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  • Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 23 maths.

Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 23 maths.

Answers (1)

6\sqrt{3}r

Hint: Using Pythagoras theorem solve the problem easily

Given: To prove the least perimeter of an isosceles triangle in which a circle of radius r

Can be inscribed in 6\sqrt{3}r

Solution:

Let ABC an isoceles triangle with AB = AC =x and BC =y and a circle with centre O and radius r inscribed in triangle ABC.

AO= 2r, OF =r

Using phythagoras theorem in triabgle ABF

AF^2+BF^2 =AB^2

(3r)^2+\left ( \frac{y}{2} \right )^2=x^2                            ..........(1)

Again for \Delta ADO,

(2r)^2=r^2+AD^2

3r^2=AD^2

AD=\sqrt{3}r

BD=BF, EC=FC

Now, AD + DB = x

\sqrt{3}r+ \left (\frac{y}{2} \right )=x

\frac{y}{2} =x =\sqrt{3}r                        ........(2)

\begin{aligned} &\therefore(3 r)^{2}+(x-\sqrt{3} r)^{2}=x^{2} \\ &9 r^{2}+x^{2}-2 \sqrt{3} r x+3 r^{2}=x^{2} \\ &12 r^{2}=2 \sqrt{3} r x \\ &6 r=\sqrt{3} x \\ &x=\frac{6 r}{\sqrt{3}} \end{aligned}

From (2)

\begin{aligned} &\frac{y}{2}=\frac{6 r}{\sqrt{3}}-\sqrt{3} r \\ &=\frac{(6 \sqrt{3}-3 \sqrt{3}) r}{3}=\frac{3 \sqrt{3}}{3} r \\ &y=2 \sqrt{3} r \end{aligned}

Perimeter = 2x +y

\begin{aligned} &=2\left(\frac{6 r}{\sqrt{3}}\right)+2 \sqrt{3} r \\ &=\frac{12 r}{\sqrt{3}}+2 \sqrt{3} r \\ &=\frac{12 r+6 r}{\sqrt{3}}=\frac{18 r}{\sqrt{3}} \\ &=\frac{18 \times \sqrt{3} r}{\sqrt{3} \times \sqrt{3}} \end{aligned}

Perimeter =6\sqrt{3}

 

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