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.

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.

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 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

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:
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 .

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

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)
∠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.

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:
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

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)

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 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):

Solution:
From the figure
(i) ABCD is a cyclic quadrilateral

Question 2.
(a) In the figure (i) given below, O is the centre of the circle. If ∠AOC = 150°, find (i) ∠ABC (ii) ∠ADC.

(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.

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.

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

(b) In the figure (if) given below, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find:

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°

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.

(b) In the figure given below, O is the centre of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC.

Solution:
(a) In the given figure, ABCD is a cyclic quadrilateral
∠ADC = 80° and ∠ACD = 52°
To find the measure of ∠ABC and ∠CBD

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:

Solution:
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°
(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.

(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

Solution:
(a) (i) ABCD is a cyclic quadrilateral

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)

(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007)

Solution:
(a) In ∆PQR,
∠PRQ = 90° (Angle in a semi circle) and ∠PQR = 58°
∠RPQ = 90° – ∠PQR = 90° – 58° = 32°
SR || PQ (given)

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.

Solution:
(a) Construction: Join BC, and AC then

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 :

Solution:
(a) ABCD is a cyclic quadrilateral

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.

(b) In the figure (ii) given below, O is the centre of the circle. If ∠OAD = 50°, find the values of x and y.

Solution:
(a) PQRS is a cyclic quadrilateral in which
PQ = QR

Question 11.
(a) In the figure (i) given below, O is the centre of the circle. If ∠COD = 40° and ∠CBE = 100°, then find :
(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)

Solution:
(a) (i) ∴ ABCD is a cyclic quadrilateral.
(ii) Arc CD subtends ∠COD at the centre
and ∠CAD at the remaining part of the circle

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.

Solution:
In the given figure,
O is the centre of the semi-circle ABCDE
and AOE is the diameter. AB = BC, ∠AEC = 50°

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.

(b) In the figure (ii) given below, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = RS.

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)

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.

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
= ∠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.

To prove: ABCD is a rectangle
Proof: ABCD is a parallelogram
∠A = ∠C and ∠B = ∠D

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

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

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

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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

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.
∠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:
(Angles in the same segment of a circle)

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.

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:
(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

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.
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.

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

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
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
(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.
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
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 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

To prove : PT × QR = PR × ST
Proof: ∠1 = ∠2
∠1 + ∠6 = ∠2 + ∠6
∠SPT = ∠QPR
In ∆PQR and ∆PST

Question 2.
In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.

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,

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.

Solution:
(a) Given : In following figure, ∠AED = ∠ABC
Required: The values of x and y.
∠AED = ∠ABC (given)
∠A = ∠A (common)
(By A.A. axiom of similarity)

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.

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

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

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

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

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

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.

Solution:
In ∆ABC and ∆ACD
∠C = ∠C (Common)
∴ ∆ABC ~ ∆ACD

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.

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.

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.

Solution:
In the given figure, ABCD is trapezium in
which AB || DC, 2AB = 3DC

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 .

(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)
∆AOD ~ ∆EOC (AA postulate)

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

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

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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°

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

Solution:
In the given figure
∆ABC ~ ∆QPR

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°

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

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

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

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

Solution:
In the given figure, AB || DE
∆ABC ~ ∆DCE

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

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

Solution:
In the given figure,
DE || BC

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

Solution:
In the given figure,
PQ || CA
Let BC = x

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

Solution:
In the given figure, MN || QR
PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm
Let PM = x cm

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

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
$$\frac { BC }{ QR } =\frac { 1 }{ 3 }$$

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
$$\frac { area\quad of\quad \Delta ABC }{ area\quad of\quad \Delta PQR }$$

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 = ?

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

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

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

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

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

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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

(By theorem 15.1)
But BC = QR =1 : 3

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,

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

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

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)

(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.

Solution:
In ∆AOQ and ∆BOP, we have
∠ OAQ = ∠ OBP [Each = 90°]
∠ AOQ= ∠BOP
[Vertically opposite angles]
∆AOQ ~ ∆BOP [A.A. similarity]

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.

(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.

Solution:
(a) In the figure,
DE || BC
∠D = ∠B and ∠E = ∠C
(Corresponding angles)

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.

(iii) If area of ∆ABC = 18cm², find the area of trapezium DBCE
Solution:
(i) Given : In ∆ABC, DE || BC.
To prove : ∆ADE ~ ∆ABC
∠A = ∠A (common)
∴ ∆ADE ~ ∆ABC. (AA axiom)

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.

Solution:
(i) To prove : ∆ABC ~ ∆DEC
In ∆ABC and ∆DEC

Question 9.
In the adjoining figure, ABC is a triangle.

(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)

Solution:

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)

Solution:
In the figure,
PQ || BC and PO is produced to Q such that CQ || BA
and AP : PB = 2 : 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
(ii) area of ∆BEF : area of ∆ABD
(iii) area of ∆ABD : area of trap. AFED
(iv) area of ∆FEO : area of ∆OBC.

Solution:
(a) In trapezium ABCD, AB || DC.
∠OAB = ∠OCD [alternate angles]
∠OBA = ∠ODC
∆AOB ~ ∆COD

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

(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

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.

(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.

Solution:
(a) In ∆ABC, DE || BC
∠A = ∠A (common)
∠D = ∠B and ∠E = ∠C
(Corresponding angles)

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.

(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.

Solution:
(a) In trapezium ABCD, DC || AB

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:
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.

Solution:
∠A = ∠A (Common)
m∠B = m∠E = 90°
Thus by angle-angle similarity, triangles,
Since they are similar triangles,
the sides are proportional Thus, we have,

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

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

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.

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)

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

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.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.2

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 13 Similarity Ex 13.2

More Exercises

Question 1.
(a) In the figure (i) given below if DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm. Find (i) AE : EC (ii) DE.
(b) In the figure (ii) given below, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm. Find CQ and BQ.
(c) In the figure (iii) given below, if XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, find PY and XY.

Solution:
(a) In the figure (i)
Given : DE || BC, AD = 3 cm, BD = 4 cm and BC = 5 cm.
To find (i) AE : EC and
(ii) DE Since DE || BC of ∆ABC

Question 2.
In the given figure, DE || BC.
(i) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.
(ii) If DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3, find the value of x.

Solution:
In the given figure, DE || BC
(i) AD = x, DB = x – 2, AE = x + 2, EC = x – 1
In ∆ABC,
∵DE || BC

Question 3.
E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 8 cm and RF = 9 cm.
(ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Solution:
(i) In ∆PQR, E and F are the points on the sides PQ and PR respectively
PE = 3.9 cm, EQ = 3 cm, PF = 8 cm,
RF = 9 cm
Is EF || QR ?

Question 4.
A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB || QR? Give reasons for your answer.
Solution:
In ∆PQR, A and B are points on the sides PQ and PR such that
PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm

Question 5.
(a) In figure (i) given below, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.
(b) In figure (ii) given below, AB || DE and BD || EF. Prove that DC² = CF x AC.
Solution:
(a) Given: In the figure,
DE || BC and BD = CE
To prove: ∆ABC is an isosceles triangle
Proof: In ∆ABC, DE || BC

Question 6.
(a) In the figure (i) given below, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm.
(b) In the given figure, ∠D = ∠E and $$\frac { AD }{ BD } =\frac { AE }{ EC }$$. Prove that BAC is an isosceles triangle.

Solution:
(a) Given : CD || LA and DE || AC
Length of BE = 4 cm
Length of EC = 2 cm
Now, in ∆BCA
DE || AC
$$\frac { BE }{ BC } =\frac { BD }{ BA }$$
(Corallary of basic proportionality theorem)

Question 7.
In the figure given below, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. show that BC || QR.

Solution:
In the given figure, A, B, C are points on
OP, OQ and OR respectively
and AB || PQ and AC || PR
To prove: BC || QR
Proof: In POQ,
AB || PQ

Question 8.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem, prove that $$\frac { AO }{ BO } =\frac { CO }{ DO }$$
Solution:
Given : ABCD is a trapezium in which AB || DC
Its diagonals AC and BD intersect each other at O

Question 9.
(a) In the figure (1) given below, AB || CR and LM || QR.
(i) Prove that $$\frac { BM }{ MC } =\frac { AL }{ LQ }$$
(ii) Calculate LM : QR, given that BM : MC = 1 : 2.
(b) In the figure (2) given below AD is bisector of ∠BAC. If AB = 6 cm, AC = 4 cm and BD = 3cm, find BC

Solution:
(a) Given: AB || CR and LM II QR.
Also BM : MC = 1:2
To Prove:
(i) $$\frac { BM }{ MC } =\frac { AL }{ LQ }$$

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.2 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

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):

Solution:
Given
(i) In ∆ABC and PQR

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.

Solution:
In the given figure, two line segments intersect each other at P.
In ∆BCP and ∆DEP
∠BPC = ∠DPE

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

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

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

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.

(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.

Solution:
(a) In the given figure,
AB||DE, AC = 3 cm, CE = 7.5 cm, BD = 14 cm

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.

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)

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.

(b) In the figure (2) given below,
(i) Prove that ∆s ABC and AED are similar.
(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.

(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.

Solution:
In the given figure,
AP = 2PB, CP = 2PD
To prove:

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.

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

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.

Question 13.
In the given figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that $$\frac { AO }{ OC } =\frac { BO }{ OD }$$

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.

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

Question 15.
In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC. Prove that $$\frac { BE }{ DE } =\frac { AC }{ BC }$$

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

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.

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

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

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.

Solution:
In the given figure, CM and RN are medians of ∆ABC and ∆PQR
respectively and ∆ABC ~ ∆PQR
To prove:
(i) ∆AMC ~ ∆PNR

>

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

Solution:
In the given figure,
AD and BE are the medians of ∆ABC
intersecting each other at G
DF || BE is drawn
To prove :

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.

(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.

Solution:
In the given figure,
AB || EF || CD
AB = 15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm
Calculate :

Question 21.
In the given figure, ∠A = 90° and AD ⊥ BC If BD = 2 cm and CD = 8 cm, find AD.

Solution:
In ∆ABC, we have ∠A = 90°
Now,
In ∆ABC, we have,
∠BAC = 90°
⇒ ∠BAD + ∠DAC = 90°…..(i)

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

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 .

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.