The area of an isosceles triangle having base `2 cm` and the length of one of the equal sides `4 cm`, is

Options

A. `sqrt15 cm^(2)`
B. `sqrt((15)/(2))cm^(2)`
C. `2sqrt15cm^(2)`
D. `4sqrt15cm^(2)`

Answer

A

Explanation

(a) Let `BAC` be an isosceles triangle in which `AB=AC =4 cm` and `BC =2 cm.`
In right angied `triangleADB`,
`AB^(2)=AD^(2)+BD^(2)`
`rArr(4)^(2)=AD^(2)+1`
`rArr AD^(2)=16-1`
`rArr AD^(2)=15`
`thereforeAD=sqrt15cm`
`["taking positive square root because length is always positive"]`
`therefore" " "Area of"triangleABC=(1)/(2)xxBCxxAD" "[because"area of triangle"=(1)/(2)("base"xx"height")]`
`=(1)/(2)xx2xxsqrt15=sqrt15 cm^(2)` Alternate Method
We know that,
`"Area of an isosceles triangle"=(a)/(4)sqrt(4b^(2)-a^(2))`
where, b is the length of equal sides and a sides and a is the length of the base.
Here, the length of side be b=4 cm and a=2 cm
`therefore"Area of an isosceles triangle"=(2sqrt(4(4)^(2)-4))/(4)=sqrt(64-4)/(2)`
`=(sqrt60)/(2)=(2sqrt15)/(2)=sqrt15cm^(2)`