Page 229 - ICSE Math 8

In right-angled ∆AOB,
2
2
AB = AO + BO 2
2
2
2
2
⇒ (50) = (40) + (BO) ⇒ (BO) = 2,500 – 1,600 ⇒ BO = 900 = 30 cm
BD = BO + OD = 30 cm + 30 cm = 60 cm
1 1
∴ Area of rhombus ABCD = × AC × BD = × 80 cm × 60 cm = 2,400 cm 2
2
2
Example 11: The area of a rhombus is equal to the area of a triangle whose base is of length 36 cm and the
corresponding altitude is 18 cm. Find the length of the other diagonal of the rhombus, if the
length of one of its diagonals is 27 cm. A
1
Solution: Area of triangle = × base × height
2
1 18 cm
= 2 × 36 cm × 18 cm = 324 cm 2
According to the question, B D C
Area of rhombus = Area of triangle 36 cm
1 D C
⇒ × AC × BD = 324 cm 2
2
1
⇒ × 27 cm × BD = 324 cm 2 O
2
324 × 2
⇒ BD = 27 cm ⇒ BD = 24 cm A B

Area of a Trapezium A a B
1
Area of trapezium = × (sum of the lengths of its parallel sides) × height h
2
1 C
= 2 × (a + b) × h D b
Here, a and b are the lengths of the parallel sides and height of the trapezium is the perpendicular distance
between its parallel sides.
2
Example 12: The area of a trapezium is 240 cm . Find the length of the other parallel side of the trapezium
if the length of one of its parallel sides is 21 cm and its height is 15 cm.
1
Solution: Area of trapezium = × (sum of the lengths of its parallel sides) × height
2
1 240 × 2 D C
∴ × (21 + CD) × 15 = 240 ⇒ 21 + CD = 15 15 cm
2
⇒ CD = 32 – 21 ⇒ CD = 11 cm A E B
∴ The length of the other parallel side of the trapezium is 11 cm.

2
Example 13: Find the height of a trapezium whose area is 4.2 m and the length of its parallel sides is
2.4 m and 3.2 m respectively.
1
Solution: Area of trapezium = × (sum of the lengths of its parallel sides) × height
2
1
∴ 4.2 = × (2.4 + 3.2) × height
2
1 4.2 × 2
⇒ 4.2 = × 5.6 × height ⇒ 5.6 = height
2
⇒ 1.5 = height
∴ The height of the trapezium is 1.5 m.

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