Question

Find the perimeter of each of the following shapes:
(i) A triangle of sides 3 cm, 4 cm and 5 cm
(ii) An equilateral triangle of side 9 cm
(iii) An isosceles triangle with equal sides 8 cm each and third side of 6 cm​

Answer 1

(i) Perimeter of Triangle ABC = Sum of all sides

= AB+BC+CA

= 4 + 3 + 5

= 12cm

Required perimeter is 12 cm

(ii) Here AB = BC = CA = a = 9cm

Perimeter of triangle ABC = 3a

= 3(9)

= 27cm.

Required perimeter is 27cm.

(iii) Here AB=AC=8CM.

BC= 6cm

Perimeter of triangle ABC = AB + BC + CA

= 8+6+8

= 22 cm.

Required Perimeter is 22cm.

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Answer 2

$\large{\underbrace{\sf{\orange{Required\:Answer:-}}}}$

(i) Find the area of A triangle of sides 3 cm, 4 cm and 5 cm.

◕  Length of the sides :-

◖a = 3 cm

◖b = 4 cm

◖c = 5 cm

To Find perimeter of the Δ :-

$:\implies\:\:\:\sf{P=a+b+c}$

$:\implies\:\:\:\sf{P=3+4+5}$

$:\implies\:\:\:\sf{P=12\:cm}$

So, Perimeter of the Δ = 12 cm

(ii) An equilateral triangle of side 9 cm.

◕  Sides of an equilateral triangle are equal.

So,

The three sides are :-

◖a = 9 cm

◖b = 9 cm

◖c = 9 cm

Perimeter of the Δ :-

$:\implies\:\:\:\sf{P=a+b+c}$

$:\implies\:\:\:\sf{P=9+9+9}$

$:\implies\:\:\:\sf{P=27\:cm}$

∴ Perimeter of the equilateral triangle = 27 cm

(iii) An isosceles triangle with equal sides 8 cm each and third side of 6 cm​.

Sides of the triangle :-

◖a = 8 cm

◖b = 6 cm

Perimeter of the triangle :-

$:\implies\:\:\:\sf{P=2(a+b)}$

$:\implies\:\:\:\sf{P=2\times 8 + 6}$

$:\implies\:\:\:\sf{P=16+ 6}$

$:\implies\:\:\:\sf{P=22\:cm}$

∴ Perimeter of the Δ = 22 cm