*Instructions:*

- Please check that this question paper consists of 11 pages.
- Code number given on the right hand side of the question paper should be written on the title page of the answer book by the candidate.
- Please check that this question paper consists of 30 questions.
**Please write down the serial number of the question before attempting it.**- 15 minutes times has been allotted to read this question paper. The question paper will be distributed at 10:15 am. From 10:15 am to 10:30 am, the students will read the question paper only and will not write any answer on the answer book during this period.

**SUMMATIVE ASSESSMENT – II**

**MATHEMATICS**

Time allowed: 3 hours Maximum Marks: 90

*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 consists of 4 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 11 questions of 4 marks each.*

*(iv) Use of calculator is not permitted.*

**SECTION – A**

*Question number 1 to 6 carry 1 mark each.*

Question 1: In Fig. 1, is a tangent at point to a circle with center . If is the diameter and , find .

Answer:

Construction: Join

Now in

(radius of the same circle)

Therefore ,

Also,

Therefore,

Question 2: For what value of will , and are the consecutive terms of an A.P.?

Answer:

If three terms , and are in A.P. then,

Since and are in A.P.

Question 3: A ladder, leaning against a wall, makes an angle of with the horizontal. If the foot of the ladder is m away from the wall, find the length of the ladder.

Answer:

Let be a ladder, is the wall and is the ground. The situation given in the question is shown in the figure.

Therefore

m. Therefore, the length of ladder is 5 m.

Question 4: A card is drawn at random from a well shuffled pack of playing cards. Find the probability of getting neither a Red card or the queen.

Answer:

There are red cards in a deck of cards

Also there are queens in a deck of cards, queens are black and are red.

Now, numbers of card which are neither red nor queen are

Therefore Probability of getting neither a red card nor a queen is

**Section – B**

*Question number 5 to 10 carry 2 mark each.*

Question 5: If is the root of the quadratic equation and the quadratic equation has equal roots, find the value of .

Answer:

If is root of

For to have equal roots,

(Determinant should be zero)

Question 6: Let and be points of trisection of the line segment joining the points and such that is nearer to . Find the coordinates of and .

Answer:

Since and be the points of trisection of the line segment joining the points and such that is nearer to . Therefore divides line segment in the rail and divides in

By section formula Coordinates of are

By section formula Coordinates of Q are

Question 7: In Fig. 2, a quadrilateral is drawn to circumscribe a circle, with center , in such a way that the sides , and touch the circle at the points and respectively. Prove that .

Answer:

Since tangents drawn from the exterior point to a circle are equal in length.

As, and are tangents from exterior point so, … … i)

Similarly,

and are tangents from exterior point so, … … … ii)

and are tangents from exterior point so, … … … iii)

and are tangents from exterior point so, … … … iv)

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

Hence proved.

Question 8: Prove that the points and are the vertices of a right angled isosceles triangle.

Answer:

If and be the vertices of the triangle. From distance formula,

Therefore we see that

This implies that is a right angled isosceles triangle.

Question 9: The term of an AP is zero. Prove that the term of the AP is three times its term.

Answer:

Let be the first term and be the common difference of an A.P. having terms.

Its term is given by,

Therefore

… … … … … i)

… … … … … ii)

From equation i) and ii)

Hence proved.

Question 10: In Fig. 3, from an external point , two tangents and are drawn to a circle with center and radius. If , show that .

Answer:

In (tangent to a circle perpendicular to the radius through the point of contact)

We know

Consider and

(radius of the same circle)

(tangents from the same external point)

is common

(by SSS criterion)

is bisector of

Also (by SAS criterion)

Therefore

In , (angles of a triangle)

Similarly,

**Section – C**

*Question number 11 to 20 carry 3 mark each.*

Question 11: In Fig. 4, is the center of the circle such that the diameter cm and cm. is joined. Find the area of shaded region. (Take )

Answer:

In (angle in a semicircle)

Therefore

cm

Area of shaded region Area of semicircle Area of triangle

Question 12: In Fig. 5, a tent is an a shape of cylinder surmounted by a conical top of same diameter. If the height of the diameter of the cylindrical part are 1 m and m respectively and the slant height of the conical part is m, find the cost of the canvas needed to make the tent if the canvas is available at the rate of Rs sq. meter. (Use )

Answer:

Canvas needed to make the tent Curved surface area of the conical part curved surface area of cylindrical part.

Radius of the conical part radius of cylindrical part m

Slant height of the conical part is 8 m.

Height of cylindrical part is m

Curved surface area of conical part

Curved surface area of cylindrical part

Total surface area

Therefore cost of canvas used

Question 13: If the point is equidistant from the points and . Prove that .

Answer:

Since the point is equidistant from the points and

Therefore

Using distance formula

Question 14: In Fig. 6, find the area of the shaded region, enclosed between two concentric circles of radii cm and cm where . (Use )

Answer:

Area of shaded region Area of circular ring Area of region ABDC

Question 15: If the ratio of the sum first terms of two APs is , find the ratio of their term.

Answer:

For first AP, let the first term be and the common difference be and Sum of the first terms be

Similarly, for Second AP, let the first term be and the common difference be and Sum of the first terms be

Therefore

For term, replace by

Hence the ratio of the term is

Question 16: Solve for

Answer:

Given

Question 17: A conical vessel with a base of radius cm and height cm, is full of water. This water is emptied into a cylindrical vessel of base radius cm. Find the height to which the water will rise in the cylindrical vessel. (Use )

Answer:

Let be the radius, height and slant height of the cone.

Let and be the radius and height of the cylinder.

Since volume will remain same

Volume of water in conical vessel Volume in cylindrical vessel

Therefore

cm

Question 18: A sphere of diameter cm, is dropped into a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by cm. Find the diameter of the cylindrical vessel.

Answer:

Diameter of sphere radius cm

Rise in water level in cylinder cm

Let the radius of cylinder is

Therefore

cm

Question 19: A man standing on the deck of the ship, which is meter above the water level, observes the angle of elevation of the top of the hill as and the angle of depression of the base of the hill as . Find the distance of the hill from the ship and the height of the ship.

Answer:

The man is standing on the deck of the ship at point .

Let be the hill with base at .

It is given that the angle of elevation of point from is and the angle of depression of point from is

So,

(alternate angles)

m

Suppose m and m

In , we have

In , we have

m

Therefore distance of the hill from the ship is m

And height of the hill is m

Question 20: Three different coins are tossed together. Find the probability of getting (i) exactly two heads (ii) at least two heads (iii) at least two tails

Answer:

If three different coins tossed together,

Total possible outcomes are TTT, TTH , THT, THH, HTT, HTH, HHT, HHH

Number of total possible outcomes

i ) Exactly two heads (favorable outcomes) TTH , THT, HTT

Number of favorable outcomes Therefore

ii ) At least two heads is (favorable outcomes) THH, HTH, HHT, HHH

Number of favorable outcomes Therefore

iii ) At least two tails is (favorable outcomes) TTT, TTH , THT, HTT

Number of favorable outcomes Therefore

**Section – D**

*Question number 21 to 31 carry 4 mark each.*

Question 21: Due to heavy flood in a state, thousands were rendered homeless. schools collectively offered to the state government to provide place and canvas for tents to be fixed by government and decided to share the whole expenditure equally. The lower part of the tent is cylindrical of base radius m and height m, with conical upper part of the same base radius but of height m. If the canvas used to make the tent costs Rs. sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem.(Use )

Answer:

Let the radius of base of both cylinder and cone be

Let the height of cone be and that of cylinder be .

Let the slant height of cone be , Then,

Area of tent CSA of cylindrical base CSA of cone

Cost of canvas used in making one tent

Cost of canvas used in making tents

Share of one school

Question 22: Prove that the lengths of the tangents drawn from an external point to the circle are equal.

Answer:

Let us draw a circle with center and two tangents and are drawn from a external point to the circle as shown in the figure given below,

Consider and

(Radius)

is common

Therefore (By SAS criterion)

Hence

Question 23: Draw a circle of radius cm. Draw two tangents to the circle inclined at an angle of to each other.

Answer:

i) Draw a circle of radius cm with center .

ii) Take a point on the circumference of circle and join .

iii) Draw a perpendicular to at

iv) Draw a radius , making an angle of with

v) Draw a perpendicular to at . Let these perpendiculars intersect at .

Question 24: In Fig. 7, two equal circles with center and , touch each other at . produced meets the circle with center at . is a tangent to the circle with center , at the point . is perpendicular to . Find the value of .

Answer:

As, circles are equal, so their radius are also equal.

That means

Consider and

is common

(By AA criterion)

Question 25: Solve for

Answer:

Given

or

Since

Question 26: The angle of elevation of the top of a vertical tower from a point on the ground is . From a point m vertically above , the angle of elevation of the top of tower is . Find the height of the tower and the distance .

Answer:

Let , therefore in

In

Therefore the height of the tower m

Now

m

Therefore horizontal distance is m

Question 27: The houses in a row are numbered consecutively from to . Show that there exists a value of such that the sum of the numbers of houses preceding the house numbered is equal to sum of the numbers of houses following .

Answer:

The number of houses preceding is

Sum of the number of houses preceding the house numbered,

Here,

Given

Substituting

Question 28: In Fig. 8, the vertices of are and . A line segment is drawn to intersect the sides and at and respectively such that . Calculate the area of and compare it with the area of .

Answer:

Area of

sq. units

Now in and

is common

Therefore

sq. units

Question 29: A number is selected randomly from the numbers and . Another number is selected at random from the numbers and . Find the probability that the product of and is less than .

Answer:

Number can be selected in four ways. Corresponding to each such way, there are four ways of selecting .

Therefore two numbers can be selected in one of the following ways,

Therefore, the two numbers can be selected in ways

The number of combinations where are

Therefore the number of possible combinations are

Therefore probability that the product is less than is

Question 30: In Fig. 9, is shown a sector of a circle with center containing . is perpendicular to the radius and meets produced at . Prove that the perimeter of shaded region is .

Answer:

In

Also

Length of arc

Perimeter of shaded region Length of arc

Question 31: A motor boat whose speed is km/h in still water takes hour more to go km upstream than to return down stream to the same spot. Find the speed of the stream.

Answer:

Let the speed of stream be .

Therefore, Speed of boat in upstream is

In downstream, speed of boat is

Given, Time taken in the upstream journey ‒ Time taken in the downstream journey hour

. Since speed cannot be negative, km/hr.

Therefore, Speed of boat in upstream is km/hr

In downstream, speed of boat is km/hr