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

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 - Mathematics

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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

Sum

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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Chapter 7: Mensuration - Exercise 7.1 [Page 253]

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Samacheer Kalvi Mathematics [English] Class 9 TN Board
Chapter 7 Mensuration
Exercise 7.1 | Q 7 | Page 253
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