Question

find the area of a quadrilateral ABCD in which AB = 3 cm BC =4 cm CD = 4cm DA= 5 cm and AC= 5 cm by heron's formula

Answer 1

$\huge \bf{ \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}}$

Let consider a quadrilateral ABCD

In ∆ABC;

AB = a = 3 cm, BC = b = 4cm, AC = c = 5 cm

Let a, b and c are the sides of triangle and s is

$<b>the semi-perimeter, then its area is given by:</b>$

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

where ;

$\implies\:S \: \: = \: \dfrac{a +b + c }{2}$

$\implies\:S \: \: = \: \: \dfrac{3 + 4 + 5}{2} \: \: = \: 6$

$\implies\:A_1 \: \: = \: \: \sqrt{6(6 - 3)(6 - 4)(6 - 5)} \\ \\ A_1 \: \: = \: \: \sqrt{6 \times 3 \times 2 \times 1 } \\ \\ A_1 \: \: = \: \: 6$

In ∆ADC;

DA = a = 5 cm, CD = 4 = 4cm, AC = c = 5 cm

Let a, b and c are the sides of triangle and s is

$<b>the semi-perimeter, then its area is given by:</b>$

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

$S \: \: = \: \: \dfrac{a + b + c}{2} \: \: = \: \: \dfrac{5 + 4 + 5}{2} = 7$

$A_2 \: \: = \: \: \sqrt{7(7 - 5)(7 - 4)(7 - 5)} \\ \\ A_2 \: \: = \: \sqrt{7(2)(3)(2)}$

$A_2 \: \: = \: 9.16 \: cm$

area of quadrilateral ABCD = $\sf\:A_1+A_2$

$\bf {\large {\boxed {area \: = \: 15.16 \: cm}}}$