SSC CGL 11 Aug 2017 Shift 1 Paper

Question : 67 of 100

Marks: +1, -0

ABCD is an isosceles trapezium such that AD∥BC,AB = 5 cm, AD = 8 cm and BC =14 cm. What is the area (in cm

^{2}

) of trapezium?

Solution:

AB =DC =5 cm

AD = 8cm
BC = 14 cm
In ΔAFB,

(AB)^2 = (AF)^2 + (BP)^2

⇒

(5)^2 = (AF)^2 + (3)^2

∴

AF = √{ 2 5 -9} = √16 =4

The area of trapezium ABCD

=1/2(8+14) ×4

=1/2 ×22 ×4=22 ×2

= 44 \cm^2

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