SELINA Solutions for Class 10 Maths Chapter 15 - Similarity (With Applications to Maps and Models)
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(A)
Question 1(i)
In the figure, given below, straight lines AB and CD intersect at P; and AC ‖ BD. Prove that:
ΔAPC and ΔBPD are similar.
Solution 1(i)
Question 1(ii)
In the figure, given below, straight lines AB and CD intersect at P; and AC ‖ BD. Prove that :
If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.
Solution 1(ii)
Question 2(i)
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that :
Δ APB is similar to Δ CPD.
Solution 2(i)
Question 2(ii)
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that :
PA x PD = PB x PC.
Solution 2(ii)
Question 3(i)
P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that :
DP : PL = DC : BL.
Solution 3(i)
Question 3(ii)
P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that :
DL : DP = AL : DC.
Solution 3(ii)
Question 4(i)
In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that :
Δ AOB is similar to Δ COD.
Solution 4(i)
Question 4(ii)
In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that :
OA x OD = OB x OC.
Solution 4(ii)
Question 5(i)
In Δ ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that :
CB : BA = CP : PA
Solution 5(i)
Question 5(ii)
In Δ ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that :
AB x BC = BP x CA
Solution 5(ii)
Question 6
Solution 6
Question 7(i)
In the given figure, DE ‖ BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.
Write all possible pairs of similar triangles.
Solution 7(i)
Question 7(ii)
In the given figure, DE ‖ BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.
Find lengths of ME and DM.
Solution 7(ii)
Question 8
In the given figure, AD = AE and AD2 = BD x EC.
Prove that: triangles ABD and CAE are similar.
Solution 8
Question 9
In the given figure, AB ‖ DC, BO = 6 cm and DQ = 8 cm; find: BP x DO.
Solution 9
Question 10
Angle BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR = 9 cm; find the length of PB.
Solution 10
Question 11
State, true or false:
(i) Two similar polygons are necessarily congruent.
(ii) Two congruent polygons are necessarily similar.
(iii) All equiangular triangles are similar.
(iv) All isosceles triangles are similar.
(v) Two isosceles-right triangles are similar.
(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.
(vii) The diagonals of a trapezium, divide each other into proportional segments.
Solution 11
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) True
(vii) True
Question 12
Given:
GHE = DFE = 90o,
DH = 8, DF = 12, DG = 3x - 1 and DE = 4x + 2.
Find: the lengths of segments DG and DE.
Solution 12
Question 13
D is
a point on the side BC of triangle ABC such that angle ADC is equal to angle
BAC. Prove that: CA2 = CB 
CD.
Solution 13
Question 14
In the given figure, ∆ABC and ∆AMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
- ∆ ABC ~ ∆ AMP.
- Find AB and BC.
Solution 14
(i) In ∆ ABC and ∆ AMP,
BAC= PAM [Common]
ABC= PMA [Each = 90°]
∆ ABC ~ ∆ AMP [AA Similarity]
(ii)
Question 15
Given:
RS and PT are altitudes of 
PQR. Prove that:
(i) 
(ii)
PQ 
QS = RQ QT.
Solution 15
(i)
(ii)
Since, triangles PQT and RQS are similar.
Question 16
Given: ABCD is a rhombus, DPR and CBR are straight lines. Prove that: DP 
CR = DC PR.
Solution 16
Hence, DP CR = DC PR
Question 17
Given:
FB = FD, AE 
FD and FC AD. Prove: 
Solution 17
Question 18
In 
PQR, 
Q = 90o
and QM is perpendicular to PR. Prove that:
(i)
PQ2 = PM 
PR
(ii)
QR2 = PR MR
(iii) PQ2 + QR2 = PR2
Solution 18
(i) In PQM and PQR,
PMQ = PQR = 90o
QPM = RPQ (Common)
(ii) In QMR and PQR,
QMR = PQR = 90o
QRM = QRP (Common)
(iii) Adding the relations obtained in (i) and (ii), we get,
Question 19
In 
ABC, 
B = 90o and BD 
AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm and AB = 7 cm; find AD.
Solution 19
(i) In CDB,
1 + 2 +3 = 180o
1 + 3 = 90o ..... (1)(Since, 2 = 90o)
3 + 4 = 90o .....(2) (Since, ABC = 90o)
From (1) and (2),
1 + 3 = 3 + 4
1 = 4
Also, 2 = 5 = 90o
Hence, AD = 6.4 cm
(iii)
Question 20
In the figure, PQRS is a parallelogram with PQ = 16 cm and 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.
Find the lengths of PN and RM.
Solution 20
Question 21
In quadrilateral ABCD, diagonals AC and BD intersect at point E such that AE: EC = BE: ED. Show that ABCD is a trapezium.
Solution 21
Given, AE: EC = BE: ED
Draw EF || AB
In 
ABD, EF || AB
Using Basic Proportionality theorem,
Thus, in DCA, E and F are points on CA and DA
respectively such that 
Thus, by converse of Basic proportionality theorem, FE || DC.
But, FE || AB.
Hence, AB || DC.
Thus, ABCD is a trapezium.
Question 22
In
triangle ABC, AD is perpendicular to side BC and AD2 = BD 
DC. Show that angle BAC = 90o.
Solution 22
Given, AD2
= BD DC
So, these two triangles will be equiangular.
Question 23
In the given figure, AB || EF || DC; AB = 67.5 cm, DC = 40.5 cm and AE = 52.5 cm.
(i) Name the three pairs of similar triangles.
(ii) Find the lengths of EC and EF.
Solution 23
(i) The three pair of similar triangles are:
BEF and BDC
CEF and CAB
ABE and CDE
(ii) Since, ABE and CDE are similar,
Since, CEF and CAB are similar,
Question 24
In
the given figure, QR is parallel to AB and DR is parallel to QB. Prove that:
PQ2 = PD 
PA.
Solution 24
Given, QR is parallel to AB. Using Basic proportionality theorem,
Also, DR is parallel to QB. Using Basic proportionality theorem,
From (1) and (2), we get,
Question 25
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E. Prove that: EL = 2BL.
Solution 25
1 = 6 (Alternate
interior angles)
2 = 3 (Vertically
opposite angles)
DM = MC (M is the mid-point of CD)
So, DE = BC (Corresponding parts of congruent triangles)
Also, AD = BC (Opposite sides of a parallelogram)
AE = AD + DE = 2BC
Now, 1 = 6 and 4 = 5
Question 26
In the figure, given below, P is a point on AB such that AP: PB = 4: 3. PQ is parallel to AC.
(i) Calculate the ratio PQ: AC, giving reason for your answer.
(ii)
In triangle ARC, 
ARC = 90o and in triangle
PQS, PSQ = 90o. Given QS = 6
cm, calculate the length of AR.
Solution 26
(i) Given, AP: PB = 4: 3.
Since, PQ || AC. Using Basic Proportionality theorem,
Now, PQB = ACB (Corresponding
angles)
QPB = CAB (Corresponding
angles)
(ii) ARC = QSP = 90o
ACR = SPQ (Alternate
angles)
Question 27
In the right-angled triangle QPR, PM is an altitude. Given that QR = 8 cm and MQ = 3.5 cm, calculate the value of PR.
Solution 27
We have:
Question 28
In the figure, given below, the medians BD and CE of a triangle ABC meet at G. Prove that:
(i) 
and
(ii) BG = 2 GD from (i) above.
Solution 28
(i) Since, BD and CE are medians.
AD = DC
AE = BE
Hence, by converse of Basic Proportionality theorem,
ED || BC
In 
EGD and CGB,
(ii) Since,
In AED and ABC,
From (1),
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(B)
Question 1(i)
In the following figure, point D divides AB in the ratio 3 : 5. Find :
Solution 1(i)
Now, DE is parallel to BC.
Then, by Basic proportionality theorem, we have
Question 1(ii)
In the following figure, point D divides AB in the ratio 3 : 5. Find :
Solution 1(ii)
Question 1(iii)
In the following figure, point D divides AB in the ratio 3 : 5. Find :
Solution 1(iii)
Question 1(iv)
In the following figure, point D divides AB in the ratio 3 : 5. Find :
DE = 2.4 cm, find the length of BC.
Solution 1(iv)
Question 1(v)
In the following figure, point D divides AB in the ratio 3 : 5. Find :
BC = 4.8 cm, find the length of DE.
Solution 1(v)
Question 2(i)
In the given figure, PQ ‖ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find :
Solution 2(i)
Question 2(ii)
In the given figure, PQ ‖ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find :
PQ
Solution 2(ii)
Question 2(iii)
In the given figure, PQ ‖ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find :
If AP = x, then the value of AC in terms
of x. 
Solution 2(iii)
Question 3
A line PQ is drawn parallel to the side BC of Δ ABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.
Solution 3
Question 4(i)
In Δ ABC, D and E are the points on sides AB and AC respectively.
Find whether DE ‖ BC, if
AB = 9cm, AD = 4cm, AE = 6cm and EC = 7.5cm.
Solution 4(i)
Question 4(ii)
In Δ ABC, D and E are the points on sides AB and AC respectively.
Find whether DE ‖ BC, if AB = 6.3 cm, EC = 11.0 cm, AD =0.8 cm and EA = 1.6 cm.
Solution 4(ii)
Question 5
In the given figure, Δ ABC ~ Δ ADE. If AE: EC = 4 : 7 and DE = 6.6 cm, find BC. If 'x' be the length of the perpendicular from
A to DE, find the length of perpendicular from A to BC in terms of 'x'.
Solution 5
Question 6
A line segment DE is drawn parallel to base BC of Δ ABC which cuts AB at point D and AC at point E. If AB = 5BD and EC = 3.2 cm, find the length of AE.
Solution 6
Question 7
In the figure, given below, AB, CD and EF are parallel lines. Given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm and CE = 3. cm, calculate the values of x and y.
Solution 7
Question 8
In the figure, given below, PQR is a right-angle triangle right angled at Q. XY is parallel to QR, PQ = 6 cm, PY = 4 cm and PX : XQ = 1 : 2. Calculate the lengths of PR and QR.
Solution 8
Question 9
In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E.
Prove that : PE = 2 PD
Solution 9
Question 10
The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE = 4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.
Solution 10
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(C)
Question 1
(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.
(ii) Areas of two similar triangles are 98 sq. cm and 128 sq. cm. Find the ratio between the lengths of their corresponding sides.
Solution 1
We know that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
(i) Required ratio = 
(ii) Required ratio = 
Question 2
A line PQ is drawn parallel to the base BC of 
ABC which meets sides AB and AC at points P and Q respectively. If AP =
PB; find the value of:
(i) 
(ii) 
Solution 2
(i) AP =PB 
(ii) 
Question 3
The perimeters of two similar triangles are 30 cm and 24 cm. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.
Solution 3
Let 
Question 4
In the given figure, AX: XB = 3: 5.
Find:
(i) the length of BC, if the length of XY is 18 cm.
(ii) the ratio between the areas of trapezium XBCY and triangle ABC.
Solution 4
Given, 
(i)
(ii)
Question 5
ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP: AB.
Solution 5
From the given information, we have:
Question 6
In the given triangle PQR, LM is parallel to QR and PM: MR = 3: 4.
Calculate the value of ratio:
(i) 
(ii) 
(iii) 
Solution 6
(i)
(ii) Since LMN and MNR have common vertex at M and their bases LN and NR are along the same straight line
(iii) Since LQM and LQN have common vertex at L and their bases QM and QN are along the same straight line
Question 7
The given diagram shows two isosceles triangles which are similar also. In the given diagram, PQ and BC are not parallel; PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ. Calculate:
(i) the length of AP,
(ii) the ratio of the areas of triangle APQ and triangle ABC.
Solution 7
(i)
(ii)
Question 8
In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP: PC = 1: 2. DP produced meets AB produces at Q. Given the area of triangle CPQ = 20 cm2. Calculate:
(i) area of triangle CDP,
(ii) area of parallelogram ABCD.
Solution 8
Question 9
In the given figure, BC is parallel to DE. Area of triangle ABC = 25 cm2, Area of trapezium BCED = 24 cm2 and DE = 14 cm. Calculate the length of BC. Also, find the area of triangle BCD.
Solution 9
Question 10
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersects at point P. If AP: CP = 3: 5, find:
(i) APB : CPB(ii) DPC : APB
(iii) ADP : APB(iv) APB : ADB
Solution 10
(i) Since APB and CPB have common vertex at B and their bases AP and PC are along the same straight line
(ii) Since DPC and BPA are similar
(iii) Since ADP and APB have common vertex at A and their bases DP and PB are along the same straight line
(iv) Since APB and ADB have common vertex at A and their bases BP and BD are along the same straight line
Question 11
In the given figure, ABC is a triangle. DE is parallel to BC and 
.
(i) Determine the ratios 
(ii) Prove that 
DEF is similar to CBF. Hence, find 
.
(iii) What is the ratio of the areas of DEF and BFC?
Solution 11
(i) Given, DE || BC and
In ADE and ABC,
A = A(Corresponding Angles)
ADE = ABC(Corresponding Angles)
(By AA- similarity)
..........(1)
Now
Using (1), we get
.........(2)
(ii) In 
DEF and CBF,
FDE = FCB(Alternate Angle)
DFE = BFC(Vertically Opposite Angle)
DEF 
CBF(By AA- similarity)
using (2)
.
(iii) Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, therefore
Question 12
In the given figure, 
B = E, ACD = BCE, AB = 10.4 cm and DE = 7.8 cm. Find the ratio between areas of the ABC and DEC.
Solution 12
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(D)
Question 1(i)
A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A' B' C' Calculate :
the length of AB, if A' B' = 6 cm.
Solution 1(i)
Question 1(ii)
A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A' B' C' Calculate :
the length of C' A' if CA = 4 cm.
Solution 1(ii)
Question 2(i)
A triangle LMN has been reduced by scale factor 0.8 to the triangle L' M' N'. Calculate:
the length of M' N', if MN = 8 cm.
Solution 2(i)
Question 2(ii)
A triangle LMN has been reduced by scale factor 0.8 to the triangle L' M' N'. Calculate:
the length of LM, if L' M' = 5.4 cm.
Solution 2(ii)
Question 3(i)
A triangle ABC is enlarged, about the point O as centre of enlargement, and the scale factor is 3. Find :
A' B', if AB = 4 cm.
Solution 3(i)
Question 3(ii)
A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find :
BC, if B' C' = 15 cm.
Solution 3(ii)
Question 3(iii)
A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find :
OA, if OA' = 6 cm.
Solution 3(iii)
Question 3(iv)
A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find : OC', if OC = 21 cm
Also, state the value of :
(a) 
(b) 
Solution 3(iv)
Given that triangle ABC is enlarged and the scale factor is m = 3 to the triangle A'B'C'.
OC = 21 cm
So, (OC)3 = OC'
i.e. 21 x 3 = OC'
i.e. OC' = 63 cm
Question 4(i)
A model of an aeroplane is made to a scale of 1 : 400. Calculate :
the length, in cm, of the model; if the length of the aeroplane is 40 m.
Solution 4(i)
Question 4(ii)
A model of an aeroplane is made to a scale of 1 : 400. Calculate :
the length, in m, of the aeroplane, if length of its model is 16 cm.
Solution 4(ii)
Question 5
The dimensions of the model of a multistorey building are 1.2 m x 75 cm x 2 m. If the scale factor is 1 : 30; find the actual dimensions of the building.
Solution 5
Question 6(i)
On a map drawn to a scale of 1 : 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and ∠ABC = 90°.
Calculate :
the actual lengths of AB and BC in km.
Solution 6(i)
Question 6(ii)
On a map drawn to a scale of 1 : 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and angle ABC = 90°.
Calculate :
the area of the plot in sq. km.
Solution 6(ii)
Question 7
A model of a ship is made to a scale 1 : 300.
i. The length of the model of the ship is 2 m. Calculate the length of the ship.
ii. The area of the deck of the ship is 180,000 m2. Calculate the area of the deck of the model.
iii. The volume of the model is 6.5 m3. Calculate the volume of the ship.
Solution 7
Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(E)
Question 1(i)
In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.
Solution 1(i)
Question 1(ii)
In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.
Solution 1(ii)
Question 1(iii)
Find XY.
Solution 1(iii)
Given that XY || BC
So, △AXY ∼ △ABC
Question 2(i)
In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5,cm, AP = 6 cm and BE = 15 cm, Calculate: CE
Solution 2(i)
Question 2(ii)
In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5,cm, AP = 6 cm and BE = 15 cm, Calculate: AF
Solution 2(ii)
Question 2(iii)
In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5,cm, AP = 6 cm and BE = 15 cm, Calculate: PE
Solution 2(iii)
Question 3
In the following figure, AB, CD and EF are perpendicular to the straight line BDF.
Solution 3
Question 4
Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: 
.
Solution 4
Question 5
Solution 5
Question 6
Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that:
.
Solution 6
Question 7
In the following figure, ∠AXY = ∠AYX. If 
show that triangle ABC is isosceles.
Solution 7
Question 8
In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown.
Prove that: 
Solution 8
Join AR.
In ACR, BX || CR. By Basic Proportionality theorem,
In APR, XQ || AP. By Basic Proportionality theorem,
From (1) and (2), we get,
Question 9
Solution 9
Question 10(i)
In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate : EF
Solution 10(i)
Question 10(ii)
In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate : AC
Solution 10(ii)
Question 11(i)
In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
Prove that ΔACD is similar to ΔBCA.
Solution 11(i)
Question 11(ii)
In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
Find BC and CD.
Solution 11(ii)
Question 11(iii)
In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
Find area of ΔACD : area of ΔABC.
Solution 11(iii)
Question 12
In the given triangle P, Q and R are mid-points of sides AB, BC and AC respectively. Prove that triangle QRP is similar to triangle ABC.
Solution 12
In 
ABC, PR || BC. By Basic proportionality theorem,
Also, in PAR and ABC,
Similarly, 
Question 13
In the following figure, AD and CE are medians of 
ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB,
(ii) AG: GD = 2: 1
Solution 13
(i)
(ii) In AFD, EG || FD. Using Basic Proportionality theorem,
… (1)
Now, AE = EB (as E is the mid-point of AB)
AE = 2EF (Since, EF = FB, by (i))
From (1),
Hence, AG: GD = 2: 1.
Question 14
Two similar triangles are equal in area. Prove that the triangles are congruent.
Solution 14
Let us assume two similar triangles as 
ABC 
PQR
Question 15
The ratio between the altitudes of two similar triangles is 3: 5; write the ratio between their:
(i) medians. (ii) perimeters. (iii) areas.
Solution 15
The ratio between the altitudes of two similar triangles is same as the ratio between their sides.
(i) The ratio between the medians of two similar triangles is same as the ratio between their sides.
Required ratio = 3: 5
(ii) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.
Required ratio = 3: 5
(iii) The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.
Required ratio = (3)2 : (5)2 = 9: 25
Question 16
The ratio between the areas of two similar triangles is 16: 25. Find the ratio between their:
(i) perimeters. (ii) altitudes. (iii) medians.
Solution 16
The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.
So, the ratio between the sides of the two triangles = 4: 5
(i) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.
Required ratio = 4: 5
(ii) The ratio between the altitudes of two similar triangles is same as the ratio between their sides.
Required ratio = 4: 5
(iii) The ratio between the medians of two similar triangles is same as the ratio between their sides.
Required ratio = 4: 5
Question 17
The following figure shows a triangle PQR in which XY is parallel to QR. If PX: XQ = 1: 3 and QR = 9 cm, find the length of XY.
Further, if the area of 
PXY = x cm2; find, in terms of x, the area of:
(i) triangle PQR. (ii) trapezium XQRY.
Solution 17
In PXY and PQR, XY is parallel to QR, so corresponding angles are equal.
Hence, 
(By AA similarity criterion)
(i) We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
(ii) Ar (trapezium XQRY) = Ar (PQR) - Ar (PXY)
= (16x - x) cm2
= 15x cm2
Question 18
On a map drawn to a scale of 1: 20000, a rectangular plot of land ABCD is measured as AB = 24 cm and BC = 32 cm. Calculate:
(i) the diagonal distance of the plot in kilometre.
(ii) the area of the plot in sq. km.
Solution 18
Scale :- 1 : 20000
1 cm represents 20000 cm= 
= 0.2 km
(i)
= 
= 576 + 1024 = 1600
AC = 40 cm
Actual length of diagonal = 40 
0.2 km = 8 km
(ii)
1 cm represents 0.2 km
1 cm2 represents 0.2 0.2 
The area of the rectangle ABCD = AB BC
= 24 32 = 768
Actual area of the plot = 0.2 0.2 768 = 30.72 km2
Question 19
The dimensions of the model of a multistoreyed building are 1 m by 60 cm by 1.20 m. If the scale factor is 1: 50, find the actual dimensions of the building. Also, find:
(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq. cm.
(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90 m3.
Solution 19
The dimensions of the building are calculated as below.
Length = 1 
50 m = 50 m
Breadth = 0.60 50 m = 30 m
Height = 1.20 50 m = 60 m
Thus, the actual dimensions of the building are 50 m 30 m 60 m.
(i)
Floor area of the room of the building 
(ii)
Volume of the model of the building
Question 20
In a triangle PQR, L and M are two points on the base QR, such that 
LPQ = QRP and RPM = RQP. Prove that:
(i) 
(ii) 
(iii) 
Solution 20
(i)
(ii)
(iii)
Question 21
A triangle ABC with AB = 3 cm, BC = 6 cm and AC = 4 cm is enlarged to 
DEF such that the longest side of DEF = 9 cm. Find the scale factor and hence, the lengths of the other sides of DEF.
Solution 21
Triangle ABC is enlarged to DEF. So, the two triangles will be similar.
Longest side in ABC = BC = 6 cm
Corresponding longest side in DEF = EF = 9 cm
Scale factor = 
= 1.5
Question 22
Two isosceles triangles have equal vertical angles. Show that the triangles are similar.
If the ratio between the areas of these two triangles is 16: 25, find the ratio between their corresponding altitudes.
Solution 22
Let ABC and PQR be two isosceles triangles.
Then, 
Also, 
A = P (Given)
Let AD and PS be the altitude in the respective triangles.
We know that the ratio of areas of two similar triangles is equal to the square of their corresponding altitudes.
Question 23
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.
Solution 23
In triangle ABC, PO || BC. Using Basic proportionality theorem,
(i)
(ii)
Question 24
The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively. Show that:
Solution 24
Question 25
In the given figure, ABC is a triangle with 
EDB = ACB. Prove that 
ABC 
EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of BED = 9 cm2. Calculate the :
(i) length of AB.
(ii) area of ABC.
Solution 25
In 
ABC and EBD,
ACB = EDB (given)
ABC = EBD (common)
(by AA- similarity)
(i) We have, 
(ii) 
Question 26
In the given figure, ABC is a right angled triangle with m∠BAC = 90°
- Prove that ∆ADB ~ ∆CDA
- If BD = 18 cm and CD = 8 cm, find AD.
- Find the ratio of the area of ∆ADB to the area of ∆CDA.
Solution 26
Question 27(i)
In the given figure, AB and DE are perpendiculars to BC.
Prove that : ΔABC ~ ΔDEC
Solution 27(i)
Question 27(ii)
In the given figure, AB and DE are perpendiculars to BC.
If AB = 6 cm, DE = 4 cm and AC = 15 cm. Calculate CD.
Solution 27(ii)
Question 27(iii)
In the given figure, AB and DE are perpendiculars to BC.
Find the ratio of the area of a ΔABC : area of ΔDEC.
Solution 27(iii)
Question 28(i)
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
ΔADE ~ ΔACB.
Solution 28(i)
Question 28(ii)
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
Solution 28(ii)
Question 28(iii)
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
Find, area of ΔADE : area of quadrilateral BCED.
Solution 28(iii)
Question 29
Given: AB || DE and BC || EF. Prove that:
(i) 
(ii) 
Solution 29
(i) In 
AGB, DE || AB , by Basic proportionality theorem,
.... (1)
In GBC, EF || BC, by Basic proportionality theorem,
.... (2)
From (1) and (2), we get,
(ii)
From (i), we have:
Question 30
PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR = ∠QPR. Given QP = 8 cm, PR = 6 cm and SR = 3 cm.
i. Prove ΔPQR ∼ ΔSPR.
ii. Find the lengths of QR and PS.
iii. 
Solution 30
i.
In ∆PQR and ∆SPR,
∠PSR = ∠QPR … given
∠PRQ = ∠PRS … common angle
⇒ ∆PQR ∼ ∆SPR (AA Test)
ii. Find the lengths of QR and PS.
Since ∆PQR ∼ ∆SPR … from (i)
iii.
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