The area of trapezium A B C D in the given figure isa 62 cm 2b 93 cm 2c 124 cm 2d 155 cm 2

(c)124 nbsp;cm2 In the right angle triangle BEC, we have: EC nbsp; nbsp;= 172 152=289 225=64=8 nbsp;cm ar(trapez. ABCD) = 12×(sum nbsp;of nbsp;para ...

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Standard VIII

Mathematics

Perimeter and Area of Trapezium

The area of t...

Question

The area of trapezium ABCD in the given figure is
(a) 62 cm²
(b) 93 cm²
(c) 124 cm²
(d) 155 cm²

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Solution

(c)124 cm²

In the right angle triangle BEC, we have:
EC = $\sqrt{17^{2} - 15^{2}} = \sqrt{289 - 225} = \sqrt{64} = 8 cm$
ar(trapez. ABCD) = $\frac{1}{2} \times (sum of parallel sides) \times distance between them = \frac{1}{2} \times 31 \times 8 = 124$ cm²

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The area of trapezium ABCD in the given figure is (a) 62 cm2 (b) 93 cm2 (c) 124 cm2 (d) 155 cm2

(c)124 cm2 In the right angle triangle BEC, we have: EC = 172-152=289-225=64=8 cm ar(trapez. ABCD) = 12×(sum of parallel sides)×distance between them=12×31×8=124 cm2

The area of trapezium ABCD in the given figure is
(a) 62 cm²
(b) 93 cm²
(c) 124 cm²
(d) 155 cm²

(c)124 cm²

In the right angle triangle BEC, we have:
EC = $\sqrt{17^{2} - 15^{2}} = \sqrt{289 - 225} = \sqrt{64} = 8 cm$
ar(trapez. ABCD) = $\frac{1}{2} \times (sum of parallel sides) \times distance between them = \frac{1}{2} \times 31 \times 8 = 124$ cm²