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A park is in the shape of a quadrilateral. The sides of the park are 15 m, 20 m, 26 m and 17 m and the angle between the first two sides is a right angle. Find the area of the park

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#### Solution

In the right angle triangle ABC ...(Given B = 90°)

AC^{2} = AB^{2} + BC^{2}

= 15^{2} + 20^{2}

= 225 + 400

AC^{2} = 625

AC = `sqrt(225)`

= 25 cm

Area of the right ΔABC = `1/2 xx "AB" xx "BC"`

= `1/2 xx 15 xx 20 "sq.m"`

= 150 sq.m

In the triangle ACD

a = 25 m b = 17 m, c = 26 m

s = `("a" + "b" + "c")/2`

= `(25 + 17 + 26)/2 "cm"`

= `62/2`

= 34 m

s – a = 34 – 25 = 9 m

s – b = 34 – 17 = 17 m

s – c = 34 – 26 = 8 m

Area of the triangle ACD

= `sqrt("s"("s" - "a")("s" - "b")("s" - "c"))`

= `sqrt(34(9)(17)(8))`

= `sqrt(2 xx 17 xx 3 xx 3 xx 17 xx 2^3)`

= `sqrt(2^4 xx 3^2 xx 17^2)`

= 4 × 3 × 17

= 204 sq.m

Area of the quadrilateral = Area of the ΔABC + Area of the ΔACD

= (150 + 204) sq.m

= 354 sq.m

Area of the quadrilateral = 354 sq.m

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