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Question
Find the area of a quadrilateral ABCD whose sides are AB = 13 cm, BC = 12 cm, CD = 9 cm, AD = 14 cm and diagonal BD = 15 cm
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Solution
In the triangle ABD,
Let a = 15 cm, b = 14 cm, c = 13 cm
s = `("a" + "b" + "c")/2`
= `(15 + 14 + 13)/2 "cm"`
= `42/2`
= 21 cm
s – a = 21 – 15 = 6 cm
s – b = 21 – 14 = 7 cm
s – c = 21 – 13 = 8 cm
Area of ΔABD
= `sqrt("s"("s" - "a")("s" - "b")("s" - "c"))`
= `sqrt(21 xx 6 xx 7 xx 8)`
= `sqrt(3 xx 7 xx 2 xx 3 xx 7 xx 2^3)`
= `sqrt(2^4 xx 3^2 xx 7^2)`
= 22 × 3 × 73
= 84 cm2
In the ΔBCD,
Let a = 15 cm, b = 9 cm, c = 12 cm
s = `("a" + "b" + "c")/2`
= `(15 + 9 + 12)/2 "cm"`
= `36/2`
= 18 cm
s – a = 18 – 15 = 3 cm
s – b = 18 – 9 = 9 cm
s – c = 18 – 12 = 6 cm
Area of the ΔBCD
= `sqrt("s"("s" - "a")("s" - "b")("s" - "c"))`
= `sqrt(18 xx 3 xx 9 xx 6)`
= `sqrt(2 xx 3xx 3 xx 3 xx 3 xx 3 xx 2 xx 3)`
= `sqrt(2^2 xx 3^6)`
= 2 × 33
= 2 × 27 sq.cm
= 54 sq. cm
Area of the quadrilateral ABCD = Area of ΔABD + Area of ΔBCD
= (84 + 54) sq.cm
= 138 sq.cm
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