Advertisements
Advertisements
प्रश्न
The perimeter of a triangle is 14a2 + 20a + 13. Two of its sides are 3a2 + 5a + 1 and a2 + 10a – 6. Find its third side.
Advertisements
उत्तर
perimeter of a triangle = 14a2 + 20a + 13
Sum of two sides
= 3a2 + 5a + 1 + a2 + 10a - 6
= 3a2 + a2 + 5a + 10a + 1 - 6
= 4a2 + 15a - 5
∴ Third side = (14a2 + 20a + 13) - (4a2 + 15a - 5)
= 14a2 + 20a + 13 - 4a2 - 15a + 5
= 14a2 - 4a2 + 20a - 15a + 13 + 5
= 10a2 + 5a + 18
APPEARS IN
संबंधित प्रश्न
Subtract: 4x2 - 5x2y + y2 from - 3y2 + 5xy2 - 7x2 - 9x2y
Find the excess of 4m2 + 4n2 + 4p2 over m2+ 3n2 – 5p2
Copy and complete the following multi-plication:
ax - b
× 2ax + 2b2
Evaluate: (a + 5)(3a - 2)(5a + 1)
Evaluate: (a + 1)(a2 - a + 1) and (a - 1)(a2 + a + 1)
Multiply: pq - pm and p2m
Multiply: x2 + 5yx - 3y2 and 2x2y
Simplify: `"k"/2 + "k"/3 + "2k"/5`
Simplify: `"y"/5 + "y" - "19y"/15`
Simplify: `"m"/5 - "m - 2"/3 + "m"`
