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KEAM 2017 Math Paper

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Question : 56 of 120

Marks: +1, -0

The area of the parallelogram with vertices (0, 0), (7, 2), (5, 9) and (12, 11) is

50
54
51
52
53

Solution:

Let

A(0,0), B(7,2), C(5,9), D(12,11)

\therefore

Area of

\triangle ABC = \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ 7 & 2 & 1 \\ 5 & 9 & 1 \end{vmatrix}

= \frac{1}{2} \cdot 1(63-10) = \frac{53}{2}

sq unit

\therefore

Area of

\triangle ACD = \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ 5 & 9 & 1 \\ 12 & 11 & 1 \end{vmatrix}

= \frac{1}{2} \cdot 1(55-108)

= \frac{53}{2}

sq unit

\therefore

Area of parallelogram

A B C D

=

Area of

\triangle ABC +

Area of

\triangle ACD

= \frac{53}{2} + \frac{53}{2}

=53

sq unit

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