SSC CHSL 7 Jan 2017 Shift 1 Paper

Question : 56 of 100

Marks: +1, -0

The base angle of an isosceles trapezium is 45°. If the shorter side and both the equal sides are 10 cm each, what is the area of the trapezium?

Solution:

In ∆ ADE, cos 45 =

{AE}/{AD}

⇒

1/√2={AE}/10

⇒ AE =

10√2=5√2

Similarly, BF =

5√2

Also, in ∆ ADE sin 45 =

{DE}/{AD}

⇒

1/√2={DE}/10

⇒ DE =

10/√2=5√2

=>

AB = AE + EF + BF =

5√2+10+5√2

=10+10√2

∴ Area of trapezium =

1/2×

sum of parallel sides × height
=

1/2×(CD+AB)×DE

1/2×(10+10+10√2)×(5√2)

(10+5√2)×5√2

50√2+50

sq. cm

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