Exercise 9.2
Exercise 9.2 Solutions
Area scaling in similar figures. Proportional relationships showing Area Ratio = (Side Ratio)².
Question 1
Area Ratios
Find the ratio of the areas of similar figures if the ratio of their corresponding lengths are:
(i) 1:3 (ii) 3:4 (iii) 2:7 (iv) 8:9 (v) 6:5
(i) 1:3 (ii) 3:4 (iii) 2:7 (iv) 8:9 (v) 6:5
Solution Step-by-Step
We know that if two figures are similar, the ratio of their areas is equal to the square of the ratio of their corresponding lengths:
A1A2
=
(l1l2)2
(i) Length ratio = 1:3
Area ratio
=
(1/3)2 = 1:9
(ii) Length ratio = 3:4
Area ratio
=
(3/4)2 = 9:16
(iii) Length ratio = 2:7
Area ratio
=
(2/7)2 = 4:49
(iv) Length ratio = 8:9
Area ratio
=
(8/9)2 = 64:81
(v) Length ratio = 6:5
Area ratio
=
(6/5)2 = 36:25
Question 2
Find Unknowns
Find the unknowns in the following figures:
(i) A1 = 240 cm², A2 = ?, l1 = 10 cm, l2 = 6 cm
(ii) A1 = 60 cm², A2 = ?, l1 = 15 cm, l2 = 20 cm
(iii) A1 = ?, A2 = 18 cm², l1 = 3.6 cm, l2 = 5.76 cm
(iv) A1 = ?, A2 = 96 cm², l1 = 15 cm, l2 = 12 cm
(v) A1 = 25/7 cm², A2 = 63 cm², l1 = ?, l2 = 12.6 cm
(i) A1 = 240 cm², A2 = ?, l1 = 10 cm, l2 = 6 cm
(ii) A1 = 60 cm², A2 = ?, l1 = 15 cm, l2 = 20 cm
(iii) A1 = ?, A2 = 18 cm², l1 = 3.6 cm, l2 = 5.76 cm
(iv) A1 = ?, A2 = 96 cm², l1 = 15 cm, l2 = 12 cm
(v) A1 = 25/7 cm², A2 = 63 cm², l1 = ?, l2 = 12.6 cm
Solution Step-by-Step
(i) Find A2
Using Area Scale Factor formula:
A2A1
=
(l2l1)2
A2240
=
(610)2 = 36100 = 0.36
A2
=
0.36 × 240 = 86.4 cm²
(ii) Find A2
Using formula:
A2A1
=
(l2l1)2
A260
=
(2015)2 = (43)2 = 169
A2
=
60 × 169 = 9609 ≈ 106.67 cm²
(iii) Find A1
Using formula:
A1A2
=
(l1l2)2
A118
=
(3.65.76)2 = (0.625)2 = 0.390625
A1
=
0.390625 × 18 = 7.03 cm²
(iv) Find A1
Using formula:
A1A2
=
(l1l2)2
A196
=
(1512)2 = (54)2 = 2516
A1
=
96 × 2516 = 6 × 25 = 150 cm²
(v) Find l1
Using formula:
A1A2
=
(l1l2)2
25/763
=
(l112.6)2
257 × 63
=
25441 = (l112.6)2
Take square root on both sides:
521
=
l112.6
l1
=
5 × 12.621 = 6321 = 3 cm
Question 3
Triangle & Trapezoid
Given that area of ΔABC = 36 cm² and AB = 6 cm, BD = 4 cm. Points D and E are such that ΔABC ~ ΔADE. Find:
(a) The area of ΔADE.
(b) The area of trapezium BCED.
(a) The area of ΔADE.
(b) The area of trapezium BCED.
Solution Step-by-Step
(a) Area of ΔADE:
The side length AD = AB + BD = 6 + 4 = 10 cm.
Since ΔABC ~ ΔADE:
The side length AD = AB + BD = 6 + 4 = 10 cm.
Since ΔABC ~ ΔADE:
Area(ΔADE)Area(ΔABC)
=
(ADAB)2
Area(ΔADE)36
=
(106)2 = 10036
Area(ΔADE)
=
36 × 10036 = 100 cm²
(b) Area of trapezium BCED:
The area of the trapezium BCED is the difference between the larger and smaller triangles:
The area of the trapezium BCED is the difference between the larger and smaller triangles:
Area(BCED)
=
Area(ΔADE) - Area(ΔABC)
=
100 - 36 = 64 cm²
Question 4
Scale Factor Area
Given that ΔABC and ΔDEF are similar, with a scale factor of k = 3. If the area of triangle ΔABC is 50 cm², what is the area of triangle ΔDEF?
Solution Step-by-Step
When two figures are similar with scale factor k, the area of the scaled figure is:
Areascaled
=
k2 × Areaoriginal
Area(ΔDEF)
=
32 × 50
=
9 × 50 = 450 cm²
Thus, the area of triangle ΔDEF is 450 cm².
Question 5
Quadrilateral Scaling
Quadrilaterals ABCD and EFGH are similar, with a scale factor of k = 4. If the area of quadrilateral ABCD is 64 cm², find the area of quadrilateral EFGH.
Solution Step-by-Step
Since the quadrilaterals are similar with scale factor k = 4:
Area(EFGH)
=
k2 × Area(ABCD)
=
42 × 64
=
16 × 64 = 1024 cm²
Thus, the area of quadrilateral EFGH is 1024 cm².
Question 6
Sides from Area
The areas of two similar triangles are 16 cm² and 25 cm². What is the ratio of a pair of corresponding sides?
Solution Step-by-Step
Let the corresponding side lengths of the two triangles be l1 and l2:
(l1l2)2
=
A1A2 = 1625
Take the square root of both sides to find the side ratio:
l1l2
=
√(1625) = 45
∴ The ratio of corresponding sides is 4:5.
Question 7
Base from Areas
The areas of two similar triangles are 144 cm² and 81 cm². If the base of the large triangle is 30 cm, find the corresponding base of the smaller triangle.
Solution Step-by-Step
Let the base of the large triangle be b1 = 30 cm and the base of the small triangle be b2. By the area similarity rule:
(b1b2)2
=
A1A2 = 14481
Take square root on both sides:
b1b2
=
129 = 43
Substitute b1 = 30:
30b2
=
43
4b2
=
90 ⇒ b2 = 904 = 22.5 cm
Thus, the corresponding base of the smaller triangle is 22.5 cm.
Question 8
Heptagon Areas
A regular heptagon is inscribed in a larger regular heptagon and each side of the larger heptagon is 1.7 times the side of the smaller heptagon. If the area of the smaller heptagon is known to be 100 cm², find the area of the larger heptagon.
Solution Step-by-Step
The side scaling ratio is k = 1.7. Since all regular heptagons are similar, the area scales quadratically:
Arealarge
=
k2 × Areasmall
Arealarge
=
1.72 × 100
=
2.89 × 100 = 289 cm²
Thus, the area of the larger regular heptagon is 289 cm².
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