The length of each of the equal sides of an isosceles triangle is $4 c m$ longer than the base. If the perimeter of the triangle is $62 c m$ , find the length of the sides of the triangles.

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Solution

Let x be length of equal sides of isoscles triangle & y be length of base
Perimeter = 2x+y
2x+y=62
Given [x=y+4]
2(y+4)+y=62
3y=62-8
3y=54
[y=18 cm] and x= y+4=22 cm
$∴$ length of equal side is 22 cm & length of base is 18 cm.

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The length of each of the equal sides of an isosceles triangle is 4 cm longer than the base. If the perimeter of the triangle is 62 cm, find the length of the sides of the triangles.

Let x be length of equal sides of isoscles triangle & y be length of basePerimeter = 2x+y2x+y=62Given [x=y+4]2(y+4)+y=623y=62-83y=54[y=18 cm] and x= y+4=22 cm∴ length of equal side is 22 cm & length of base is 18 cm.

The length of each of the equal sides of an isosceles triangle is $4 c m$ longer than the base. If the perimeter of the triangle is $62 c m$ , find the length of the sides of the triangles.

Let x be length of equal sides of isoscles triangle & y be length of base
Perimeter = 2x+y
2x+y=62
Given [x=y+4]
2(y+4)+y=62
3y=62-8
3y=54
[y=18 cm] and x= y+4=22 cm
$∴$ length of equal side is 22 cm & length of base is 18 cm.