Question

1 consists of a circle C with centre O. OA and OB are perpendicular to each other. The area of the triangle AOB is 32 square units.
If AD =BD, the area of the triangle ABD is

• A. $32\pi$ Square units
• B. $32(1+\sqrt{2})$ Square units
• C. $64\sqrt{2}$ square units
• D. 64 square units.

Answer 1

The correct option is B $32(1+\sqrt{2})$ Square units


OA =OB & $\angle AOB={90}^{\circ }$
$\Rightarrow \Delta OAB$ is a right angle isoceless triangle.
$\therefore {\perp }^{r}fromObisechAB$
(even otherwise since AB is achard ${\perp }^r$ from O will meet AB)
$\Delta O\times B$ is abo an isocels right angled triangle since $\angle XBO=45$ & $\angle O\times B={90}^{\circ }$
$\therefore OX=BX=\frac{1}{2}AB$ =ht of triangle AB =h (say)
Area of $\Delta OAB=\frac{1}{2},AB\times h=(B\times )h=h^2=32$
$\therefore h=\sqrt{32}cm$
$\Delta ABD$ is isoceles since AD=AB
${\perp }^r$ from D to AB will AB & hence is pass thw o
Let ration of circle =r =OA =OB =OD
$\therefore$ Height of $\Delta ABD=rth$.
Area of $\Delta ABD=\frac{1}{2}AB\times DX$
=(BX)(rth)=h(r+h)-(1)
Now from $\Delta O\times B$, it can be see that $r^2=d{h}^2\Rightarrow r=\sqrt{2}h$
$\therefore A\Delta ABD=h(\sqrt{2}h+h)=h^2(\sqrt{2}+1)=32(\sqrt{2}+5)sq.units$