Question

The area of a quadrilateral ABCD in which AB=3cm, BC = 4 cm , CD = 4 cm , DA = 5 cm AC = 5 cm . Solve it by using herons formula​

Answer 1

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. ∴ ΔABC is right angled with ∠B = 90°. = 2 x 4.6 cm2 (approx.) = 9.2 cm2 (approx.)

Answer 2

$\underline{\underline{\huge{\blue{\tt{\textbf Answer :-}}}}}$

Area of Quadrilateral = ar of ∆ABC + ar of ∆ADC

• Area of ∆ABC

• heron's formula

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

{where, s is semi perimeter}

$s = \frac{a + b + c}{2} \\ \: = \frac{3 + 4 + 5}{2 } \\ = \frac{12}{2} \\ = 6$

area of ∆ABC

$= \sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} \\ = \sqrt{6 \times 3 \times 2 \times 1 } \: {cm}^{2} \\ = \sqrt{6 \times 6} \: {cm}^{2} \\ = 6 {cm}^{2}$

Area of∆ADC

$= \sqrt{s(s - a)(s - b)(s - c)} \\ \: s = \frac{a + b + c}{2} \\ \frac{5 + 4 + 5}{2} = \frac{14}{2} = 7cm \\ \\ ar \: of \: triangle \: \\ = \sqrt{7(7 - 5)(7 - 4)(7 - 5)} \\ = \sqrt{7 \times 2 \times 3 \times 2} \\ = \sqrt{2 \times 2 \times 7 \times 3} \\ = 2\sqrt{21} \\ = 2 \times 4.58 = 9.16 {cm}^{2}$

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Area of Quadrilateral = ar of ∆ABC + ar of ∆ADC

$= 6 + 9.16 \\ = 15.16 {cm}^{2} \\ = 15.2 {cm}^{2} \: (approx.)$

So the Area of Quadrilateral ABCD = 15.2cm .sq

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