SSC CGL Tier II 12 Apr 2015 Paper

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Question : 35 of 100

Marks: +1, -0

If the length of each of two equal sides of an isosceles triangle is 10 cm. and the adjacent angle is 45°, then the area of the triangle is

20 $√2$ square cm.
12 $√2$ square cm.
25 $√2$ square cm.
15 $√2$ square cm

Solution:

(3)

AD = AB sin 45° = 10 ×

1/√2

=5√{2} {cm} .

∴

Area of

\Δ{ABC}

= {1}/{2}×{BC} ×{AD}

= {1}/{2} × 10 × 5√{2}

=25 √{2}

square cm.

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