Question

Find the area of a quadrilateral whose sides are $AB=3$ $cm$, $BC=4$ $cm$, $CD=6$ $cm$ and $DA=5$ $cm$ and diagonal $AC=5$ $cm$.

• A. $16c{m}^2$
• B. $18c{m}^2$
• C. $20c{m}^2$
• D. $22c{m}^2$

Answer 1

The correct option is B $18c{m}^2$

Area $ABCD$=Area $ABC$+Area $ACD$

Area$ABC$ $=\sqrt{S(S-a)(S-b)(S-c)},S=\frac{a+b+c}{2}=\frac{3+4+5}{2}=6cm\sqrt{6(6-3)(6-4)(6-5)}=\sqrt{6\times 3\times 2\times 1}=6c{m}^2$

Area$ACD$ $=\sqrt{S(S-a)(S-b)(S-c)},S=\frac{a+b+c}{2}=\frac{5+5+6}{2}=8cm\sqrt{8(8-5)(8-5)(8-6)}=\sqrt{8\times 3\times 3\times 2}=12c{m}^2$

$\therefore$ Area $ABCD=6+12=18c{m}^2$