Question

In $\vartriangle ABC$ if ${\sin ^2}A + {\sin ^2}B = {\sin ^2}C$ and $l\left( {AB} \right) = 10,$ then the maximum value of the area of $\vartriangle ABC$ is
(A)$50$
(B)$10\sqrt 2 $
(C)$25$
(D)$25\sqrt 2 $

Answer 1

Hint: In this question we are finding the maximum value of area of $\vartriangle ABC$. For finding, you should have to use the Sine rule. It is used when we are given either two angles and one side or two sides and a non-included angle, this law is very useful for solving these types of triangles.
i.e. $\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}$

Complete step by step solution:
First of all we have to draw diagram

${\sin ^2}A + {\sin ^2}B = {\sin ^2}C$
Use sine rule,
${a^2} + {b^2} = {c^2}$
Area of $\vartriangle ABC$$ = \dfrac{1}{2}ab$ -- $1$
From Sine rule
$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}$$ = k$
$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{{10}}{1}$
$\dfrac{a}{{\sin A}} = 10$ And $\dfrac{b}{{\sin B}} = 10$
$a = 10\sin A$ And $b = 10\sin B$
Now, use the equation$1$,
Area of $\vartriangle ABC$$ = \dfrac{1}{2}ab$
                           $ = \dfrac{1}{2} \times \left( {10\sin A} \right) \times \left( {10\sin B} \right)$
Area of $\vartriangle ABC$$ = $$50\sin A\sin B$
But the maximum value of $\sin A\sin B = \dfrac{1}{2}$
$ \Rightarrow $Maximum value of area of $\vartriangle ABC$$ = \dfrac{1}{2} \times 50$
                                                                    $ = 25$
So the correct option is (C).

Note: In this question students can face problems in using the Sine rule so be careful while solving the solution. One should know the properties of Sine rule. They should have a knowledge of values of different angles for different trigonometric functions. With the help of Sine rule we can solve these types of questions in an easier way.