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.
1) Draw a line segment AB = 8 cm.
2) Draw a ray AX making an acute angle with AB.
3) Draw a ray BY parallel to AX by making an acute angle .
4) Mark four points on AX and five points on BY in such a way that .
6) Let this line intersect AB at a point P.
Thus, P is the point dividing the line segment AB internally in the ratio of 4 : 5.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.
The th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.
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" .
Show that ΔABC, where A(–2, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 2) are similar triangles.
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.
The ratio of the sums of first m and first n terms of an A. P. is m2 : n2. Show that the ratio of its mth and nth terms is (2m−1):(2n−1).
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.
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
So, area of ΔOPQ
Area of minor segment PRQ
= Area of sector OPRQ − Area of ΔOPQ
= 52.33 − 43.30
= 9.03 cm2
Area of major segment PSQ
= Area of circle − Area of minor segment PRQ
The angle of elevation of a cloud from a point 60 m above the surface of the water of a lake is 30° and the angle of depression of its shadow in water of lake is 60°. Find the height of the cloud from the surface of water.
Side of square = 28 cm and radius of each circle = cm
Area of the shaded region
= Area of the square + Area of the two circles − Area of the two quadrants
Therefore, the area of the shaded region is 1708 cm2.
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.
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