Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(A)

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 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 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 = 90^o, 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.

1. ∆ ABC ~ ∆ AMP.
2. 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 = 90^o 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 = 90^o

QPM = RPQ (Common)

(ii) In QMR and PQR,

QMR = PQR = 90^o

QRM = QRP (Common)

(iii) Adding the relations obtained in (i) and (ii), we get,

Question 19

In ABC, B = 90^o 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 = 180^o

1 + 3 = 90^o ..... (1)(Since, 2 = 90^o)

3 + 4 = 90^o .....(2) (Since, ABC = 90^o)

From (1) and (2),

1 + 3 = 3 + 4

1 = 4

Also, 2 = 5 = 90^o

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 = 90^o.

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 = 90^o and in triangle PQS, PSQ = 90^o. 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 = 90^o

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),