Question

The perimeter of a triangle is 8y2 - 9y + 4 and
its two sides are 3y2 - 5y and 4y2 + 12.
Find its third side.​

Answer 1

Answer:

Perimeter of the triangle = Sum of three sides

= 8y2 – 9y + 4

= 8y2 – 9y + 4Sum of two sides = 3y2 – 5y + 4y2 + 12

= 8y2 – 9y + 4Sum of two sides = 3y2 – 5y + 4y2 + 12= 7y2 – 5y + 12

= 8y2 – 9y + 4Sum of two sides = 3y2 – 5y + 4y2 + 12= 7y2 – 5y + 12∴ (8y2 – 9y + 4) – (7y2 – 5y + 12)

= 8y2 – 9y + 4Sum of two sides = 3y2 – 5y + 4y2 + 12= 7y2 – 5y + 12∴ (8y2 – 9y + 4) – (7y2 – 5y + 12)= 8y2 – 9y + 4 – 7y2 + 5y – 12

= 8y2 – 9y + 4Sum of two sides = 3y2 – 5y + 4y2 + 12= 7y2 – 5y + 12∴ (8y2 – 9y + 4) – (7y2 – 5y + 12)= 8y2 – 9y + 4 – 7y2 + 5y – 12= y2 – 4y – 8

= 8y2 – 9y + 4Sum of two sides = 3y2 – 5y + 4y2 + 12= 7y2 – 5y + 12∴ (8y2 – 9y + 4) – (7y2 – 5y + 12)= 8y2 – 9y + 4 – 7y2 + 5y – 12= y2 – 4y – 8Hence third side = y2 – 4y - 8