Frank Solutions for Chapter 15 Similarity Class 10 ICSE Mathematics
1. In ΔABC, DE is parallel to BC and AD/DB = 2/7. If AC = 5.6, find AE.
Answer
Consider ΔABC,
DE is parallel to BC,
Let us assume AE = y
From Basic Proportionality Theorem (BPT) AD/DB = AE/EC
2/7 = y/(5.6 – y)
⇒ 2(5.6 – y) = 7y
⇒ 11.2 – 2y = 7y
⇒ 11.2 = 7y + 2y
⇒ 11.2 = 9y
⇒ y = 11.2/9
⇒ y = 1.24
Therefore, AE is 1.24
2. In ΔPQR, MN is drawn parallel to QR. If PM = x, MQ = (x – 2) and NR = (x – 1), find the value of x.
Answer
Consider ΔPQR,
MN is drawn parallel to QR,
From Basic Proportionality Theorem (BPT) PM/MQ = PN/NR
x/(x – 2) = (x + 2)/( x- 2)
⇒ x(x – 2) = (x + 2) (x – 2)
⇒ x2 – 2x = x2 – 4
⇒ x2 – x2 – 2x = – 4
⇒ -2x = – 4
⇒ x = – 4/-2
⇒ x = 2
Therefore, PM = x = 2
MQ = x – 2 = 2 – 2 =0
NR = x – 1 = 2 – 1 = 1
3. ΔABC is similar to ΔPQR. If AB = 6 cm, BC = 9 cm, PQ = 9 cm and PR = 10.5 cm, find the lengths of AC and QR.
Answer
AB = 6 cm, BC = 9 cm, PQ = 9 cm and PR = 10.5 cm
Now, consider ΔABC and ΔPQR
AB/PQ = BC/QR = AC/PR
6/9 = 9/M = L/10.5
Take first two fraction, 6/9 = 9/q
6M = 81
⇒ M = 81/6
⇒ M = 27/2
Then, QR = 13.5 cm
Now consider second and third fraction, 9/M = L/10.5
9/13.5 = L/10.5
⇒ 9(10.5) = 13.5L
⇒ 94.5 = 13.5L
⇒ L = 94.5/13.5
⇒ L = 7 cm
So, AC = 7 cm
4. ABCD and PQRS are similar figures. AB = 12 cm, BC = x cm, CD = 15 cm, AD = 10 cm, PQ = 8 cm, QR = 5 cm, RS = m cm and PS = n cm. Find the values of x, m and n.
Answer
Then, AB/PQ = BC/QR = DC/SR = AD/SR
12/8 = x/5 = 15/m = 10/n
Consider, 12/8 = x/5
By cross multiplication we get,
12 × 5 = 8 × x
⇒ 60 = 8x
⇒ x = 60/8
⇒ x = 7.5 cm
Then, consider x/5 = 15/m
7.5/5 = 15/m
By cross multiplication we get,
7.5 × m = 15 × 5
⇒ 7.5m = 75
⇒ m = 75/7.5
⇒ m = 750/75
⇒ m = 10 cm
Now, consider 15/m = 10/n
15/10 = 10/n
By cross multiplication we get,
15 × n = 10 × 10
⇒ 15n = 100
⇒ n = 100/15
⇒ n = 6.67 cm
Therefore, the value of x is 7.5 cm, value of m is 10 cm and value of n is 6.67 cm.
5. AD and BC are two straight lines intersecting at O. CD and BA are perpendiculars from B and C on AD. If AB = 6 cm, CD = 9 cm, AD = 20 cm and BC = 25 cm, find the lengths of AO, BO, CO and DO.
Answer
We have to find the lengths of AO, BO, CO and DO.
Consider ΔAOB and ΔCOD,
∠OAB = ∠ODC … [because both angles are equal to 90o]
∠AOB = ∠DOC … [because vertically opposite angles are equal]
Therefore, ΔAOB ~ ΔDOC [from AA corollary]
Then, AO/DO = OB/OC = AB/DC
x/(20 – x) = y/(25 – y) = 6/9
Consider, x/(20 – x) = 6/9
x/(20 – x) = 2/3
By cross multiplication we get,
3x = 2(20 – x)
⇒ 3x = 40 – 2x
⇒ 3x + 2x = 40
⇒ 5x = 40
⇒ x = 40/5
⇒ x = 8
Now, consider y/(25 – y) = 6/9
y/(25 – y) = 2/3
By cross multiplication we get,
3y = 2(25 – y)
⇒ 3y = 50 – 2y
⇒ 3y + 2y = 50
⇒ 5y = 50
⇒ y = 50/5
⇒ y = 10
Therefore, AO = x = 8 cm
OD = 20 – x = 20 – 8 = 12 cm
BO = y = 10 cm
OC = 25 – y = 25 – 10 = 15 cm
6. In ΔABC, DE is drawn parallel to BC. If AD: DB = 2: 3, DE = 6 cm and AE = 3.6 cm find BC and AC.
Answer
Now consider the ΔABC,
DE ∥ BC … [given]
From Basic Proportionality Theorem (BPT) AD/DB = AE/EC
2/3 = 3.6/m
⇒ 2m = 3 × 3.6
⇒ 2m = 10.8
⇒ m = 10.8/2
⇒ m = 5.4 cm
So, EC = 5.4 cm
Therefore, AC = 3.6 + x
= 3.6 + 5.4
= 9 cm
Now, consider the ΔADE and ΔABC,
∠ABC = ∠ADE … [because corresponding angles are equal]
∠ACB = ∠AED … [because corresponding angles are equal]
Therefore, ΔABC ~ ΔADE … [from AA corollary]
AE/AC = DE/BC
3.6/9 = 6/n
By cross multiplication we get,
3.6n = 9 × 6
⇒ 3.6n = 54
⇒ n = 54/3.6
⇒ n = 15
So, BC = 15 cm
7. D and E are points on the sides AB and AC respectively of ΔABC such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, show that DE is parallel to BC.
Answer
We have to show that, DE ∥ BC
Consider the ΔABC,
AD/DB = 1.4/4.2
= 14/42
⇒ AD/DB = 1/3 …(i)
AE/EC = 1.8/5.4
= 18/54
⇒ AE/EC = 1/3 …(ii)
From (i) and (ii)
AD/DB = AE/EC
Hence, it is proved that DE ∥ BC by converse BPT.
8. In ΔABC, D and E are points on the sides AB and AC respectively. If AD = 4 cm, DB = 4.5 cm, AE = 6.4 cm and EC = 7.2 cm, find if DE is parallel to BC or not.
Answer
We have to show that, DE ∥ BC
Consider the ΔABC,
AD/DB = 4/4.5
= 40/45
⇒ AD/DB = 8/9 … (i)
AE/EC = 6.4/7.2
= 64/72
⇒ AE/EC = 8/9 … (ii)
From (i) and (ii)
AD/DB = AE/EC
Hence, it is proved that DE ∥ BC by converse BPT.
9. ΔABC ~ ΔPQR such that AB = 1.8 cm and PQ = 2.1 cm. Find the ratio of areas of ΔABC and ΔPQR.
Answer
We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.
So, Area of ΔABC/Area of ΔPQR = AB2/PQ2
= (1.8)2/(2.1)2
= (18/21)2
= (6/7)2
= 36/49
Therefore, the ratio of areas of ΔABC and ΔPQR is 36: 49.
10. ΔABC ~ ΔPQR, AD and PS are altitudes from A and P on sides BC and QR respectively. If AD: PS = 4: 9, find the ratio of the areas of ΔABC and ΔPQR.
Answer
We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.
So, Area of ΔABC/Area of ΔPQR = AB2/PQ2 … [equation (i)]
Now, consider the ΔBAD and ΔQPS
∠ABD = ∠PQS … [because ΔABC ~ ΔPQR]
∠AOB = ∠PSQ … [because both angles are equal to 90o]
Therefore, ΔBAD ~ ΔQPS … [from AA corollary]
So, AB/PQ = AD/PS … [equation (ii)]
From equation (i) and equation (ii),
Area of ΔABC/Area of ΔPQR = AD2/PS2
= (4)2/(9)2
= 16/81
Therefore, the ratio of areas of ΔABC and ΔPQR is 16: 81.
11. ΔABC ~ ΔDEF. If BC = 3 cm, EF = 4 cm and area of ΔABC = 54 cm2, find the area of ΔDEF.
Answer
We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.
So, Area of ΔABC/Area of ΔDEF = BC2/EF2
⇒ 54/Area of ΔDEF = (3)2/(4)2
⇒ 54/Area of ΔDEF = 9/16
⇒ Area of ΔDEF = (54 × 16)/9
⇒ Area of ΔDEF = 96 cm2
Therefore, area of ΔDEF is 96 cm2.
12. ΔABC ~ ΔXYZ. If area of ΔABC is 9 cm2and area of ΔXYZ is 16 cm2and if BC = 2.1 cm, find the length of YZ.
Answer
We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.
So, Area of ΔABC/Area of ΔXYZ = BC2/YZ2
9/16 = (2.1)2/(YZ)2
Taking square root on both sides we get,
3/4 = 2.1/YZ
⇒ YZ = (2.1 × 4)/3
⇒ YZ = 2.8
Therefore, YZ is 2.8 cm.
13. Prove that the area of ΔBCE described on one side BC of a square ABCD is one half the area of the similar ΔACF described on the diagonal AC.
Answer
We know that, Pythagoras theorem, AC2 = AB2 + BC2
AC2 = 2BC2 … [because AB = BC]
From the question it is given that, ΔBCE ~ ΔACF
We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.
So, Area of ΔBCE/Area of ΔACF = BC2/AC2
= ½
Therefore, the ratio of areas of ΔBCE and ΔACF is 1: 2.
14. In ΔABC, MN ∥ BC
Answer
(a) From the it is given that, AN: AC = 5: 8
We have to find the ratio of area of ΔAMN: area of ΔAB
So, consider the ΔAMN and ΔABC
∠AMN = ∠ABC … [because corresponding angles are equal]
∠ANM = ∠ACB … [because corresponding angles are equal]
Therefore, ΔAMN ~ ΔABC … [from AA corollary]
We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.
So, Area of ΔAMN/Area of ΔABC = AN2/AC2
= (5/8)2
= 25/64
Therefore, the ratio of areas of ΔAMN and ΔABC is 25: 64.
(b) From the it is given that, AB/AM = 9/4
In the above solution we prove that, ΔAMN ~ ΔABC
So, Area of ΔAMN/Area of ΔABC = AM2/AB2
= 42/92
= 16/81
(Area of ΔABC – Area of trapezium MBCN)/area of ΔABC = 16/81
By cross multiplication we get,
81(Area of ΔABC – Area of trapezium MBCN) = area of ΔABC × 16
⇒ (81 × Area of ΔABC) – (81 × area of trapezium MBCN) = 16 × area of ΔABC
⇒ 64 area of ΔABC = 81 area of trapezium MBCN
Then, area of trapezium MBCN/area of ΔABC = 65/81
Therefore, the ratio of areas of trapezium MBCN and ΔABC is 65: 81.
(c) From the it is given that, BC = 14 cm and MN = 6 cm
In the above solution (a) we prove that, ΔAMN ~ ΔABC
So, Area of ΔAMN/Area of ΔABC = MN2/BC2
= 62/142
= 36/196
= 9/49
(Area of ΔAMN/Area of ΔAMN + Area of trapezium MBCN) = 9/49
By cross multiplication we get,
49(Area of ΔAMN) = 9 (Area of ΔAMN + Area of trapezium MBCN)
⇒ 49 Area of ΔAMN = 9 Area of ΔAMN + 9 Area of trapezium MBCN
⇒ 49 area of ΔAMN = 9 area of trapezium MBCN
Then, area of ΔABC/area of trapezium MBCN = 9/40
Therefore, the ratio of area of ΔABC and area of trapezium is 9: 40.
15. In ΔABC, DE ∥ BC; DC and EB intersect at F, if DE/BC = 2/7, find area of ΔFDE/area of ΔFBC
Answer
Consider the ΔFDE and ΔFCB
∠FDE = ∠FCB … [because interior alternate angles are equal]
∠ANM = ∠ACB … [because interior alternate angles are equal]
Therefore, ΔFDE ~ ΔFCB … [from AA corollary]
We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.
So, Area of ΔFDE/Area of ΔFBC = DE2/BC2
= (2/7)2
= 4/49
Therefore, the ratio of areas of ΔFDE and ΔFBC is 4: 49.
16. Figure shows ΔPQR in which ST || QR and SR and QT intersect each other at M. IF PT/TR = 5/3, find Ar.(ΔMTS)/Ar.(ΔMQR).
17. Figure shows ΔKLM, P and T on KL and KM respectively such that ∠KLM = ∠KTP. If KL/KT = 9/5, find (Ar. ΔKLM)/(Ar. ΔKTP).
18. In figure, DEF is a right-angled triangle with ∠E = 90 ͦ. FE is produced to G and GH is drawn perpendicular to DF. If DE = 8 cm, DH = 8 cm, DH = 6 cm and HF = 4 cm, find Ar.(ΔDEF)/Ar.(ΔGHF).
19. In the figure, ABCD is a quadrilateral. F is a point on AD such that AF = 2. 1 cm and FD = 4.9 cm. E and G are points on AC and AB respectively such that EF || CD and GE || BC. Find Ar.(ΔBCD)/Ar.(ΔGEF).
Exercise 15.2
1. The dimensions of a building are 50 m long, 40 m wide and 70 m high. A model of the same building is made with a scale factor of 1 : 500. Find the dimensions of the model.
Answer
2. A ship is 400 m long and 100 m wide. The length of its model is 20 cm. find the surface area of the deck of the model.
Answer
3. A model of a ship is made with a scale factor of 1 : 500. Find
(i) The length of the ship, if the model length is 60 cm.
(ii) The deck area of the model, if the deck area of the ship is 1500000 m2.
(iii) The volume of the ship, if the volume of its model is 200 cm2.
Answer
4. An aeroplane is 30 m long and its model is 15 cm long. If the total outer surface area of the model is 150 cm2, find the cost of painting the outer surface of the aerolplane at Rs 120 per m2, if 50 m2 is left out for windows.
Answer
5. The length of a river in a map is 54 cm. If 1 cm on the map represents 12500 m on land, find the length of the river.
Answer
6. The scale of a map is 1 : 200000. A plot of land of area 20 km2 is to be represented on the map. Find:
(i) The number of kilometers on the ground represented by 1 cm on the map.
(ii) The area in km2 that can be represented by 1 cm2.
(iii) The area on the map that represents the plot of land.
Answer
7. The actual area of an island is 1872 km2. On a map, this area is 117 cm2. If the length of the coastline is 44 cm on the map, find the length of its actual coastline.
Answer
8. On a map drawn to a scale of 1 : 25000, a triangular plot of a land is marked as ABC with AB = 6 cm, BC = 8 cm and ∠ABC = 90 ͦ. Calculate the actual length of AB in km and the actual area of the plot in km2.
Answer
9. On a map drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD is measured as AB = 12 cm and BC = 16 cm. Calculate the diagonal distance of the plot in km and the plot area in km2.
Answer
