Question

Find the area of a quadrilateral ABCD whose sides in metres are 9, 40, 28 and 15 respectively and the angle between first two sides is a right angle.

Solution

For ΔABC Area of right triangle ABC

equals 1 half cross times Base cross times Height equals 1 half cross times 9 cross times 40 equals 180 space straight m squared

For ΔACD
a = 28 m b = 41 m c = 15 m
$therefore space space space straight s equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction equals fraction numerator 28 plus 41 plus 15 over denominator 2 end fraction equals 84 over 2 equals 42 space straight m therefore space space Area space of space increment ACD equals square root of straight s left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight b right parenthesis left parenthesis straight s minus straight c right parenthesis end root space space space space space space space space space space space space space equals square root of 42 left parenthesis 42 minus 28 right parenthesis left parenthesis 42 minus 41 right parenthesis left parenthesis 42 minus 15 right parenthesis end root space space space space space space space space space space space space equals square root of 42 left parenthesis 14 right parenthesis left parenthesis 1 right parenthesis left parenthesis 27 right parenthesis end root space space space space space space space space space space space space equals space 14 cross times 3 cross times 3 equals 126 space straight m squared$

Area of the quadrilateral ABCD
= Area of ΔABC + Area of ΔACD
= 180 m²+ 126 m² = 306 m².

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Statistics

Find the area of a quadrilateral ABCD whose sides in metres are 9, 40, 28 and 15 respectively and the angle between first two sides is a right angle.

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