I want to know the best answer of the question from Heron’s Formula chapter of class 9^{th} ncert math. The question from exercise 12.2 of math. Give me the easy way for solving this question of 2 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. Q.2

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First, construct a diagram with the given parameter.

Now, apply Pythagorean theorem in ΔABC,

AC

^{2}= AB^{2}+BC^{2 }⇒ 5

^{2}= 3^{2}+4^{2 }⇒ 25 = 25

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

So, area of ΔBCD = (½ ×3×4) = 6 cm

^{2}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

= 2√21 cm

^{2 }= 9.17 cm^{2}(approximately)Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD = 6 cm

^{2 }+9.17 cm^{2}= 15.17 cm^{2 }