SSC MTS Model Paper 2

Question : 42 of 150

Marks: +1, -0

In the trapezium ABCD, ∠ BAE = 30°, ∠ CDF = 45°, BC = 6 cm and AB = 12 cm, then find the area of trapezium.

Solution:

(3)

{ { BE } } /{ { AB } } = \sin 30 ^ {°} = { 1 }/ { 2 }

{ BE } = { 1 }/ { 2 } × { AB } = 6 { cm } = { CF }

and

{ { CF } }/ { { DF } } = \tan 45 ^ { ° } = 1

∴ { DF } = { CF } = 6 { cm }

∴ { AE } = √ { 12 ^ { 2 } - 6 ^ { 2 } } = 6 √ { 3 } { cm }

{ AD } = 6 + 6 + 6 √ { 3 } = 6 ( 2 + √ { 3 } )

Area of trapezium ADCB
=

1/2×(AD+BC)×BE

= { 1 } /{ 2 } × [ 6 ( 2 + √ { 3 } ) + 6 ] × 6

= 3 ( 2 + √ { 3 } + 1 ) × 6 = 18 ( 3 + √ { 3 } ) { cm } ^ { 2 }

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