# 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.**

**Answer**

**(i)**In △ADE and △ABC

∠D = ∠B and ∠C = ∠E (DE || BC)

⇒ △ADE ∼ △ABC

Hence,

AD/DB = AE/EC

⇒ {4/(x – 4)} = {8/(3x – 19)}

⇒ 4 (3x – 19) = 8 (x – 4)

⇒ 12 x – 76 = 8x – 32

⇒ 4x = 44

We get,

x = 11

**(ii)** In △ADE and △ABC

∠D = ∠B and ∠C = ∠E (DE || BC)

⇒ △ADE ∼ △ABC

Hence,

AD / DB = AE / EC

⇒ 4/5 = AE/2.5

⇒ AE = (4×2.5)/5

We get,

AE = 2 cm

**(iii)** In △ADE and △ABC

∠D = ∠B and ∠C = ∠E (DE || BC)

⇒ △ADE ∼ △ABC

Hence,

AD / DB = AE / EC

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

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

⇒ 20x^{2} – 15x – 12x + 9 = 24x^{2} – 21x – 8x + 7

⇒ 4x^{2} – 2x – 2 = 0

By splitting middle term, we get

4x^{2} – 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

**(iv)** In △ADE and △ABC

∠D = ∠B and ∠C = ∠E (DE || BC)

⇒ △ADE ∼ △ABC

Hence,

AD/DB = AE/EC

DB = AB – AD

⇒ 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**

**Answer**

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

AD/AB = (1.4)/(5.6)

⇒ AD/AB = (7/28)

⇒ AD/AB = (1/4)

⇒ AE/AC = (1.8)/(7.2)

⇒ AE/AC = (2/8)

⇒ AE/AC = (1/4)

⇒ AD/AB = AE/AC

Hence,

△ADE ∼ △ABC

⇒ ∠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 = AB – BD

⇒ AD = 10.8 – 4.5

⇒ AD = 6.3 cm

AD/AB = 6.3/10.8

⇒ AD/AB = 21/36

⇒ AD/AB = 7/12

⇒ AE/AC = 2.8/4.8

⇒ AE/AC = 14/24

⇒ AE / AC = 7/12

⇒ AD/AB = AE/AC

Hence,

△ADE ∼ △ABC

⇒ ∠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

AD/BD = 5.7/9.5

⇒ AD/BD = 0.6

⇒ AE/EC = 3.3/5.5

⇒ AE/EC = 3/5

⇒ AE/EC = 0.6

⇒ AD/BD = AE/EC

Hence,

△ADE ∼ △ABC

⇒ ∠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**

**Answer**

**(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:**

**(i) AD: DB**

**(ii) AE, if AC = 16**

**Answer**

**(i)** Since DE || BC

DE/BC = AD/AB

⟹ 3/8 = AD/AB

⟹ AD/AB = 3/8

Since DB = AB – AD

⟹ DB = 8 – 3

⟹ DB = 5

Hence,

AD: DB = 3: 5

**(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**

**(ii) AD/AB**

**(iii) AE/AC**

**(iv) BC, if DE = 2.5 cm**

**(v) DE, if BC = 4.8 cm**

**Answer**

**(i)** AD/DB = AE/EC

⟹ AE/EC = AD/DB

⟹ AE/EC = 5/7

**(ii) **AD/DB = 5/7

Since, AB = AD + DB

⟹ AB = 5 + 7

⟹ AB = 12

Therefore,

AD/AB = 5/12

**(iii)** AD/DB = AE/EC

⟹ AE/EC = AD/DB

⟹ AE/EC = 5/7

Since, AC = AE + EC

⟹ AC = 5 + 7

⟹ AC = 12

Therefore,

AE/AC = 5/12

**(iv)** Since, DE || BC

AD/AB = DE/ BC

⟹ 5/12 = 2.5/BC

⟹ BC = (2.5×12)/5

We get,

BC = 6 cm

**(v)** Since, DE || BC

AD/AB = 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.**

**Answer**

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**

**Answer**

**(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’.**

**Answer**

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**

**Answer**

DE || BC

AD/AB = AE/AC

⇒ 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**

**Answer**

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?**

**Answer**

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?**

**Answer**

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 cm ^{2}and 9 cm^{2}respectively. If the altitude of the smaller triangle is 1.8 cm, find the length of the altitude corresponding to the larger triangle.**

**Answer**

Therefore,

Area (△ABC)/Area (△DEF) = (AL^{2}/DM^{2})

⇒ (16/9) = (AL^{2}/1.8^{2})

⇒ AL^{2} = (16×3.24)/9

On further calculation, we get,

AL^{2} = 5.76

Hence,

AL = 2.4 cm

**14. ****The areas of two similar triangles are 169 cm ^{2 }and 121 cm^{2 }respectively. If one side of the larger triangle is 26 cm, find the length of the corresponding side of the smaller triangle.**

**Answer**

Therefore,

Area (△ABC)/Area (△DEF) = AB^{2}/DE^{2}

(169/121) = (26^{2}/DE^{2})

⇒ DE^{2} = (121×26^{2})/169

⇒ DE^{2} = (121×676)/169

On simplification, we get,

DE^{2} = 81796/169

⇒ DE^{2} = 484

Hence,

DE = 22 cm

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

**(i) area (****△****ADE)/area (****△****ABC)**

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

**Answer**

AD: DB = 2: 3

We know,

AB = AD + DB

⇒ AB = 2 + 3

⇒ AB = 5

**(i)** area (△ADE)/area (△ABC) = AD^{2}/AB^{2}

⇒ area (△ADE)/area (△ABC) = 2^{2}/5^{2}

⇒ 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. **

**Answer**

**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.**

**Answer**

**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) OC ^{2}/OB^{2} = (PC **

**× PO)/(QB × QO)**

**Answer**

**(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.**

**Answer**

**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.**

**Answer**

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

**Answer**

**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. **

**Answer**

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

**Answer**

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

**Answer**

**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. **

**Answer**

**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 **

**Answer**

**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. **

**Answer**

**(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 **

**Answer**

**(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.**

**Answer**

**Exercise 16.2 **

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

**Answer**

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

**Answer**

**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. **

**Answer**

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

**Answer**

**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. **

**Answer**

**(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 m ^{2}, Model area = 100 cm^{2 }**

**(v) Model area = 75 cm ^{2}, Actual area = 3 m^{2 }**

**(vi) Model volume = 200 cm ^{3}, Actual volume = 8 m^{3} **

**Answer**

**(i)**

**(ii)**

**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 **

**Answer**

**(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 **

**Answer**

**(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’. **

**Answer**

**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. **

**Answer**

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. **

**Answer**

**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. **

**Answer**

Scale factor = 1 : 50000

**(i)**

**(ii)**

**13. A plot of land of area 20 km ^{2} 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 cm ^{2} on the map. **

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

**Answer**

**(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 **

**Answer**

Scale = 1 : 20000

**(i)**

**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 m ^{2} **

**Answer**

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 m ^{2} **

**(iii) The volume of the model when the volume of the ship is 1 km ^{3} **

**Answer**

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 **

**Answer**

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. **

**Answer**

Scale = 1 : 25000

**(i)**

**(ii)**

**19. In a triangle ABC, AB = 4 cm, BC = 4.5 cm and CA = 5 cm. Construct ****△****ABC. Find the image A’B’C of the ****△****ABC obtained by enlarging it by a scale factor 2. Measure the sides of the image A’B’C and show that **

**AB : A’B’ = AC: B’C’ = CA : C’A’**

**Answer**