP = 2 cm, PC = 5 cm, find PD.
(iii) AP = 5 cm, PB = 6 cm and CD = 13 cm, find CP.
Solution:
In a circle, two chords AB and CD intersect
each other at P internally.

Question 30.
(a) In the figure (i) given below, PT is a tangent to the circle. Find TP if AT = 16 cm and AB = 12 cm.

(b) In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find (i) AB. (ii)the length of tangent PT.

Solution:
(a) PT is the tangent to the circle and AT is a secant.
PT² = TA × TB
Now TA = 16 cm, AB = 12 cm
TB = AT – AB = 16 – 12 = 4 cm
∴ PT² = 16 + 4 = 64 = (8)²
⇒ PT = 8 cm or TP = 8 cm
(b) PT is tangent and PDC is secant out to the circle
∴ PT² = PC × PD

Question 31.
PAB is secant and PT is tangent to a circle
(i) PT = 8 cm and PA = 5 cm, find the length of AB.
(ii) PA = 4.5 cm and AB = 13.5 cm, find the length of PT.
Solution:
∵ PT is the tangent and PAB is the secant of the circle.

Question 32.
In the adjoining figure, CBA is a secant and CD is tangent to the circle. If AB = 7 cm and BC = 9 cm, then
(i) Prove that ∆ACD ~ ∆DCB.
(ii) Find the length of CD.

Solution:
In ∆ACD and ∆DCB
∠C = ∠C (common)
∠CAD = ∠CDB

Question 33.
(a) In the figure (i) given below, PAB is secant and PT is tangent to a circle. If PA : AB = 1:3 and PT = 6 cm, find the length of PB.
(b) In the figure (ii) given below, ABC is an isosceles triangle in which AB = AC and Q is mid-point of AC. If APB is a secant, and AC is tangent to the circle at Q, prove that AB = 4 AP.

Solution:
(a) In the figure (i),
PAB is secant and PT is the tangent to the circle.
PT² = PA × PB

Question 34.
Two chords AB, CD of a circle intersect externally at a point P. If PA = PC, Prove that AB = CD.
Solution:
Given: Two chords AB and CD intersect
each other at P outside the circle. PA = PC.

Question 35.
(a) In the figure (i) given below, AT is tangent to a circle at A. If ∠BAT = 45° and ∠BAC = 65°, find ∠ABC.

(b) In the figure (ii) given below, A, B and C are three points on a circle. The tangent at C meets BA produced at T. Given that ∠ATC = 36° and ∠ACT = 48°, calculate the angle subtended by AB at the centre of the circle. (2001)

Solution:
(a) AT is the tangent to the circle at A
and AB is the chord of the circle.

Question 36.
In the adjoining figure ∆ABC is isosceles with AB = AC. Prove that the tangent at A to the circumcircle of ∆ABC is parallel to BC.

Solution:
Given: ∆ABC is an isosceles triangle with AB = AC.
AT is the tangent to the circumcircle at A.

Question 37.
If the sides of a rectangle touch a circle, prove that the rectangle is a square.
Solution:
Given: A circle touches the sides AB, BC, CD and DA
of a rectangle ABCD at P, Q, R and S respectively.

To Prove : ABCD is a square.

Question 38.
(a) In the figure (i) given below, two circles intersect at A, B. From a point P on one of these circles, two line segments PAC and PBD are drawn, intersecting the other circle at C and D respectively. Prove that CD is parallel to the tangent at P.

(b) In the figure (ii) given below, two circles with centres C, C’ intersect at A, B and the point C lies on the circle with centre C’. PQ is a tangent to the circle with centre C’ at A. Prove that AC bisects ∠PAB.

Solution:
Given: Two circles intersect each other at A and B.
From a point P on one circle, PAC and PBD are drawn.
From P, PT is a tangent drawn. CD is joined.

Question 39.
(a) In the figure (i) given below, AB is a chord of the circle with centre O, BT is tangent to the circle. If ∠OAB = 32°, find the values of x and y.

(b) In the figure (ii) given below, O and O’ are centres of two circles touching each other externally at the point P. The common tangent at P meets a direct common tangent AB at M. Prove that:
(i) M bisects AB (ii) ∠APB = 90°.

Solution:
AB is a chord of a circle with centre O.
BT is a tangent to the circle and ∠OAB = 32°.
∴ In ∆OAB,
OA = OB (radii of the same circle)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.3 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.

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

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Chapter Test

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.
   

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.
   

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
   

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.
   

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.
   

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
   

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.
   

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.
   

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

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

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4
• 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 MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

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

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 = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi \times { \left( 9 \right) }^{ 3 }{ cm }^{ 3 }\)

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 = \(\frac { 2 }{ 3 } \pi { r }^{ 3 }=\frac { 2 }{ 3 } \pi \times { \left( 8 \right) }^{ 3 }{ cm }^{ 3 }\)

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

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

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

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 = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

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

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

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

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 = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi \times { \left( 6 \right) }^{ 3 }{ cm }^{ 3 }\)

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

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

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

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

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

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

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

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 = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

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 =\(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

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,

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

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

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

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

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

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

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

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

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