Std 8 Mathematics · Chapter 15

15. Area

Solutions to Practice Sets 15.1 to 15.6 — parallelogram, rhombus, trapezium, Herons formula, irregular plots and circles, with figures. Maharashtra State Board (Balbharati), Standard Eight Mathematics.

Practice Set 15.1

Formula: Area of a parallelogram = base × height

1If the base of a parallelogram is 18 cm and its height is 11 cm, find its area.

Area = base × height = 18 × 11 = 198

Answer: 198 sq cm

2If the area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.

Area = base × height ∴ 29.6 = 8 × height
∴ height = 29.68 = 3.7

Answer: 3.7 cm

3Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its base.

83.2 = base × 6.4
∴ base = 83.26.4 = 83264 = 13

Answer: 13 cm

Practice Set 15.2

Formula: Area of a rhombus = 12 × product of the lengths of the diagonals. The diagonals of a rhombus are perpendicular bisectors of each other.

1Lengths of the diagonals of a rhombus are 15 cm and 24 cm. Find its area.

Area = 12 × 15 × 24 = 15 × 12 = 180

Answer: 180 sq cm

2Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm. Find its area.

Area = 12 × 16.5 × 14.2 = 16.5 × 7.1 = 117.15

Answer: 117.15 sq cm

3If the perimeter of a rhombus is 100 cm and the length of one diagonal is 48 cm, what is the area of the quadrilateral?

Side = 100 ÷ 4 = 25 cm. Half of the given diagonal = 48 ÷ 2 = 24 cm.
The half-diagonals and a side form a right angled triangle. Other half-diagonal = √25² − 24² = √625 − 576 = √49 = 7 cm
∴ Other diagonal = 2 × 7 = 14 cm
Area = 12 × 48 × 14 = 24 × 14 = 336

Answer: 336 sq cm

4If the length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.

240 = 12 × 30 × d ∴ 240 = 15d ∴ d = 16 cm
Half-diagonals: 30 ÷ 2 = 15 cm and 16 ÷ 2 = 8 cm
Side = √15² + 8² = √225 + 64 = √289 = 17 cm
Perimeter = 4 × 17 = 68

Answer: 68 cm

Practice Set 15.3

Formula: Area of a trapezium = 12 × sum of the lengths of parallel sides × height

1In □ABCD, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm. Find the area of □ABCD.

AB ∥ DC and AD ⊥ DC, so AD is the height.

Area = 12 × (13 + 9) × 8 = 12 × 22 × 8 = 88

Answer: 88 sq cm

2The lengths of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm and its height is 4.2 cm. Find its area.

Area = 12 × (8.5 + 11.5) × 4.2 = 12 × 20 × 4.2 = 42

Answer: 42 sq cm

The textbook key prints this answer as “42 cm”; an area is measured in square units, so it is 42 sq cm.

3★□PQRS is an isosceles trapezium, l(PQ) = 7 cm, seg PM ⊥ seg SR, l(SM) = 3 cm, and the distance between the two parallel sides is 4 cm. Find the area of □PQRS.

In an isosceles trapezium the two slanting parts cut off at the ends are equal.

Draw seg QN ⊥ seg SR. Since the trapezium is isosceles, l(NR) = l(SM) = 3 cm and l(MN) = l(PQ) = 7 cm.
∴ l(SR) = 3 + 7 + 3 = 13 cm
Area = 12 × (7 + 13) × 4 = 12 × 20 × 4 = 40

Answer: 40 sq cm

Practice Set 15.4

Heron’s formula: A = √s(s − a)(s − b)(s − c), where s = a + b + c2

1Sides of a triangle are 45 cm, 39 cm and 42 cm. Find its area.

s = 45 + 39 + 422 = 1262 = 63
Area = √63 × (63 − 45) × (63 − 39) × (63 − 42) = √63 × 18 × 24 × 21
= √(3 × 3 × 7) × (2 × 3 × 3) × (2 × 2 × 2 × 3) × (3 × 7) = √2⁴ × 3⁶ × 7²
= 2² × 3³ × 7 = 4 × 27 × 7 = 756

Answer: 756 sq cm

2Look at the measures shown in the figure and find the area of □PQRS.

Diagonal PR divides □PQRS into two triangles.

△PSR is right angled at S. l(PR) = √36² + 15² = √1296 + 225 = √1521 = 39 m
A(△PSR) = 12 × 36 × 15 = 270 sq m
△PQR: sides 56, 25, 39. s = 56 + 25 + 392 = 60
A(△PQR) = √60 × 4 × 35 × 21 = √176400 = 420 sq m
A(□PQRS) = 270 + 420 = 690

Answer: 690 sq m

3Some measures are given in the figure. Find the area of □ABCD.

Diagonal DB divides □ABCD into △DAB and △DBC.

A(△DAB) = 12 × AB × AD = 12 × 40 × 9 = 180 sq m (right angle at A)
A(△DBC) = 12 × DC × BT = 12 × 60 × 13 = 390 sq m (BT is the height)
A(□ABCD) = 180 + 390 = 570

Answer: 570 sq m

Practice Set 15.5

1Find the areas of the given plots. (All measures are in metres.)

(1)

Plot PQRST divided into triangles and a trapezium (all measures in metres).

A(△PAQ) = 12 × 30 × 50 = 750
A(trapezium AQRC) = 12 × (50 + 25) × 60 = 2250 (AC = 30 + 30 = 60)
A(△RCS) = 12 × 60 × 25 = 750
A(△PTS) = 12 × 150 × 30 = 2250 (PS = 30 + 30 + 30 + 60 = 150)
Total area = 750 + 2250 + 750 + 2250 = 6000

Answer: 6000 sq m

(2)

Plot ABCDE divided into three triangles by BE and CE.

△BAE is right angled at A: l(AE) = √30² − 24² = √324 = 18
A(△BAE) = 12 × 24 × 18 = 216 sq m
△BCE: sides 30, 26, 28. s = 842 = 42; A = √42 × 12 × 16 × 14 = √112896 = 336 sq m
△CDE: base CE = 28, height DF = 16. A = 12 × 28 × 16 = 224 sq m
Total area = 216 + 336 + 224 = 776

Answer: 776 sq m

In the textbook figure the label 16 is printed near seg CD, but it must be the height DF (seg DF ⊥ seg CE) for the area of △CDE; this gives the key’s answer of 776 sq m.

Practice Set 15.6

Formula: Area of a circle = πr², taking π = 227.

1Radii of circles are given below; find their areas. (1) 28 cm (2) 10.5 cm (3) 17.5 cm

(1) r = 28: Area = 227 × 28 × 28 = 22 × 4 × 28 = 2464 sq cm
(2) r = 10.5: Area = 227 × 10.5 × 10.5 = 22 × 1.5 × 10.5 = 346.5 sq cm
(3) r = 17.5: Area = 227 × 17.5 × 17.5 = 22 × 2.5 × 17.5 = 962.5 sq cm

Answer: (1) 2464 sq cm (2) 346.5 sq cm (3) 962.5 sq cm

2Areas of some circles are given below; find their diameters. (1) 176 sq cm (2) 394.24 sq cm (3) 12474 sq cm

(1) 176 = 227 × r² ∴ r² = 176 × 722 = 56 ∴ r = √56. Diameter = 2√56 cm (= 4√14 ≈ 14.97 cm)
(2) 394.24 = 227 × r² ∴ r² = 394.24 × 722 = 125.44 ∴ r = 11.2. Diameter = 22.4 cm
(3) 12474 = 227 × r² ∴ r² = 12474 × 722 = 3969 ∴ r = 63. Diameter = 126 cm

Answer: (1) 2√56 cm (2) 22.4 cm (3) 126 cm

3The diameter of a circular garden is 42 m. There is a 3.5 m wide road around the garden. Find the area of the road.

The road (brown ring) lies between the garden circle and the outer circle.

Radius of the garden = 42 ÷ 2 = 21 m. Radius of the outer circle = 21 + 3.5 = 24.5 m.
Area of the road = area of outer circle − area of garden
= 227 × (24.5² − 21²) = 227 × (600.25 − 441) = 227 × 159.25 = 500.5

Answer: 500.50 sq m

4Find the area of the circle if its circumference is 88 cm.

Circumference = 2πr ∴ 88 = 2 × 227 × r ∴ r = 88 × 744 = 14 cm
Area = 227 × 14 × 14 = 22 × 2 × 14 = 616

Answer: 616 sq cm

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