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.
What is the common difference of an A.P. in which a21 – a7 = 84?
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
Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.
Consider the A.P 63, 65, 67,...
Here, First term (a) = 63 and common difference (d) = 2
Consider another A.P 3, 10, 17,...
Here, First term (a') = 3 and common difference (d') = 7
According to question,
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.
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 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.
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.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
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.
Two different dice are thrown together. Find the probability that the numbers obtained have
Given: XYand X'Y' are two parallel tangents to the circle with centre O touching the circle at P and Q, respectively. AB is a tangent at the point C, which intersects XY at A and X'Y' at B.
To prove: AOB = 90°
Construction: Join OC.
In ΔOAP and ΔOAC,
OP = OC (Radii of the same circle)
AP = AC (Length of tangents drawn from an external point to a circle are equal)
AO = OA (Common side)
ΔOAP ≅ ΔOAC (SSS congruence criterion)
∴ AOP = COA (C.P.C.T) .....(1)
Similarly, ΔOBQ ≅ ΔOBC
∴ BOQ = COB .....(2)
POQ is a diameter of the circle. Hence, it is a straight line.
∴ AOP + COA + BOQ + COB = 180º
2COA + 2COB = 180º [From (1) and (2)]
⇒ COA + COB = 90º
⇒ AOB = 90°
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.
If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27), then find the ratio of their 9th terms.
Suppose B alone can finish the work in x days.
So, A alone can finish the work in (x − 6) days.
It is given that A and B together can finish the work in 4 days.
∴ Work finished by A and B working together in 1 day = .....(1)
Work finished by B in 1 day =
Work finished by A in 1 day =
∴ Work finished by A in 1 day work + Work finished by B in 1 day = .....(2)
From (1) and (2), we have
But x cannot be less than 6 because when x is less than 6, then the time taken by A to complete the work is negative, which is not possible.
Therefore, x = 12.
Thus, B alone can finish the work in 12 days.
From the top of a tower, 100, high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression 30° and 45°. Find the distance between the cars. [Take = 1.732]
BC is the diameter of the circle passing through the centre O.
∆ABC is a right angles triangle, right angled at A. (Angle subtended by the diameter on the circumference of the circle is 90)
In right ∆ABC,
BC2 = AB2 + AC2
= (7)2 + (24)2
= 49 + 576
∴ BC = 25 cm
COD + BOD = 180° (Linear pair angles)
⇒COD = 180° − 90° = 90°
Area of the shaded region
= Area of sector OCABDO − Area of ∆ABC
Hence, the area of shaded region is approximately 283.97 cm2.
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