Question

the perimeter of a triangle is 8+13a+7a2 nd two of its side are 2a2+ 3a+2 nd 3a2-4a-1 find the third side of the triangle

Answer 1

Let X , Y, Z are three sides of triangle & P be the perimeter of triangle
where X=2a^2+3a+2
Y=3a^2-4a-1 Z=?
P=7a^2+13a+8
Now
P= X+Y+Z
7a^2+13a+8=2a^2+3a+2+3a^2-4a-1 +Z
7a^2+13a+8=5a^2-a+1+1
Z=7a^2-5a^2+13a+a+8-1
Z=2a^2+14a+7
Hence , the third side of triangle =2a^2+14a+7.
Hope it helps

Answer 2

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