﻿ The points A (4, 7), B (p, 3) and C (7, 3) are the vertices a ri

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## CBSE Class 10 Mathematics Solved Question Paper 2015

1.

If the quadratic equation has two roots, then find the value of p.

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2.

In Given figure, a tower AB is 20 m high and BC, its shadow on the ground, is m long. Find the sun's altitude.

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3.

The different dice are tossed together. Find the probability that the product of the two number on the top of the dice is 6.

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4.

In the given figure, PQ is a chord of a circle with centre O and PT is tangent. If ∠QPT = 60o, find ∠ PRQ.

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5.

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120o, then prove that OR = PR + RQ.

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6.

In the given figure, a triangle ABC is drawn to circumscribe of radius 3 cm, such that the segments RD and DC are respectively of length 6 cm and 9 cm. If the area of ΔABC is 54 cm2, then find lengths of sides AB and AC.

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7.

Solve the following quadratic equation for x,

4x2 + 4bx - (a2 - b2) = 0

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8.

In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP,  Where Sn denotes the sum of its first n terms.

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# 9.The points A (4, 7), B (p, 3) and C (7, 3) are the vertices a right triangle, right-angled at B find the value of p.

Given ,
The vertices of a right triangle, such as,
A (4, 7), B (p, 3) and C (7, 3)

In right ΔABC, using Pythagoras theorem

(AB)2 + (BC)2 = (AC)2

⇒ [(3 - 7)2 + (p - 4)2] + [(3-3)2 +(7 - p)2] = [(3-7)2 + (7 - 4)2]

⇒ (p - 4)2 + (7 - p)2 = 9

⇒ p2 + 16 -8p + 49 + p2 - 14p = 9

⇒ 2p2 - 22p + 56 = 0

⇒ p- 11p + 28 = 0

⇒ p2 - 4p -7p + 28 =0

⇒ p(p - 4) -7(p - 4)=0

⇒ (p - 7 )(p - 4) = 0

⇒ p - 7 = 0 or p - 4 = 0

⇒ p = 7 or p = 4

Hence, the value of p is 4 or 7

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10.

Find the relation between x and y if the points A (x, y), B (-5, 7) and C (-4, 5) are collinear.

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