Question

The base of an isosceles triangle is 4/5 cm The perimeter of the triangle is 68/13 what is the length of either of the remaining equal sides​

Answer 1

Given :

• Base of an Isosceles triangle = 4/5cm

• Perimeter of the triangle = 68/13cm

To Find :

• Length of remaining equal sides

Solution :

Base = BC = $\sf \dfrac{4}{5}cm$

Let, AB and AC be "x"

(Two sides of isosceles triangle are equal)

As we know that,

$\boxed{\sf{Perimeter = Sum\:of \:all \:sides \:of \:triangle}}$

$:⟹\sf \dfrac{68}{13} = AB + BC + AC$

$:⟹\sf \dfrac{68}{13} = x + \dfrac{4}{5}+x$

$:⟹\sf \dfrac{68}{13} = 2x + \dfrac{4}{5}$

$:⟹\sf 2x + \dfrac{4}{5} = \dfrac{68}{13}$

On transposing 4/5 to R.H.S we get,

$:⟹\sf 2x= \dfrac{68}{13} - \dfrac{4}{5}$

$:⟹\sf2x = \dfrac{340 - 52}{65}$

$:⟹\sf2x = \dfrac{288}{65}$

$:⟹\sf \: x = \dfrac{288}{65} \times \dfrac{ 1}{2}$

$:⟹\sf \: x = \dfrac{144}{65}$

$:⟹{\sf{\red{\ x = 2 \dfrac{14}{65}}}}$

∴ length of remaining 2 sides = $\sf 2 \dfrac{14}{65}$

Verification :

As we know that,

Perimeter of triangle = Sum of all sides

$:⟹\sf \dfrac{68}{13} = x +\dfrac{4}{5}+x$

By substituting the value of 'x' we get,

$:⟹{\sf {\dfrac{68}{13} = {\dfrac{144}{65} + \dfrac{4}{5}+{\dfrac{144}{65}}}}}$

$:⟹\sf \dfrac{68}{13} = \dfrac{340}{65}$

$:⟹{\bf{\purple{ \dfrac{68}{13} = \dfrac{68}{13}}}}$

Hence, Verified

Answer 2

5/7

base = 3/4

cm

Let the equal sides be x.

Perimeter=S

1

+S

2

+S

3

⇒

15

62

=x+x+

3

4

⇒

15

62

=2x+

3

4

⇒2x=

15

62

−

3

4

⇒2x=

15

62−20

⇒2x=

15

42

⇒x=

15

21

∴x=

5

7

Thus,two equal sides =x=

5

7

cm