Question

The perimeter of an isosceles triangle is 32 cm and each of the equal sides is 5/6 times of the base. What is the area (in cm2) of the triangle ?

Answer 1

Let the base = x cm
Then each of side = 5x/6

X+ 5x/6=32
6x+5x+5x/6=32

16x/6=32
16x=32x6
X=12


Hence base = 12 cm


Eachside= 5/6*12

5*2=10
Now

Area= 1/2*12*root(10^2-12/2^2)^2

After solving this we get
48 cm2

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

Base of isosceles ∆ = p

Equal sides of ∆

$\tt{\rightarrow\dfrac{5}{6}p}$

Peimeter = 32

Hence,

$\tt{\rightarrow p+\dfrac{5}{6}p+\dfrac{5}{6}p=32}$

$\tt{\rightarrow\dfrac{16}{6}p=32}$

p = 12

So,

$\tt{\rightarrow\dfrac{5}{6}p}$

$\tt{\rightarrow\dfrac{5}{6}\times 12}$

= 10 cm

Sides = 12 , 10 and 10

Perimeter = 2s = 32

$\tt{\rightarrow s=\dfrac{32}{2}}$

= 16 cm

Area of ∆

$\tt{\rightarrow\sqrt{s(s-a)(s-b)(s-c)}}$

$\tt{\rightarrow\sqrt{16(16-12)(16-10)(16-10)}}$

$\tt{\rightarrow\sqrt{16\times 4\times 6\times 6}}$

= 4 × 2 × 6

= 48 cm²