Question

A field in the form of a quadrilateral $ABCD$ whose sides taken in order are respectively equal to $192dm,576dm,288dm$ and $480dm$ has the diagonal equal to $672dm$. Its area to the nearest square metre is $1011$.

• A. True
• B. False

Answer 1

The correct option is B False

Area of $\square ABCD=ar\Delta ABC+ar\Delta ACDar\Delta ABC:s=\sqrt{s(s-a)(s-b)(s-c)},s=\frac{a+b+c}{2}=\frac{672+192+576}{2}=720dmar\Delta ABC=\sqrt{720(720-672)(720-192)(720-576)}=\sqrt{720\times 48\times 28\times 144}=51260.76{dm}^{2}ar\Delta ACD=\sqrt{s(s-a)(s-b)(s-c)},s=\frac{a+b+c}{2}=\frac{672+288+480}{2}=720dmar\Delta ACD=\sqrt{720(720-672)(720-288)(720-480)}=\sqrt{720\times 48\times 432\times 240}=59859.68d{m}^2\therefore ar\square ABCD=51260.76+59859.68=111120.44d{m}^2=1111.20m^2$
Hence, given statement is false.