## ICSE Class 10 Similarity With Applications to Maps and Models Solution By Clarify Knowledge

*ICSE Class 10 Similarity With Applications to Maps and Models Solution Solution New Pattern 2022*

## Chapter 15 - ICSE Class 10 Similarity With Applications to Maps and Models Solution 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 : PASolution 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 AD^{2} = 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) TrueQuestion 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: CA^{2} = 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 PRQuestion 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) PQ^{2} = PM PR

(ii) QR^{2} = PR MR

(iii) PQ^{2} + QR^{2} = PR^{2}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 AD^{2} = BD DC. Show that angle BAC = 90^{o}.Solution 22

Given, AD^{2} = 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: PQ^{2} = 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),

## 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 cm^{2}. 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 cm^{2}, Area of trapezium BCED = 24 cm^{2} 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 m^{2}. Calculate the area of the deck of the model.

iii. The volume of the model is 6.5 m^{3}. 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: 25Question 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: 5Question 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 cm^{2}; 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) cm^{2}

= 15x cm^{2}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 km^{2}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 m^{3}.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)