Question

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.
​

Answer 1

$\huge {\boxed {\underline {\pink{question}}}}$

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.

$\huge {\boxed {\underline {\blue{answer}}}}$

$\red☞$Area of ABCD=Area of ΔADC+Area of ΔABC

IN ΔABC

AREA of ΔABC

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

$\pink☞a = 3 \: \: b = 4 \: and \: c = 5$

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

$\orange ☞ \sqrt{6(6 - 3)(6 - 4)(6 - 5)}$

$\red☞ \sqrt{6 \times 3 \times 2 \times 1}$

$\green☞ \sqrt{6 \times 6}$

$\pink☞ = 6 {cm}^{2}$

IN ΔADC

AREA OF ΔADC

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

$\orange ☞ a = 5 \: \: \: \: b = 4 \: and \: c = 5$

$\green ☞s = \frac{5 + 4 + 5}{2} = \frac{14}{2} = 7cm$

$\red ☞ \sqrt{7(7 - 5)(7 - 4)(75 - 5)}$

$\green ☞ \sqrt{7 \times 2 \times 3 \times 2}$

$\pink ☞ \sqrt{2 \times 2 \times 3 \times 7}$

$\blue ☞2 \times \sqrt{21}$

$\orange ☞2 \times 4.58$

$\purple ☞9.16$

$AREA \: OF \: ABCD = (6 + 9.16) {cm}^{2}$

$\orange ☞15.16 {cm}^{2}$

SEE IN ATTACHMENT ALSO

HOPE IT'S HELPS YOU ❣️