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P(4,-2), Q(7,2), R(0,9), S(-3,5) forms a parallelogram
Let the height of parallelogram taking PQ as base be h.
{tex}\therefore \quad PQ = \sqrt { ( 7 - 4 ) ^ { 2 } + ( 2 + 2 ) ^ { 2 } }{/tex}
{tex}= \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } = \sqrt { 9 + 16 }{/tex}
{tex}=\sqrt{25}{/tex}
= 5 units
Area of triangle PQR
{tex}= \frac { 1 } { 2 } \left[ x _ { 1 } \left( y _ { 2 } - y _ { 3 } \right) + x _ { 2 } \left( y _ { 3 } - y _ { 1 } \right) + x _ { 3 } \left( y _ { 1 } - y _ { 2 } \right) \right]{/tex}
{tex}= \frac { 1 } { 2 } [ 4 ( 2 - 9 ) + 7 ( 9 + 2 ) + 0 ( 2 - 2 ) ]{/tex}
{tex}= \frac{1}{2} [4(-7)+7(11)]{/tex}
{tex}= \frac { 1 } { 2 } \times 49 = \frac { 49 } { 2 } {/tex}sq. units
Now, {tex}\frac { 1 } { 2 } \times PQ \times h = \frac { 49 } { 2 }{/tex}
{tex}\Rightarrow{/tex} {tex}\frac { 1 } { 2 } \times 5 \times h = \frac {49}{2}{/tex}
{tex}\Rightarrow{/tex} h = 9.8 units
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Priya Dharshini ? 7 years, 9 months ago
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