# Frank Solutions for Chapter 16 Similarity Class 9 Mathematics ICSE

### Exercise 16.1

1. In ABC, D and E are the mid-points on AB and AC such that DE || BC

(i) If AD = 4, AE = 8, DB = x – 4 and EC = 3x – 19, find x.

(ii) If AD: BD = 4: 5 and EC = 2.5 cm, find AE.

(iii) If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find x.

(iv) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.

∠D = ∠B and ∠C = ∠E (DE || BC)

Hence,

⇒ {4/(x – 4)} = {8/(3x – 19)}

⇒ 4 (3x – 19) = 8 (x – 4)

⇒ 12 x – 76 = 8x – 32

⇒ 4x = 44

We get,

x = 11

∠D = ∠B and ∠C = ∠E (DE || BC)

Hence,

AD / DB = AE / EC

⇒ 4/5 = AE/2.5

⇒ AE = (4×2.5)/5

We get,

AE = 2 cm

∠D = ∠B and ∠C = ∠E (DE || BC)

Hence,

AD / DB = AE / EC

⇒ (4x – 3)/(3x – 1) = (8x – 7)/(5x – 3)

⇒ (4x – 3) (5x -3) = (8x – 7) (3x – 1)

⇒ 20x2 – 15x – 12x + 9 = 24x2 – 21x – 8x + 7

⇒ 4x2 – 2x – 2 = 0

By splitting middle term, we get

4x2 – 4x + 2x – 2 = 0

⇒ 4x (x -1) + 2 (x-1) = 0

⇒ x = -2/4 or x = 1

We know that, side of triangle can never be negative

Hence, x = 1

∠D = ∠B and ∠C = ∠E (DE || BC)

Hence,

⇒ DB = 12 – 8

⇒ DB = 4

⟹ 8/4 = 12/ EC

⇒ 8×EC = 12×4

⇒ EC = (12×4)/8

We get,

⇒ EC = 6 cm

2. In ABC, D and E are points on AB and AC. Show that DE || BC for each of the following case or not:

(i) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm

(ii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm

(iii) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm

(i) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm

⇒ AE/AC = (1.8)/(7.2)

⇒ AE/AC = (2/8)

⇒ AE/AC = (1/4)

Hence,

⇒ ∠D = ∠B, ∠E = ∠C

But these are corresponding angles

Therefore,

DE || BC

(ii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm

⇒ AD = 10.8 – 4.5

⇒ AE/AC = 2.8/4.8

⇒ AE/AC = 14/24

⇒ AE / AC = 7/12

Hence,

⇒ ∠D = ∠B, ∠E = ∠C

But these are corresponding angles

Therefore,

DE || BC

(iii) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm

⇒ AE/EC = 3.3/5.5

⇒ AE/EC = 3/5

⇒ AE/EC = 0.6

Hence,

⇒ ∠D = ∠B, ∠E = ∠C

But these are corresponding angles

Therefore,

DE || BC

3. In the figure, PQ is parallel to BC, AP: AB = 2: 7. If QC = 10 and BC = 21,

Find:

(i) AQ

(ii) PQ

(i) Since PQ || BC

AP/PB = AQ/QC

⟹ {AP/(AB – AP)} = (AQ/QC)

⟹ (2/5) = (AQ/10)

⟹ AQ = (2×10)/5

We get,

AQ = 4

(ii) Since PQ || BC

AP/AB = PQ/BC

⟹ 2/7 = PQ/21

⟹ PQ = (2×21)/7

We get,

PQ = 6

4. In ABC, DE is parallel to BC and DE: BC = 3: 8

Find:

(ii) AE, if AC = 16

(i) Since DE || BC

Since DB = AB – AD

⟹ DB = 8 – 3

⟹ DB = 5

Hence,

(ii) DE: BC = 3: 8

Since DE || BC

DE/BC = AE/AC

⟹ 3/8 = AE/16

⟹ AE = (3×16)/8

We get,

AE = 6

5. In ABC, point D divides AB in the ratio 5: 7. Find:

(i) AE/EC

(iii) AE/AC

(iv) BC, if DE = 2.5 cm

(v) DE, if BC = 4.8 cm

Considering DE || BC

⟹ AE/EC = 5/7

Since, AB = AD + DB

⟹ AB = 5 + 7

⟹ AB = 12

Therefore,

⟹ AE/EC = 5/7

Since, AC = AE + EC

⟹ AC = 5 + 7

⟹ AC = 12

Therefore,

AE/AC = 5/12

(iv) Since, DE || BC

⟹ 5/12 = 2.5/BC

⟹ BC = (2.5×12)/5

We get,

BC = 6 cm

(v) Since, DE || BC

⟹ 5/12 = DE/4.8

⟹ DE = (5×4.8)/12

We get,

DE = 2 cm

6. In PQR, AB is drawn parallel to QR. If PQ = 9 cm, PR = 6 cm and PB = 4.2 cm, find the length of AP.

In △PQR

AB || QR

AP/PQ = PB/PR

⟹ AP/9 = 4.2/6

⟹ AP = (4.2×9)/6

We get,

AP = 6.3 cm

7. In ABC, MN is drawn parallel to BC. If AB = 3.5 cm, AM: AB = 5: 7 and NC = 2 cm, find:

(i) AM

(ii) AC

(i) AM/AB = 5/7

AB = 3.5 cm (given)

Hence,

AM = (5×AB)/7

⟹ AM = (5×.5)/7

On further calculation, we get,

AM = 2.5 cm

(ii) Since in △ABC, MN || BC and

AM/MB = AN/NC

Since AB = 3.5 cm and AM = 2.5 cm

Hence,

MB = AB – AM

⇒ MB = 3.5 – 2.5

⇒ MB = 1 cm

⇒ AM/MB = AN/NC

⇒ 2.5/1 = AN/2

⇒ AN = (2.5×2)/1

We get,

AN = 5 cm

We know,

AC = AN + NC

⇒ AC = 5 + 2

We get,

AC = 7 cm

8. The sides PQ and PR of the PQR are produced to S and T respectively. ST is drawn parallel to QR and PQ: PS = 3:4. If PT = 9.6 cm, find PR. If p be the length of the perpendicular from P to QR, find the length of the perpendicular from P to ST in terms of ‘p’.

Since QR is parallel to ST

By Basic Theorem of Proportionality

PQ/PS = PR/PT

⇒ 3/4 =(PR/9.6

⇒ PR = (9.6×3)/4

On simplification, we get,

PR = 7.2 cm

Since QR is parallel to ST,

QM || SD

By Basic Theorem of Proportionality,

PQ/PS = PM/PD

⇒ (3/4) = (p/PD)

⇒ PD = 4p/3

Therefore, the length of the perpendicular from P to ST in terms of p is 4p/3

9. In ABC, DE || BC such that AD = 1.5 cm, DB = 3 cm and AE = 1 cm. Find AC

Given in △ABC,

DE || BC

⇒ 1.5/4.5 = 1/AC

On further calculation, we get,

AC = 3 cm

10. Given is a triangle with sides 3 cm, 5 cm and 6 cm. Find the sides of a triangle which is similar to the given triangle and its shortest side is 4.5 cm

Since the two triangles are similar,

Hence, the ratio of the corresponding sides are equal

Consider x and y be the sides of the triangle where y is the longest side

(3/5) = (4.5/x)

On simplification, we get,

⇒ x = 7.5 cm

⇒ (5/6) = (7.5/y)

On further calculation, we get

⇒ y = 9 cm

Hence, the sides of the triangles are 4.5 cm, 7.5 cm and 9 cm

11. Two figures are similar. If the ratio of their perimeters is 8: 16. What will be the ratio of the corresponding sides?

We know that,

For two similar triangles, ratio of the corresponding sides is equal to the ratio of the perimeters of the triangles

Hence,

Ratio of the corresponding sides = 8/16 = 1/2

That is, ratio of the corresponding sides is 1:2

12. Harmeet is 6 feet tall and casts a shadow of 3 feet long. What is the height of a nearby pole if it casts a shadow of 12 feet long at the same time?

Harmeet and the pole will be perpendicular to the ground

So,

PQ || ST

In △PQR and △STR,

∠PQR = ∠STR …(Both are right angles)

∠PRQ = ∠SRT …(common angle)

△PQR ∼ △STR …(AA criterion for similarity)

PQ/ST = QR/TR

⟹ h/6 = 12/3

On simplification, we get,

h = 24 feet

Therefore, the height of the pole is 24 feet

13. The areas of two similar triangles are 16 cm2and 9 cm2respectively. If the altitude of the smaller triangle is 1.8 cm, find the length of the altitude corresponding to the larger triangle.

The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding altitudes.

Therefore,

Area (△ABC)/Area (△DEF) = (AL2/DM2)

⇒ (16/9) = (AL2/1.82)

⇒ AL2 = (16×3.24)/9

On further calculation, we get,

AL2 = 5.76

Hence,

AL = 2.4 cm

14. The areas of two similar triangles are 169 cmand 121 cmrespectively. If one side of the larger triangle is 26 cm, find the length of the corresponding side of the smaller triangle.

The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides

Therefore,

Area (△ABC)/Area (△DEF) = AB2/DE2

(169/121) = (262/DE2)

⇒ DE2 = (121×262)/169

⇒ DE2 = (121×676)/169

On simplification, we get,

DE2 = 81796/169

⇒ DE2 = 484

Hence,

DE = 22 cm

15. In ABC, DE is drawn parallel to BC cutting AB in the ratio 2: 3. Calculate:

(ii) area (trapezium EDBC)/area (ABC)

Given

We know,

⇒ AB = 2 + 3

⇒ AB = 5

⇒ area (△ADE)/area (△ABC) = 22/52

⇒ area (△ADE)/area (△ABC) = 4/25

(ii) area (trapezium EDBC)/area (△ABC) = {area (△ABC) – area (△ADE)}/ area of (△ABC)}

⟹ area (trapezium EDBC)/area (△ABC) = (25–4)/25

⟹ area (trapezium EDBC)/area (△ABC) = (21/ 25)

16. In the figure, PR || SQ. If PR = 10 cm, PT = 5 cm, TQ = 6 cm and ST = 9 cm, Calculate RT and SQ.

17. ABCD is a parallelogram whose sides AB and BC are 18 cm and 12 cm respectively. G is a point on AC such that CG: GA = 3 : 5. BG is produced to meet CD at Q and AD produced at P. Prove that CGB AGP. Hence, find AP.

18. In ABC, BP and CQ are altitudes from B and C on AC and AB respectively. BP and CQ intersect at O. Prove that

(i) PC × OQ = QB × OP

(ii) OC2/OB2 = (PC × PO)/(QB × QO)

(i)

(ii)

19. In the figure, PQR is a straight line and PS || RT. If QS = 12 cm, QR = 15 cm, QT = 10 cm and RT = 6 cm, find PQ and PS.

20. The diagonal AC of a parallelogram ABCD intersects DP at the point Q, where P is any point on side AB. Prove that CQ × PQ = QA × QD.

21. AM and DN are the altitudes of two similar triangles ABC and DEF. Prove that : AM : DN = AB : DE.

22. Prove that the external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

23. In the figure, AB || RQ and BC || SQ, prove that PC/PS = PA/PR.

24. In the figure, DE || AC and DC ||AP. Prove that BE/EC = BC/CP

25. PQ is perpendicular to BA and BD is perpendicular to AP. PQ and BD intersect at R. Prove that ABD APQ and AB/AP = BD/PQ.

26. Through the vertex S of a parallelogram PQRS, a line is drawn to intersect the sides QP and QR produced at M and N respectively. Prove that SP/PM = MQ/QN = MR/SR

27. (a) In a quadrilateral PQRS, the diagonals PR and QS intersect each other at the point T. If PT : TR = QT : TS = 1 : 2, show that PTQ – DRTS.

(b) In a quadrilateral PQRS, the diagonals PR and QS intersect each other at the point T. If PT : TR = QT : TS = 1 : 2, show that TP : TQ = TR : TS.

(a)

(b)

28. (a) In the given figure, PB is the bisector of ABC and ABC = ACB. Prove that:

(a) BC × AP = PC × AB

(b) AB : AC = BP : BC

(a)

(b) Note : It is not possible to prove this part due to inadequate data.

29. In a right-angled triangle ABC, B = 90°, P and Q are the points on the sides AB and AC such as PQ || BC, AB = 8 cm, AQ = 6 cm and PA : AB = 1 : 3. Find the lengths of AC and BC.

### Exercise 16.2

1. Given that ABC DPR, Q name the corresponding angles and the corresponding sides.

2. In ABC, DE || BC such that AD = 1.5 cm, DB = 3 cm and AE = 1 cm. Find AC.

3. Given is a triangle with sides 3 cm, 5 cm and 6 cm. Find the sides of a triangle which is similar to the given triangle and its shortest sides is 4.5 cm.

4. Two figures are similar. If the ratio of their perimeters is 8 : 16. What will be the ratio of the corresponding sides?

5. In ABC, AB = 8 cm, AC = 10 cm and B = 90°. P and Q are the points on the sides AB and AC respectively such that PQ = 3 cm and PQA = 90°. Find

(i) The area of AQP

(i) Area of quadrilateral PBCQ : area of ABC.

(i)

(ii)

6. Find the scale factor in each of the following and state the type of size transformation:

(i) Image length = 6 cm, Actual length = 4 cm.

(ii) Actual length = 12 cm, Image length = 15 cm

(iii) Image length = 8 cm, Actual length = 20 cm.

(iv) Actual area = 64 m2, Model area = 100 cm2

(v) Model area = 75 cm2, Actual area = 3 m2

(vi) Model volume = 200 cm3, Actual volume = 8 m3

(i)

(ii)

(iii)
(iv)
(v)
(iv)

7. ABC has been reduced by a scale factor 0.6 to A'B'C'. Calculate:

(i) length of B’C’, if BC = 8 cm

(ii) Length of AB, if A’B’ = 5.4 cm

(i)

(ii)

8. ABC is enlarged, with a scale factor 5. Find:

(i) A’B’, if AB = 4 cm

(ii) BC, if B’C’ = 16 cm

(i)

(ii)

9. XYZ is enlarged to X’Y’Z’. If XY = 12 cm, YZ = 8 cm and XZ = 14 cm and the smallest side of X’Y’Z’ is 12 cm, find the scale factor and use it to find the length of the other sides of the image X’Y’Z’.

10. 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, ABC = 90°. Calculate:

(i) The actual length of AB in km.

(ii) The area of the plot in sq. km.

Scale = 1 : 25000

(i)

(ii)

11. The dimensions of the model of a building are 1.2 m × 75 cm × 2 m. If the scale factor is 1 : 20; find the actual dimensions of the building.

12. The scale of a map is 1 : 50000. The area of a city is 40 sq. km which is to be represented on the map. Find:

(i) The area of land represented on the map.

(ii) The length of a side in km represented by 1 cm on the map.

Scale factor = 1 : 50000

(i)

(ii)

13. A plot of land of area 20 km2 is represented on the map with a scale factor of 1 : 200000. Find:

(i) The number of km represented by 2 cm on the map.

(ii) The ground are in km2 that is represented by 2 cm2 on the map.

(iii) The area on the map that represented the plot of land.

(i)

(ii)

(iii)

14. A map is drawn to a scale of 1: 20000. Find:

(i) The distance covered by 6 cm on the map

(ii) The distance on the map representing 4 km

(iii) The area of the lake on the map which has an actual area of 12 km2

Scale = 1 : 20000

(i)

(ii)
(iii)

15. A model of cargo truck is made to a scale of 1 : 40. The length of the model is 15 cm. Calculate:

(i) The length of the truck

(ii) The volume of the model if the volume of the truck is 64 m3

(iii) The base area of the truck, if the base area of the model is 30 m2

Scale = 1 : 40

(i)

(ii)
(iii)

16. A model of a ship is made to a scale of 1 : 500. Find:

(i) The length of the ship, if length of the model is 1.2.

(ii) The area of the deck of the ship, if the area of the deck of its model is 1.6 m2

(iii) The volume of the model when the volume of the ship is 1 km3

Scale = 1 : 500

(i)

(ii)
(iii)

17. On a map drawn to a scale of 1 : 25000, a rectangular plot of land has sides 12 cm × 16 cm. Calculate

(i) The diagonal distance of the plot in km

(ii) The area of the plot in sq km

Scale = 1 : 25000

(i)

(ii)

18. On a map drawn to a scale of 1: 25000, a triangular plot of land is right angled and the sides forming the right angle measure 225 cm and 64 cm.

(i) The actual length of the sides in km.,

(ii) The area of the plot in sq. km.