SSC CGL Tier II 25 Oct 2015 Paper 1

Question : 57 of 100

Marks: +1, -0

The area of an isosceles trapezium is 176 cm

^{2}

and the height is

2 /11

th of the sum of its parallel sides. If the ratio of the length of the parallel sides is 4 : 7, then the length of a diagonal (in cm) is

Solution:

(3)

BC = 7x cm.
AD = 4x cm.
AB = DC ; AE ⊥ BC ; DF ⊥ BC
Area of trapezium ABCD

=1/2×(\AD+\BC)×\AE

⇒ 176= {1}/{2} × 11 x ×{2}/{11} × 11 x

⇒176=11 x^{2}

⇒ x^{2}= {176}/{11}=16

⇒ x=√{16}=4

∴ BC = 7 × 4 = 28 cm.
AD = 4 × 4 = 16 cm.
∴ BE = FC

= {1}/{2}(28-16) {\cm}

= 6 cm
∴ BF = 16 + 6 = 22 cm.

∴{DF}= {2}/{11} × 11 x=2 x= 8 cm.

∴ Diagonal

{BD}=√{ {BF}^{2}+ {FD}^{2}}

=√{22^{2}+8^{2}}

=√{484+64}

=√{548}=√{4 × 137}

=2√{137} {\cm} .

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