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CBSE Class 10 Mathematics Solved Question Paper 2012

Short Answer Type

11.

The volume of a hemisphere is 2425 $\frac{1}{2}$ cm³^. Find its curved surface area.

$[Use π = \frac{22}{7}]$

12.

Tangents PA and PB are drawn from an external point P to two concentric circle with centre O and radii 8 cm and 5 cm respectively, as shown in Fig., If AP = 15 cm, then find the length of BP.

Long Answer Type

13.

In fig., an isosceles triangle ABC, with AB =AB, circumscribes a circle. Prove that the point of contact P bisects the base BC.

In fig., the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

14.

In fig., OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O, then find the area of the shaded region. $[Use π = \frac{22}{7}]$

15.

Find the sum of all three digit natural numbers, which are multiples of 7.

16.

Find the values (s) of k so that the quadratic equation 3x² - 2kx + 12 = 0 has equal roots.

17.

A point P divides the line segment joining the points A(3,-5) and B(-4, 8) such that $\frac{AP}{PB} = \frac{K}{1}$ . If P lies on the line x + y = 0, then find the value of K.

18.

If the vertices of a triangle are (1, -3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value (s) of p.

19.
Prove that the parallelogram circumscribing a circle is a rhombus.

OR

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Let ABCD be a parallelogram such that its sides touching a circle with centre O.

We know that the tangents to a circle from an exterior point are equal in length.

$∴$ AP = AS .......[Tangnts from point A] .......(i)

BP = BQ .......[Tangents from point B] .......(ii)

CR = CQ .......[Tangents from point C] .......(iii)

and, DR = DS .......[Tangents from point D] ........(iv)

Adding (i), (ii), (iii), and (iv), we get

AP + BP + CR + DR = AS + BQ + CQ + DS

$\Rightarrow$ (AP + BP) + (CR + DR) = ( AS + DS) + (BQ + CQ)

$\Rightarrow$ AB + CD = AD + BC

$\Rightarrow$ 2AB = 2BC [ $∵$ ABCD is a parallelogram $∴$ AB = CD and BC = AD]

$\Rightarrow$ AB = BC

Thus, AB = BC = CD = AD

Hence, ABCD is a rhombus.

A circle with centre O touches the sides AB, BC, CD and DA of a quadrilateral

ABCD at the points P, Q, R and R respectively.

To prove: $∠ AOB + ∠ COD = 180 ° and, ∠ AOD + ∠ BOC = 180 °$

Construction: Join OP, OQ, OR and OS.

Proof: Since the two tangents drawn from a external point to a circle

subtend equal angles at the centre.

$∴ ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, ∠ 5 = ∠ 6 and ∠ 7 = ∠ 8 Now, ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360 ° [Sum of all the angles subtended at a point is 360 °] \Rightarrow 2 (∠ 2 + ∠ 3 + ∠ 6 + ∠ 7) = 360 ° and 2 (∠ 1 + ∠ 4 + ∠ 5 + ∠ 8) = 360 ° \Rightarrow (∠ 2 + ∠ 3) + (∠ 6 + ∠ 7) = 180 ° and (∠ 1 + ∠ 8) + (∠ 4 + ∠ 5) = 180 ° [∴ ∠ 2 + ∠ 3 = ∠ AOB, ∠ 6 + ∠ 7 = ∠ COD ∠ 1 + ∠ 8 = ∠ AOD and ∠ 4 + ∠ 5 = ∠ BOC] \Rightarrow ∠ AOB + ∠ COD = 180 ° and ∠ AOD + ∠ BOC = 180 °$

Hence proved.

20.

From a solid cylinder of height 7 cm and base diameter 12 cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid. $[Use π = \frac{22}{7}]$

OR

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, then find the radius and slant height of the heap.