CBSE [Delhi]_X_Mathematics_2015_Set II
To Access the full content, Please Purchase

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.

Q2
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

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:

Q4Marks:1
Answer:

Q5
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 
Q6
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.

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

Q8
Solve the following quadratic equation for x:
4x^{2} – 4a^{2}x + (a^{4} – b^{4}) = 0
Marks:2Answer:

Q9Marks:2
Answer:

Q10
Find the middle term of the A.P. 213, 205, 197, …, 37.
Marks:2Answer:
Since, l = a + (n – 1)d
Here, a = 213, d = 205 – 213 =– 8 and l = 37So, 37 = 213 + (n – 1)( – 8)
37 = 213 – 8n + 8
n = 23