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प्रश्न
Each of the equal sides of an isosceles triangle measure 2 cm more than its height, and the base of the triangle measure 12 cm. Find the area of the triangle.
उत्तर
Let the height of the triangle be h cm.
Each of the equal sides measures `a=(h+2)cm and b=12 cm(base)`
Now, Area of the triangle = Area of the isosceles triangle
=`1/2xxbasexxheight=1/4xxbsqrt(4a^2-b^2)`
⇒ `1/2xx12xxh=1/4xx12xxsqrt(4(h+2)^2-144)`
⇒`6h=3sqrt(4h^2+16h-144)`
⇒`2h=sqrt(4h^2+16h+16-144)`
On squaring both the sides, we get:
⇒`4h^2=4h^2+16h+16-144`
⇒`16h-128=0`
⇒`h=8`
Area of the triangle=`1/2xxbxxh`
=`1/2xx12xx8`
=`48cm^2`
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