Question

The area of quadrilateral KITE in which $KI=3cm,IT=4cm,TE=4cm,KE=5cmand\angle KIT={90}^{\circ }$ is

$Use\sqrt{21}=4.58$

• A.
  $16.16c{m}^2$
• B.
  $15.16c{m}^2$
• C.
  $16.15c{m}^2$
• D.
  $15.15c{m}^2$

Answer 1

The correct option is B

$15.16c{m}^2$
Join K to T.



In $\Delta KIT,$ by Pythagoras theorem,

$(KT{)}^2=(KI{)}^2+(IT{)}^2\Rightarrow (KT{)}^2=(3{)}^2+(4{)}^2=9+16\Rightarrow (KT{)}^2=(5{)}^2\Rightarrow KT=5cm\therefore Areaof\Delta KIT=\frac{1}{2}\times KI\times IT=\frac{1}{2}\times 3\times 4=6c{m}^{2}In\Delta KET,semi-perimeter,\left(s\right)=\frac{KE+ET+KT}{2}=\frac{5+4+5}{2}=7cm\therefore Areaof\Delta KET=\sqrt{s(s-KE)(s-ET)(s-KT)}=\sqrt{7\times 2\times 3\times 2}=2\sqrt{21}2\times 4.58=9.16c{m}^2\therefore \text{Area of KITE}=Ar\left(\Delta KIT\right)+Ar\left(\Delta KET\right)=6+9.16=15.16c{m}^2$

Hence, the correct answer is option (b).