The area of trapezium ABCD in the given figure is

(a) $62 c m^{2}$

(b) $93 c m^{2}$

(c) $124 c m^{2}$

(d) $155 c m^{2}$

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Solution

(c)124 cm^2

In the right angle triangle BEC, we have:
EC = √17^2-15^2=√289-225 =√64 =8 cm
ar(trapez. ABCD) = 1/2×(sum of parallel sides)×distance between them=1/2×31×8=124cm^2

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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 cm^2 In the right angle triangle BEC, we have: EC = √17^2-15^2=√289-225 =√64 =8 cm ar(trapez. ABCD) = 1/2×(sum of parallel sides)×distance between them=1/2×31×8=124cm^2

The area of trapezium ABCD in the given figure is

(a) $62 c m^{2}$

(b) $93 c m^{2}$

(c) $124 c m^{2}$

(d) $155 c m^{2}$

(c)124 cm^2

In the right angle triangle BEC, we have:
EC = √17^2-15^2=√289-225 =√64 =8 cm
ar(trapez. ABCD) = 1/2×(sum of parallel sides)×distance between them=1/2×31×8=124cm^2