Question

$ABCD$ is a quadrilateral such that $AB=5cm,BC=4cm;CD=7cm,AD=6cm$ and diagonal $BD=5cm$, then the $A\left(\square ABCD\right)$ is $4(3+\sqrt{6})c{m}^2$.

• A. True
• B. False

Answer 1

The correct option is A True

$ar\square ABCD=ar\Delta ABD+ar\Delta BCDar\Delta ABD=\sqrt{s(s-a)(s-b)(s-c)}s=\frac{a+b+c}{2}=\frac{6+5+5}{2}=8cmar\Delta ABD=\sqrt{8(8-6)(8-5)(8-5)}ar\Delta ABD=\sqrt{8\times 2\times 3\times 3}=4\times 3=12{cm}^{2}ar\Delta BCD=\sqrt{s(s-a)(s-b)(s-c)}s=\frac{a+b+c}{2}=\frac{7+4+5}{2}=8cmar\Delta BCD=\sqrt{8(8-7)(8-4)(8-5)}=\sqrt{8\times 1\times 4\times 3}=\sqrt{4\times 2\times 4\times 3}=4\sqrt{6}{cm}^2\therefore ar\square ABCD=(12+4\sqrt{6}){cm}^2=4(3+\sqrt{6}){cm}^2$
Hence, statement is true.