The base of an isosceles triangle is 4/3cm. The perimeter of the triangle is 62/15 cm. What is the length of either of the remaining equal sides?
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perimeter= sum of sides
62/15=4/3+2equal sides
62/15-4/3=2equal sides
now to subtract we should make the denominator common
lets make the denominator of 4/3 as 15 so that it becomes common
so to change the denominator 3 into 15 we should multiply 3 by 5
so as we are multiplying 3 we should multiply 4 also with 5(rule)
so it becomes4×5/3×5=20/15
62/15-20/15= 2 equal sides
42/15=2x
cross multiply
42=30x
x=42/30=7/5
perimeter= sum of sides
62/15=4/3+2equal sides
62/15-4/3=2equal sides
now to subtract we should make the denominator common
lets make the denominator of 4/3 as 15 so that it becomes common
so to change the denominator 3 into 15 we should multiply 3 by 5
so as we are multiplying 3 we should multiply 4 also with 5(rule)
so it becomes4×5/3×5=20/15
62/15-20/15= 2 equal sides
42/15=2x
cross multiply
42=30x
x=42/30=7/5
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base of triangle=4/3cm
perimeter of triangle=62/15
let the length of the side be x.therefore
perimeter of isosceles triangle=2*side + base
62/15=2x+4/3
62/15-4/3=2x
62/15-4*5/3*5=2x
62-20/15=2x
42/15=2x
x=42/30
x=7/5
hope it helps you: -D
perimeter of triangle=62/15
let the length of the side be x.therefore
perimeter of isosceles triangle=2*side + base
62/15=2x+4/3
62/15-4/3=2x
62/15-4*5/3*5=2x
62-20/15=2x
42/15=2x
x=42/30
x=7/5
hope it helps you: -D
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