Question

The sides of a triangle are $51,35,26$.The sides of another triangle are $a,b,41.$ If the two triangles have same perimeter and same area, then the ratio of their circumradii is

• A. $\frac{175}{201}$
• B. $\frac{221}{205}$
• C. $\frac{220}{203}$
• D. $\frac{175}{204}$

Answer 1

The correct option is B $\frac{221}{205}$

Let $PQR$ be a triangle whose sides are $51,35,26$(given)
Perimeter of $△PQR=\frac{51+35+26}{2}=56$
Let $ABC$ be a triangle whose sides are $a,b,41$(given)
Two triangles have same perimeter and area.
$\Rightarrow$Perimeter of $△ABC=$Perimeter of $△PQR$
$\Rightarrow \frac{a+b+41}{2}=56$
$\Rightarrow a+b=71$ or $b=71-a$
Area$△=\sqrt{s(s-a)(s-b)(s-c)}$
Area of $△PQR=$Area of $△ABC$
$\Rightarrow 56(56-51)(56-35)(56-26)=56(56-a)(56-b)(56-41)$
$\Rightarrow 5\times 21\times 30=(56-a)(56-71+a)(56-41)$ where $b=71-a$ shown above
$\Rightarrow 5\times 21\times 30=(56-a)(-15+a)(56-41)$
On simplifying we get
$\Rightarrow 210=(56-a)(-15+a)$
$\Rightarrow a^2-71a+1050=0$
Solving $a=50,21$
The sides are $50,21,$ and $41$(given)
The ratio of their circumradii is $\frac{51\times 35\times 26}{50\times 21\times 41}=\frac{221}{205}$