Question

Find the area of an isosceles triangle each of whose equal sides measures 9 cm and whose base is 12 cm​

Answer 1

Answer:

Step-by-step explanation:

The three sides of the isosceles triangle are 9cm, 9cm , 12cm.

 

   

                      $\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A }\put(0.5,-0.3){ \bf C }\put(5.2,-0.3){ \bf B }\end{picture}$          

AC = AB= 9cm and CB = 12cm.

We can find the area by the usage of Heron's formula. For that first we need to find the value of semi-perimeter of the triangle.

$Semi-Perimeter = \dfrac{a+b+c}{2}$

Here,

a = b = 9cm and c = 12cm

$Semi- Perimeter = \dfrac{9+9+12}{2} = \dfrac{30}{2} = 15$

Heron's Formula

$Area = \sqrt{s(s - a)(s - b)(s - c)}$

Here,

     s = 15cm

         AND

a = b = 9cm and c = 12cm

Area Of the Triangle

$Area = \sqrt{15(15-9)(15-9)(15-12)}$

$Area = \sqrt{15 * 6*6*3}$

$Area = \sqrt{1620}$

Area = 40.25 cm² ( Approximately)