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find the area of the quadrilateral abcd whose vertices are a(3 -1) b(9 -5) c(14 0) and d(9 19)

Posted by Gursagar Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Join A and C
Now, Area of quadrilateral ABCD = Area of {tex} \Delta{/tex}ABC + Area of {tex}\Delta{/tex}ACD
Area of {tex} \Delta A B C = \frac { 1 } { 2 } | 3 ( - 5 - 0 ) + 9 ( 0 + 1 ) + 14 ( - 1 + 5 ) |{/tex}
{tex}= \frac { 1 } { 2 } | - 15 + 9 + 56 |{/tex}
{tex}= \frac { 1 } { 2 } \times 50{/tex}
= 25 sq. units
Area of {tex}\Delta ACD = \frac{1}{2}|3(0 - 19) + 14(19 + 1) + 9( - 1 - 0)|{/tex}
{tex}= \frac { 1 } { 2 } | - 57 + 280 - 9 |{/tex}
{tex}= \frac { 1 } { 2 } \times 214{/tex}
= 107 sq. units
Area of quad. ABCD = Area of {tex}\Delta{/tex}ABC + Area of {tex}\Delta{/tex}ACD
= (25 + 107) sq. units
=132 sq. units

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