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.
Given: TP and TQ are two tangents of a circle with centre O and P and Q are points of contact
Now by theorem, "The lengths of a tangents drawn from an external point to a circle are equal".
So, TPQ is an isoceles triangle.
Also by theorem "The tangents at any point of a circle is perpendicular to the radius through the point of contact" .
Let OC be the tower of height H m. Suppose AB be the building of height 7 m. The angle of elevation of top of tower is and an angle of depression from the its foot is .
Suppose BC = x m and OD = h m.
∴ Height of the tower OC, H = h + 7 =
In a hospital used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of hospital whose length is 25 m and breadth is 20 m. If tank is filled completely then what will be the height of standing water used for irrigating the park. Write your views on recycling of water.
Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylinderical tank, Vc = =
Length of the park (l) = 25 m
Breadth of park (b) = 20 m
height of standing water in the park = h
Volume of water in the park = lbh =
Now water from the tank is used to irrigate the park. So,
Volume of cylinderical tank = Volume of water in the park
Through recycling of water, better use of the natural resource occurs without wastage. It helps in reducing and preventing pollution.
It thus helps in conserving water. This keeps the greenery alive in urban areas like in parks gardens etc.
In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where O and O' are centres of the circles. Find the area of shaded region.
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.
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Let the speed of the stream be x km/h.
It is given that the speed of a boat in still water is 15 km/h.
Speed of the boat upstream = Speed of the boat in still water − Speed of the stream = (15 − x) km/h
Speed of the boat downstream = Speed of the boat in still water + Speed of the stream = (15 + x) km/h
We know that
According to question,
Time taken for upstream journey + Time taken for the downstream journey = 4 h 30 min
Since speed can not be negative, therefore, x = 5.
Thus, the speed of the stream is 5 km/h.
If , prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.
Let a be the first term and d be the common difference of the AP.
It is given that the sum of first m terms is same as the sum of its first n terms.
Sum of first (m + n) terms
Thus, the sum of first (m + n) terms of the AP is zero.
Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15 m, then find the distance between the points.