CBSE [Delhi]_X_Mathematics_2015_Set III
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Q1
In Fig. 1, PA and PB are tangents to the circle with centre O such that ∠APB=50^{o}. Write the measure of∠OAB.
Marks:1Answer:
Join OC.

Q2Marks:1
Answer:

Q3
The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x:y.
Marks:1Answer:

Q4
A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a constant.
Marks:1Answer:
Number of English alphabet = 26
Number of constants = 21Probability of choosing a constant = 21/26

Q5
From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ.
Marks:2Answer:
Given: TP and TQ are tangents to circle and OT intersects chord PQ at R.
To Prove: OT bisects PQ at right angle.

Q6
Find the middle term of the A.P.
6, 13, 20, …, 216.Marks:2Answer:

Q7
In Fig. 2, AB is the diameter of a circle with centre O and AT is a tangent. If .
Marks:2Answer:

Q8
If A(5, 2), B(2,–2) and C(–2, t) are the vertices of a right angled triangle with angle B = 90°, then find the value of t.
Marks:2Answer:
In right angled triangle, ABC,
By Pythagoras Theorem,
AC^{2} = AB^{2} + BC^{2}
(5 + 2)^{2} +(2 – t)^{2} =(5 – 2)^{2} +(2 +2)^{2} +(2 + 2)^{2}
+(–2 – t)^{2}7^{2} + 4 – 4t + t^{2} = 9 + 16 + 16 + 4 + 4t + t^{2}
53 – 4t = 45 + 4t
53 – 45 = 8t
t = 8/8 = 1 
Q9Marks:2
Answer:

Q10
Solve the following quadratic equation for x:
9x^{2} – 6b^{2}x – (a^{4} – b^{4}) = 0
Marks:2Answer: