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JEE Main 2018
Properties of Triangle
Properties of Triangle
Easy

Question

The sides of a triangle are 3x+4y,3x + 4y, 4x+3y4x + 3y and 5x+5y5x + 5y where xx, y>0y>0 then the triangle is :

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Solution

Let a=3x+4y,b=4x+3y\,\,\,\,a = 3x + 4y,b = 4x + 3y and c=5x+5yc = 5x + 5y as x,y>0,c=5x+5y\,\,\,\,x,y > 0,c = 5x + 5y is the largest side \therefore CC is the largest angle. Now cosC=(3x+4y)2+(4x+3y)3(5x+5y)22(3x+4y)(4x+3y)\cos \,C = {{{{\left( {3x + 4y} \right)}^2} + {{\left( {4x + 3y} \right)}^3} - {{\left( {5x + 5y} \right)}^2}} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}} =2xy2(3x+4y)(4x+3y)<0 = {{ - 2xy} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}} < 0 \therefore CC is obtuse angle ΔABC \Rightarrow \Delta ABC is obtuse angled

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