The perimeter of a triangle is 8y^2-9y+4 and its two sides are 3y^2-5y and 4y^2+12.Find the third side.
Answers
Answered by
28
Step-by-step explanation:
Perimeter of the triangle = Sum of three sides
= 8y2 – 9y + 4
Sum 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
Answered by
5
Answer: HERE IS UR ANS
Step-by-step explanation: Perimeter of the triangle = Sum of three sides
= 8y2 – 9y + 4
Sum 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
Hence third side = y2 – 4y - 8
Similar questions