In `DeltaABC`, if `sin^(2)A+sin^(2)B=sin^(2)C and l(AB)=10,` then the maximum value of the area of `DeltaABC` is

Options

A. 50
B. `10sqrt(2)`
C. `25`
D. `25sqrt(2)`

Answer

C

Explanation

Given , `sin^(2)A + sin^(2)B = sin^(2)C`
`rArr a^(2)+ b^(2) = c^(2)` (By sine rule)
Which is result of Pythagoras theorem.
`:.DeltaABC` is right angled triangle and right angled at C.
Now, area of `(DeltaACB) = 1/2 ab "…."(i)`

[ `:'` Area = `1/2 "Base" xx "height"`]
From sine rule's we have
`(a)/(sin A) = (b)/(sinB) = (c )/(sinC)`
`rArr a/(sinA) = (b)/(sinB) = (10)/(1)`
`rArr a = 10 sin A`
`b = 10 sin B`
On substituting `a = 10sin A and b = 10 sin B `in Eq. (i) , we get
Area of `Delta ACB = 1/2(10 sin A) (10 sin B)`
`= 50 sin A sin B"..."(ii)`
But maximum valueof `sin A sin B = 1/2"...."(iii)`
`:.` Maximum value of area of `Delta ACB`
`= 50 xx 1/2 = 25` [from Eqs. (ii) and (iii)]