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Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. - CBSE Class 9 - Mathematics

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Question

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution

For ΔABC,

AC2 = AB2 + BC2

(5)2 = (3)2 + (4)2

Therefore, ΔABC is a right-angled triangle, right-angled at point B.

Area of ΔABC`= 1/2xxABxxBC=1/2xx3xx4=6 cm^2`

For ΔADC,

Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm

s = 14/2 = 7 cm

By Heron’s formula,

`"Area of triangle "=sqrt(s(s-a)(s-b)(s-c))`

`"Area of "triangle ADC=[sqrt(7(7-5)(7-5)(7-4))]cm^2`

                           `=(sqrt(7xx2xx2xx3))cm^2`

                           `=2sqrt21 cm^2`

                            = (2 x 4.583) cm2

                            = 9.166 cm2

Area of ABCD = Area of ΔABC + Area of ΔACD

                    = (6 + 9.166) cm2

                    = 15.166 cm2

                    = 15.2 cm2 (approximately)

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Mathematics Class 9 (2018 to Current)
Chapter 12: Heron's Formula
Ex.12.20 | Q: 2 | Page no. 206

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Solution Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. Concept: Application of Heron’S Formula in Finding Areas of Quadrilaterals.
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