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SSC MTS Model Paper 6

Question : 43 of 150

Marks: +1, -0

If ABCD is a rectangle and E, F are the mid-points of BC and AD respectively and G is any point on EF, then ∆GAB equals: 1 1

$\frac{1}{6}$ (rectangle ABCD)
$\frac{1}{4}$ (rectangle ABCD)
$\frac{1}{3}$ (rectangle ABCD)
$\frac{1}{2}$ (rectangle ABCD)

Validate

Solution:

AB || EF || CD. So ABEF is a rectangle

∴ ∆AGB =

\frac{1}{2}

(area of rectangle ABEF)

\frac{1}{2}

× (

\frac{1}{2}

area of rectangle ABCD )

\frac{1}{4}

(area of rectangle ABCD)

or, If a triangle and a parallelogram are on the same base and between the same parallels then the areaof the triangle is equal to half the area of the parallelogram.

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