Question

perimeter of a triangle is 49cm.one side is 7cm longer than another side and 5cm shorter than the third side. find the sides.

Answer 1

Answer:

let sides of the triangle be 'x' , 'y' , 'z'

according to the problem

x= 7+y => y = x-7

x= z-5 => z= x+5

perimeter = x+y+z

x+x-7+x+5 = 49

3x=51

x= 17cm

y= 10cm

z=22cm

Answer 2

Answer:

Given :-

• Perimeter of a triangle is 49 cm.
• One side is 7 cm longer than another side and 5 cm shorter than the third side.

To Find :-

• What is the sides.

Solution :-

Let,

❒ First side of a triangle = x cm

❒ Second side of a triangle = (x - 7) cm

❒ Third side of a triangle = (x + 5) cm

According to the question,

➟ x + (x - 7) + (x + 5) = 49

➟ x + x - 7 + x + 5 = 49

➟ x + x + x - 7 + 5 = 49

➟ 3x - 2 = 49

➟ 3x = 49 + 2

➟ 3x = 51

➟ x = 51/3

➠ x = 17 cm

Hence, the required sides of a triangle are :

❒ First side of a triangle :

↦ First side of a triangle = x cm

➲ First side of a triangle = 17 cm

❒ Second side of a triangle :

↦ Second side of a triangle = (x - 7) cm

↦ Second side of a triangle = (17 - 7) cm

➲ Second side of a triangle = 10 cm

❒ Third side of a triangle :

↦ Third side of a triangle = (x + 5) cm

↦ Third side of a triangle = (17 + 5) cm

➲ Third side of a triangle = 22 cm

∴ The sides of a triangle is 17 cm, 10 cm and 22 cm respectively.

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VERIFICATION :-

⇒ x + (x - 7) + (x + 5) = 49

By putting x = 17 we get,

⇒ 17 + (17 - 7) + (17 + 5) = 49

⇒ 17 + 10 + 22 = 49

➤ 49 = 49

Hence, Verified.