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
∠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)
⇒ 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
(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
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,
(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
(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
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 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.
Answer
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 cm2 and 121 cm2 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) = 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:
(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) = AD2/AB2
⇒ 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.
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.
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.
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
(a)
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
(ii)
(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
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)
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)
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)
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 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.
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)
(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
Answer
Scale = 1 : 40
(i)
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
Answer
Scale = 1 : 500
(i)
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)
(i) The actual length of the sides in km.,
(ii) The area of the plot in sq. km.
Answer
Scale = 1 : 25000
(i)
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
