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

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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 imes 3 imes 2 imes 1 } \: {cm}^{2} \\ = \sqrt{6 imes 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 imes 2 imes 3 imes 2} \\ = \sqrt{2 imes 2 imes 7 imes 3} \\ = 2\sqrt{21} \\ = 2 imes 4.58 = 9.16 {cm}^{2}
::::::::::::::::::::::: ◆

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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