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प्रश्न
The perimeter of a triangle is 8y2 – 9y + 4 and its two sides are 3y2 – 5y and 4y2 + 12. Find its third side.
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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
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संबंधित प्रश्न
Evaluate :
−7x2 + 18x2 + 3x2 − 5x2
Evaluate :
7x − 9y + 3 − 3x − 5y + 8
Add : 5a + 3b, a − 2b, 3a + 5b
Add : 8x − 3y + 7z, −4x + 5y − 4z, −x − y − 2z
Subtract : 3a − 5b + c + 2d from 7a − 3b + c − 2d
Subtract : x3 − 4x − 1 from 3x3 − x2 + 6
Subtract : 6a + 3 from a3 − 3a2 + 4a + 1
Take away – 3x3 + 4x2 – 5x+ 6 from 3x3 – 4x2 + 5x – 6
The sides of a triangle are x2 – 3xy + 8, 4x2 + 5xy – 3 and 6 – 3x2 + 4xy. Find its perimeter.
How much bigger is 5x2y2 – 18xy2 – 10x2y than –5x2 + 6x2y – 7xy?
