Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9

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 - Mathematics

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Sum

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°)

AC2 = AB2 + BC2

= 152 + 202

= 225 + 400

AC2 = 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

  Is there an error in this question or solution?
Chapter 7: Mensuration - Exercise 7.1 [Page 253]

APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 9th Mathematics Answers Guide
Chapter 7 Mensuration
Exercise 7.1 | Q 8 | Page 253

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