Question

If $\angle B_{1}O{A}_1={60}^{\circ }$ & radius of biggest circle is r. According to figure trapezium $A_1{B}_1{D}_1{C}_1,C_1{D}_1{D}_2{C}_2,C_2{D}_2{D}_3{C}_3.........$ and so on are obtained. Sum of areas of all the trapezium is -

• A. $\frac{r^2}{2\sqrt{3}}$
• B. $\frac{9r^2}{2\sqrt{3}}$
• C. $\frac{9r^2}{\sqrt{3}}$
• D. $\frac{r^2}{9\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{9r^2}{\sqrt{3}}$

The amplitude of trapezium A , B , c , D =2r

Let O be the centre of inscribed circle of the trapezium

Let O, K be a radius of the circle , where k is a tangent point to OA, $\angle O,OK={30}^0$ , we get oo, =2 r

Let $O{O}_1\cap D{C}_1=\left\{M\right\}$

Thus , M is a tangent point between this circle are $D_1{C}_1$ gives that

$OM=O{O}_1-M{O}_1=2r-r=r$

Thus $D_1{C}_1=2M{C}_1=2r\tan 30=\frac{2r}{\sqrt{3}}$

And the altitude to $A_1{B}_1$ of $\Delta O{A}_1{B}_1=3r$

We obtain ,$A_1{B}_1=2.3ran{30}^0=\frac{6r}{\sqrt{3}}$

Thus , $S_{A_1{B}_1{D}_1{C}_1}=\frac{\left(2r/\sqrt{3}+6r/\sqrt{3}\right).3r}{2}$

$=\frac{8r^2}{\sqrt{3}}$

Now , let x be a radius of the inscribed circle of $D_1{C}_1{C}_2{D}_2$ let $O_2$ be centre of the circle thus, $\angle O_{1}OK={30}^0$

$S_{D_1{C}_1{D}_2{C}_2}=\frac{1}{9}{S}_{A_1{B}_1{C}_1{D}_1}$

$\frac{8r^2/\sqrt{3}}{1-49}=\frac{9r^2}{\sqrt{3}}$