Advertisements
Advertisements
प्रश्न
The perimeter of a triangle is 8y2 – 9y + 4 and its two sides are 3y2 – 5y and 4y2 + 12. Find its third side.
उत्तर
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
APPEARS IN
संबंधित प्रश्न
Evaluate :
−7x2 + 18x2 + 3x2 − 5x2
Evaluate :
7x − 9y + 3 − 3x − 5y + 8
Add : 3b − 7c + 10, 5c − 2b − 15, 15 + 12c + b
Add : a − 3b + 3; 2a + 5 − 3c; 6c − 15 + 6b
Subtract : −2x2y + 3xy2 from 8x2y
Subtract : cab − 4cad − cbd from 3abc + 5bcd − cda
What must be added to x4 – x3 + x2 + x + 3 to obtain x4 + x2 – 1 ?
The sides of a triangle are x2 – 3xy + 8, 4x2 + 5xy – 3 and 6 – 3x2 + 4xy. Find its perimeter.
What must be subtracted from 19x4 + 2x3 + 30x – 37 to get 8x4 + 22x3 – 7x – 60 ?
How much bigger is 5x2y2 – 18xy2 – 10x2y than –5x2 + 6x2y – 7xy?
