General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into four sections – A, B, C and D. (iii) Section A contains 4 questions of 1 mark each, Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each. (iv) Use of calculated is not permitted.
Let PA and PB be the two tangents drawn to the circle with centre O and radius a such that
In ∆OPB and ∆OPA
OB = OA = a (Radii of the circle)
(Tangents are perpendicular to radius at the point of contact)
BP = PA (Lengths of tangents drawn from an external point to the circle are equal)
So, ∆OPB ≌ ∆OPA (SAS Congruence Axiom)
Thus, the length of OP is 2a.
A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, –5) is the mid-point of PQ, then find the coordinates of P and Q.
ABCD is a quadrilateral. Suppose a circle touches the sides AB, BC, CD and DA of the quadrilateral ABCD at P, Q, R and S, respectively.
We know that the length of tangents drawn from an external point to a circle are equal.
DR = DS .....(1)
CR = CQ .....(2)
BP = BQ .....(3)
AP = AS .....(4)
Adding (1), (2), (3) and (4), we get
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
Or, AB + CD = BC + DA
Which term of the A.P. 8, 14, 20, 26, ... will be 72 more than its 41st term?
Let the length of the pipe be h cm.
Volume of the solid iron cuboid = 4·4 m × 2·6 m × 1·0 m =
Internal radius of the pipe, r = 30 cm
External radius of the pipe, R = 30 + 5 = 35 cm
Volume of iron in the pipe =
Volume of iron in the pipe = Volume solid iron cuboid
Hence, the length of the pipe is 112 m.
In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm. If ∠AOB = 60°, find the area of the shaded region.
Width of the canal = 5.4 m
Depth of the canal = 1.8 m
Height of the standing water needed for irrigation = 10 cm = 0.1 m
Speed of the flowing water = 25 km/h = m/min
Volume of water flowing out of the canal in 1 min
= Area of opening of canal ×
= 4050 m3
∴ Volume of water flowing out of the canal in 40 min = 40 × 4050 m3 = 162000 m3
Thus, the area irrigated in 40 minutes is 162 hectare.
In what ratio does the point divide the line segment joining the points P(2, –2) and Q(3, 7)? Also find the value of y.
Let the point P divide the line PQ in the ratio k : 1.
Then, by the section formula:
On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.
Let AB be the tower of height h m. Suppose the angle of elevation of the top of the tower from point C on the ground be . Then the angle of elevation of the top of the tower from point D is .
Here, BC = 4 m and BD = 16 m.
Multiplying (1) and (2), we get
Thus, the height of the tower is 8 m.
A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.
In the given AP, let the first term be a and the common difference be d.
Then, Tn = a + (n − 1)d
Now, we have:
T10 = a + (10 − 1)d
⇒ a + 9d = 52 ...(1)
T13 = a + (13 − 1)d = a + 12d ...(2)
T17 = a + (17 − 1)d = a + 16d ...(3)
But, it is given that T17 = 20 + T13
i.e., a + 16d = 20 + a + 12d
⇒ 4d = 20
⇒ d = 5
On substituting d = 5 in (1), we get: a + 9 ⨯ 5 = 52
⇒ a = 7
Thus, a = 7 and d = 5
∴ The terms of the AP are 7, 12, 17, 22,...
If the roots of the equation (c2 – ab) x2 – 2 (a2 – bc) x + b2 – ac = 0 in x are equal, then show that either a = 0 or a3 + b3 + c3 = 3abc.
Steps of construction: Step 1. Draw line BC = 7 cm. Step 2. At B, construct and at C, construct Step 3. The point of intersection of BX and CY gives A. Thus, ∆ABC is obtained. Step 4. Draw any ray BZ making an acute angle with BC on the side opposite to the vertex A. Step 5. Locate four points B1, B2, B3 and B4 on BZ such that BB1 = B1B2 = B2B3 = B3B4. Step 6. Join B4C and draw a line through B3 parallel to B4C to intersect BC at C'. Step 7. Draw a line through C′ parallel to the line CA to intersect BA at A′.
Here, ΔA′BC′ is the required triangle similar to the ΔABC.
In a rain-water harvesting system, the rain-water from a roof of 22 m × 20 m drains into a cylindrical tank having diameter of base 2 m and height 3·5 m. If the tank is full, find the rainfall in cm. Write your views on water conservation.
Total amount of rainfall = Volume of rain water collected in the cylindrical vessel.
Water conservation is very important for sustainable development. Different methods can be used for conservation of water. One such method is rain water harvesting which not only avoids wastage of water but also helps in meeting the water demand during summers.
Prove that the lengths of two tangents drawn from an external point to a circle are equal.
Given: TP and TQ are two tangent drawn from an external point T to the circle C(O, r). To prove: TP = TQ Construction: Join OT. Proof:
We know that a tangent to the circle is perpendicular to the radius through the point of contact.
∴ OPT = OQT = 90°
In ΔOPT and ΔOQT,
OT = OT (Common)
OP = OQ (Radius of the circle)
OPT = OQT (90°)
∴ ΔOPT ΔOQT (RHS congruence criterion)
⇒ TP = TQ (CPCT)
Hence, the lengths of the tangents drawn from an external point to a circle are equal.
In the given figure, XY and X'Y' are two parallel tangents to circle with centre O and another tangent AB with point of contact C, is intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°.
Let constant speed of the train be x km/hr.
Thus, time taken to travel 300 km = hours.
Now when the speed is increased then time is reduced by 2 hours.
Time taken to cover 300 km with speed x km/hr − Time taken to cover 300 km with increased speed = 2 hours
Since speed cannot be negative so, the original speed of the train = 25 km/hour.
A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time taken by the car now to reach the tower.
Suppose AB be the tower of height h meters. Let C be the initial position of the car and let after 12 minutes the car be at D. It is given that the angles of depression at C and D are 30º and 45º respectively.
Let the speed of the car be v meter per minute. Then,
CD = distance travelled by the car in 12 minutes
CD = 12v meters
Suppose the car takes t minutes to reach the tower AB from D. Then DA = vt meters.
Substituting the value of h from equation (i) in equation (ii), we get
In the given figure, ∆ ABC is a right-angled triangle in which ∠ A is 90°. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.