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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.

  

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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.

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First, construct a diagram with the given parameter.

Now, apply Pythagorean theorem in ΔABC,

AC2 = AB2 + BC2

⇒ 52 = 32+42

⇒ 25 = 25

Thus, it can be concluded that ΔABC is a right angled at B.

So, area of ΔBCD = (1/2 × 3 × 4) = 6 cm2

The semi perimeter of ΔACD (s) = (perimeter/2)

= (5+5+4)/2 cm = 14/2 cm

= 7 m

Now, using Heron’s formula,

Area of ΔABD

= \(\sqrt{s(s - a)(s - b)(s - c)}\)

= \(\sqrt{7(7 - 5)(7 - 5)(7 - 4)}\)cm2

= \(\sqrt{7 \times 2 \times 2 \times 3}\)cm2

= 2√21 cm

= 9.17 cm2 (approximately)

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD

= 6 cm+ 9.17 cm2 

= 15.17 cm2

This post was modified 1 week ago by Samar shah
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