Mathematics

CBSE Class 10

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1.

In Fig. , PQ is tangent at point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.

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3.

A ladder leaning against a wall makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.

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4.

A card is drawn at random from a well -shuffled fled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.

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5.

If -5 is a root of the quadratic equation 2x^{2} + px – 15 = 0 and the quadratic equation p(x^{2} + x)k = 0 has equal roots, find the value of k.

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6.

Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

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7.

In Fig.2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.

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8.

Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle.

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9.

The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

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In Fig, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP=2r, show that ∠ OTS = ∠ OST = 30°.

In the given figure,

Given

OP= 2r

∠OTP = 90° (radius drawn at the point of contact is perpendicular to the tangent)

Now ,

In ΔOTP,

Sin∠ OPT = OT/OP = 1/2 = Sin 30

⇒ ∠ OPT = 30

therefore,

∠ TOP=60

∴ ΔOTP is a 30

In ΔOTS,

OT = OS … (Radii of the same circle)

therefore,

ΔOTS is an isosceles triangle.

∴∠OTS = ∠OST … (Angles opposite to equal sides of an isosceles triangle are equal)

In ΔOTQ and ΔOSQ

OS = OT … (Radii of the same circle)

OQ = OQ ...(side common to both triangles)

∠OTQ = ∠OSQ … (angles opposite to equal sides of an isosceles triangle are

equal)

∴ ΔOTQ ≅ ΔOSQ … (By S.A.S)

∴ ∠TOQ = ∠SOQ = 60° … (C.A.C.T)

∴ ∠TOS = 120° … (∠TOS = ∠TOQ + ∠SOQ = 60° + 60° = 120°)

∴ ∠OTS + ∠OST = 180° – 120° = 60°

∴ ∠OTS = ∠OST = 60° ÷ 2 = 30°

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