Question

2. Two equal sides of an isoscles triangle are each 2 cm more than thrice the third side. If
the perimeter of triangle is 67 cm, find the lengths of its sides.
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Answer 1

$\bf \: let \: one\: side \: be \: \red{x}$

$\bf \: let \: two \: side \: be \: \: \red{3x + 2}$

$\bf \: perimeter \: = \: \red{67cm}$

$\bf \: from \: question \: we \: get$

$\sf \: 3x + 2 + x + 3x + 2 = 67$

$\sf \implies \: 7x + 4 = 67$

$\sf \implies7x = 67 - 4$

$\sf \implies7x = 63$

$\sf \implies \: x = \frac{ \cancel{63}}{ \cancel{7}}$

$\sf \boxed{ \implies \sf x = 9}$

$\bf \: one \: side \: \blue{9cm}$

$\bf \: other \: two \: sides \: \: \red{3x + 2}$

$\sf \implies\blue{3x + 2}$

$\sf \implies \blue{3 \times 9 + 2}$

$\sf \implies \blue{27 + 2}$

$\boxed{ \sf \implies \blue{29cm}}$

Answer 2

ANSWER:

Given:

• The equal sides of an isosceles triangle are 2cm more than thrice thenthird side
• Perimeter = 67cm

To Find:

• Length of all sides

Solution:

$\text{Let the third side be x.}\\\\\text{So,}\\\\:\longrightarrow\text{Equal Sides}=3x + 2\\\\\text{We are given that,}\\\\:\longrightarrow Perimeter=67cm\\\\\text{We know that, perimeter is sum of all sides. So,}\\\\:\implies Perimeter=x+(3x+2)+(3x+2)\\\\:\implies 67=x+3x+2+3x+2\\\\:\implies67=7x+4\\\\:\implies7x=67-4\\\\:\implies7x=63\\\\:\implies x=\dfrac{63\!\!\!\!\!/^{\:\:\:9}}{7\!\!\!/}\\\\:\implies x=9$

$\text{So, length of each side is:}\\\\:\implies\sf Equal\:Sides=3x+2=3(9)+2==27+2=29cm\\\\:\implies\sf Third\:Side=x=9cm\\\\\text{\bf{Hence, the length of sides is 29cm, 29cm and 9cm respectively.}}$