maths class 10 ka important questions for half yearly
Answers
Answer:
Q. Find the length of the tangent from a point M which is at a distance of 17 cm from the centre O of the circle of radius 8 cm
Sol.
Consider the figure:
Jagranjosh
Since, MN is the tangent of the circle,
∠MNO = 90⁰
⟹ MO2 = MN2 + ON2
⟹ 172 = MN2 + 82
⟹ 289 = MN2 + 64
⟹ 289 – 64 = MN2
⟹ MN2 = 225
⟹ MN = 15
Thus, the length of the tangent is 15 cm.
Q. If the common difference of an A.P. is 3, then find a20 – a15.
Sol.
Let the first term of the AP be a.
an = a(n − 1)d
a20 – a15 = [a + (20 – 1)d] – [a + (15 – 1)d]
= 19d – 14d
= 5dQ. Solve the following system of linear equations by substitution method:
2x – y = 2
x + 3y =15
Sol.
Here, 2x – y = 2
⟹ y = 2x – 2
⟹ x + 3y = 15
Substituting the value of y from (i) in (ii), we get
x + 6x – 6 = 15
⟹ 7x = 21
⟹ x = 3
From (i), y = 2 × 3 – 2 = 4
∴ x = 3 and y = 4
Q. From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semi-circular portion with BC as diameteris cut off. Find the area of remining paper (use π = 22/7)
Sol.
Given situation can be represented as the following diagram:
Jagranjosh
Length of paper, AB = l = 40cm
Width of paper, AD = b = 28cm
Area of paper = l × b = 40 × 28 = 1120 cm2
Diameter of semi-circle = 28cm
∴ Radius of semi-circle, r = 14cm
Thus, area of semi-circle = 1/2. πr2
= 1/2 × 22/7 × 14 × 14
= 308cm2
∴ Area of remaining paper = 1120 – 308 =
812 cm2
Answer:
The hypotenuse of a right angled triangle is 5m. If the smaller side is doubled and the lower side is tripled, the new hypotenuse is 6√5m. Find all the sides of triangle