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.
Advertisements
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 :
3x2 + 5xy − 4y2 + x2 − 8xy − 5y2
Add : 13ab − 9cd − xy, 5xy, 15cd − 7ab, 6xy − 3cd
Subtract : −2x2y + 3xy2 from 8x2y
Subtract : cab − 4cad − cbd from 3abc + 5bcd − cda
Subtract : a2 + ab + b2 from 4a2 − 3ab + 2b2
Take away – 3x3 + 4x2 – 5x+ 6 from 3x3 – 4x2 + 5x – 6
Subtract the sum of 5y2 + y – 3 and y2 – 3y + 7 from 6y2 + y – 2.
What must be added to x4 – x3 + x2 + x + 3 to obtain x4 + x2 – 1 ?
How much more than 2x2 + 4xy + 2y2 is 5x2 + 10xy – y2 ?
