SSC CGL Tier II 12 Jan 2017 Paper 1

Question : 80 of 100

Marks: +1, -0

PQRA is a rectangle, AP = 22 cm, PQ = 8 cm. ΔABC is a triangle whose vertices lie on the sides of PQRA such that BQ = 2 cm and QC = 16 cm. Then the length of the line joining the mid points of the sides AB and BC is

$4 √2$ cm.
5 cm.
6 cm.
10 cm.

Validate

Solution:

(2)

QC = 16 cm.
∴ CR = 22 – 16 = 6 cm.

{BC}=√{ {BQ}^{2}+ {QC}^{2}}

=√{2^{2}+16^{2}} =√{4+256} =√{260} {cm}

A C=√{C R^{2}+A R^{2}}

=√{6^{2}+8^{2}}=√{36+64}=√100

=10 cm
BD = DC
BE = EA
∴ DE || AC and DE =

1/ 2

AC =

10 /2 = 5

cm.

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