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Page No S.A.-2.10:
Question 25:
A and B are friends. A is elder to B by 5 years. B's sister C is half the age of B while A's father D is 8 years older than twice the age of B. If the present age of D is 48 years, they find the present ages of A, B and C.
Answer:
Page No S.A.-2.10:
Question 31:
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Answer:
Hence, the circle drawn with any side of rhombus as diameter passes through the point of intersection of its diagonals.
Page No S.A.-2.2:
Question 1:
The two consecutive class marks of a distribution are 52 and 57. Find the class limits of the two intervals.
Answer:
As, the two consecutive class marks of the distribution are 52 and 57.
So, the class size or class width of the distribution = 57 52 = 5
So, the class limits of the two intervals are: 49.5-54.5 and 54.5-59.5
Page No S.A.-2.2:
Question 2:
Three cubes each of volume 125 cm3 are joined end-to-end to form a cuboid. Find the total surface area of cuboid.
Answer:
Let the edge of each cube be a.
So, the total surface area of the cuboid so formed is 350 cm2.
Page No S.A.-2.2:
Question 3:
For the question given below,four alternative choices have been provided of which only one is correct. You have to select the correct choice:
A triangle and a rhombus are on the same base and between the same parallels. Then, the ratio of area of triangle to that of rhombus is:
(a) 1 : 1
(b) 1 : 2
(c) 1 : 3
(d) 1 : 4
Answer:
We have to find the ratio of area of triangle to that of rhombus, when they lie on the same base.
Let the triangle be ABC and the rhombus be ABDC.
Therefore, by using the properties of a rhombus that the diagonal divides the rhombus in two equal areas, we get
So, the ratio of the area of triangle to that of rhombus is 1 : 2.
Hence, the correct choice is option (b).
Page No S.A.-2.2:
Question 4:
For the question given below,four alternative choices have been provided of which only one is correct. You have to select the correct choice:
In Fig. 1, O is the centre of the circle and ∠OBA = 60°. Then ∠ACB equals:
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer:
We have, ∠OBA = 60°
In AOB,
As, OA = OB (Radii)
So, ∠OAB = ∠OBA (Angles opposite to equal sides are equal)
Therefore, ∠OAB = 60°
Also, using angle sum property of triangle
Now,
Hence, the correct choice is option (c).
Page No S.A.-2.3:
Question 5:
For the question given below, four alternative choices have been provided of which only one is correct. You have to select the correct choice:
The diameter and height of a right circular cone are 7 cm and 12 cm, respectively. The volume of the cone (in cm3) is:
(a) 88
(b) 112
(c) 154
(d) 616
Answer:
Hence, the correct choice is option (c).
Page No S.A.-2.3:
Question 6:
For the question given below, four alternative choices have been provided of which only one is correct. You have to select the correct choice:
A fair coin tossed 100 times and the Head occurs 58 times and tail 42 times. The experimental probability of getting a Head is:
(a)
(b)
(c)
(d)
Answer:
We have,
The number of times a coin is tossed = 100
The number of times Head occurs = 58 and
The number of times Tail occurs = 42
Let the event of getting a Head be E.
Hence, the correct choice is option (c).
Page No S.A.-2.3:
Question 7:
If the radius and height of a cone both are increased by 10%, then the volume of the cone is increased by:
(a) 10%
(b) 21%
(c) 33.1%
(d) 100%
Answer:
Hence, the correct answer is option (c).
Page No S.A.-2.3:
Question 8:
Eleven bags of wheat flour, each marked 10 kg actually contained the following weights (in kg) of flour:
10.05, 10.20, 10.00, 9.75, 10.00, 10.03, 9.95, 10.35, 9.90, 10.00, 10.08
Find the probability that any of these bags chosen at random contains:
(i) more than 10 kg of wheat flour.
(ii) less than 9.5 kg of wheat flour.
Answer:
We have,
The total number of bags = 11
(i) As, the number of bags containing more than 10 kg of wheat flour = 5
So, the probability of choosing a bag containing more than 10 kg of wheat flour =
(ii) As, the number of bags containing less than 9.5 kg of wheat flour = 0
So, the probability of choosing a bag containing less than 9.5 kg of wheat flour = = 0
Page No S.A.-2.3:
Question 9:
In the following figure, a circle with centre O is drawn and ∠BAC = 50°. Find x.
Answer:
Page No S.A.-2.3:
Question 10:
The following observations have been arranged in ascending order. If the median of the data is 63, then find the value of x.
29, 32, 48, x − 2, x, x + 2, 72, 78, 84, 95.
Answer:
Page No S.A.-2.3:
Question 11:
Express y in terms of x in the equation 2x − 3y = 12. Find the points where the line represented by the equations 2x − 3y = 12 cuts the x-axis and y-axis.
Answer:
Page No S.A.-2.3:
Question 12:
D is the mid-point of side BC of ΔABC and E is the mid-point BD. If O is the mid point of AE, then prove that ar(ΔBOE) = ar(ΔABC).
Answer:
Page No S.A.-2.3:
Question 13:
Construct a ΔABC with perimeter 11 cm and base angles 90° and 60°.
Answer:
Given: In ΔABC, AB + BC + CA = 11 cm, B = 60° and C = 90°.
To construct: ΔABC
Steps of construction:
1. Draw a line segment XY = 11 cm.
2. At point X, make PXY = 30°.
3. At point Y, make QYX = 45° and let PX and QY meet at point A.
4. Draw perpendicular bisector ST of AX and let ST meets XY at point B.
5. Draw perpendicular bisector UV of AY and let UV meets XY at point C.
6. Join AB and AC.
Thus, ABC is the required triangle.
Page No S.A.-2.4:
Question 11:
A three-wheeler scooter charges Rs. 10 for the first kilometer and Rs. 4.50 each for every subsequent kilometer. For a distance of x km, an amount of Rs. y is paid. Write the linear equation representing the above information.
Answer:
As,
For every first kilometer the scooter charges Rs 10.
For every subsequent kilometer the scooter charges Rs 4.50 each.
To find a distance of x km, an amount of Rs y is paid.
Therefore,
x km can be written as
x = 1 + (x 1)
The price paid for x km is y can be written as
So, the linear equation representing the given information is 9x 2y + 11 = 0.
Page No S.A.-2.4:
Question 12:
ABCD is parallelogram. The angle bisectors of ∠A and ∠D intersect at O. Find the measures of ∠AOD.
Answer:
Page No S.A.-2.4:
Question 13:
In Fig. 2, ABCD is a quadrilateral in which P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that PQRS is a parallelogram.
Answer:
Given: In a quadrilateral ABCD, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.
To prove: PQRS is a parallelogram.
Construction: Join AC.
Proof:
In ACD,
As, R and S are mid-points of DC and DA, respectively.
So, by using the Mid-point theorem − The segment connecting the mid-points of two sides of a triangle is parallel to the third side and is half the length of the third side, we get
RSAC and RS = AC .....(1)
Similarly, in ABC, by using the mid-point theorem, we get
PQAC and PQ = AC .....(2)
Therefore, from (1) and (2), we get
PQRS and PQ = RS
But this is a pair of opposite sides of the quadrilateral PQRS.
Therefore, PQRS is a parallelogram.
Page No S.A.-2.4:
Question 14:
What length of canvas 3 m wide will be required to make a conical tent of height 8 m and radius of base 6 m? (use π = 3.14)
Answer:
So, the length of the canvas required to make the conical tent is 62.8 cm.
Page No S.A.-2.4:
Question 15:
Heights in cm of 12 students are given as:
154, 158, 155, 160, 165, 145, 148, 150, 147, 151, 166, 152
Make a frequency table taking 145-150 as first class interval.
Answer:
The frequency distribution table of the given data is as follows:
Class Interval | Tally Marks | Frequency |
145-150 | ||| | 3 |
150-155 | |||| | 4 |
155-160 | || | 2 |
160-165 | | | 1 |
165-170 | || | 2 |
Total | 12 |
Page No S.A.-2.4:
Question 16:
Write as a linear equation of the form ax + by + c = 0. Also, write the values corresponding to a, b, c. Does the graph of this linear equation pass through origin? Give your answer in 'yes' or 'no'.
Answer:
Yes, the graph of the given linear equation passes through origin.
Page No S.A.-2.4:
Question 17:
A sphere and a cube have the same surface area. Show that the ratio of the volume of the sphere to that of the cube is .
Answer:
So, the ratio of the ratio of the volume of the sphere to that of the cube is .
Page No S.A.-2.4:
Question 18:
In the following figure, ABCD is a parallelogram. If E is the mid-point of BC and AE is the bisector of ∠A, then prove that AB = AD.
Answer:
Page No S.A.-2.4:
Question 19:
E and F are mid-points of sides AB and CD, respectively of a parallelogram ABCD. AF and CE intersect diagonal BD in P and Q, respectively. Prove that diagonal BD is trisected at P and Q.
Answer:
Page No S.A.-2.4:
Question 20:
Consider the following frequency distribution which gives the weight of 38 students of a class:
Weights (in kg) | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 | 56-60 | 61-65 | 66-70 | Total |
No. students | 9 | 5 | 14 | 3 | 1 | 2 | 2 | 2 | 38 |
(i) Find the probability that the weight of a student in the class lies between 36-45 kg.
(ii) Give 2 events in the context, one having probability 0 and the other having probability 1.
Answer:
We have,
Total number of students = 38
(i) As, the number of students in the class whose weight lies between 36-45 kg = 5 + 14 = 19
So, the probability that the weight of a student in the class lies between 36-45 kg =
(ii) As, the number of students in the class whose weight is less than 31 kg = 0
So, the probabilty that the weight of a student in the class is less than 31 kg =
Also, the number of students in the class whose weight lies between 31-70 kg = 38
So, the probabilty that the weight of a student in the class lies between 31-70 kg =
Hence, the two events, one having probabilty 0 is " A student whose weight is less than 31 kg" and the other having probability 1 is " A student whose weight lies between 31-70 kg".
Page No S.A.-2.5:
Question 16:
The curved surface area of a cylinder is 176 cm2 and its base area is 38.5 cm2. Find the volume of the cylinder and justify your answer.
[Use π = ]
Answer:
So, the volume of the cylinder is 308 cm3.
Page No S.A.-2.5:
Question 17:
A hemispherical bowl is made of steel 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. [Use π = ]
Answer:
So, the outer curved surface area of the hemispherical bowl is 173.25 cm2.
Page No S.A.-2.5:
Question 18:
Find the mean of the first ten prime numbers.
Answer:
Hence, the mean of the first ten prime numbers is 12.9.
Page No S.A.-2.5:
Question 19:
Draw the graph of two lines, whose equations are 3x − 2y + 6 = 0 and x + 2y − 6 = 0 on the same graph paper. Find the area of triangle formed by the two lines and x-axis.
Answer:
The points satisfying the equation 3x − 2y + 6 = 0 are:
And, the points satisfying the equation x + 2y − 6 = 0 are:
The graph of the given two lines is as follows:
The vertices of the triangle formed by the given lines and x-axis are B(6, 0), A(−2, 0) and C(0, 3).
Now,
So, the area of the triangle so formed is 12 square units.
Page No S.A.-2.5:
Question 20:
If the number of hours for which a labourer works is x and y are his wages (in rupees) and y = 2x − 1, then draw the graph of work-wages equation. From the graph, find the wages of the labourer if he works for 6 hours.
Answer:
The values of x and y satisfying the given equation y = 2x − 1 are:
x |
0 |
1 |
2 |
3 |
4 |
y |
–1 |
1 |
3 |
5 |
7 |
The graph of the given equation y = 2x − 1 is as follows:
In the graph, a point A(6, 11) lies on the line.
So, at x = 6, y = 11
Hence, the wages of the labourer if he works for 6 hours is Rs 11.
Page No S.A.-2.5:
Question 22:
Construct a ΔABC in which BC = 4.5 cm, ∠C = 45° and sum of sides AB and AC is 8 cm. Justify your construction.
Answer:
Given: In ΔABC, BC = 4.5 cm, ∠C = 45° and AB + AC = 8 cm
To construct: ΔABC
Steps of construction:
1. Draw a line segment BC = 4.5 cm.
2. At point C, make ∠XCB = 45°.
3. With C as centre, draw an arc of radius 8 cm to cut the ray CX at point D.
4. Join BD.
5. At point B, make ∠YBD = ∠BDC and let the ray BY intersect CD at point A.
Thus, ABC is the required triangle.
Justification:
In ΔABD,
As, ∠ABD = ∠BDC (By construction)
So, AB = AD (Sides opposite to equal angles are equal)
Now,
CD = 8 cm (By construction)
AC + AD = 8 cm
So, AC + AB = 8 cm (Proved above)
Page No S.A.-2.5:
Question 21:
ABCD is a parallelogram. AP the bisector of ∠A and CQ the bisector of ∠C meet the opposite sides in P and Q, respectively. Prove that
AP || CQ.
Answer:
Page No S.A.-2.5:
Question 23:
Write a linear equation in two variables to represent the statement ''Four times the cost of a table is equal to seven times the cost of a chair''. From the graph of above linear equation, find the cost of one chair if the cost of the table is â¹1050.
Answer:
Let the cost of a table and a chair be x and y, respectively.
The graph of the linear equation is as follows:
Since, at x = 1050, y = 600
So, the cost of one chair is â¹600.
Page No S.A.-2.5:
Question 24:
Rain water which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius
20 cm. What will be the height of water in the cylindrical vessel if the rainfall is 1 cm. Give your answer to the nearest whole number.
(Use π = 3.14)
Answer:
So, the height of water in the cylindrical vessel is 191 cm.
Page No S.A.-2.5:
Question 25:
Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Answer:
Theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Given: Arc AB of a circle with centre O, subtending AOB at the centre and APB at a point P on the remaining part of the cirlce.
To prove: AOB = 2APB
In the given condition, clearly we have three cases:
Case I: Arc AB is a minor arc
Case II: Arc AB is a semicircle
Case III: Arc AB is a major arc
Construction: Join PO and produce it to a point Q.
Proof:
In AOP, AOQ is an exterior angle
As, an exterior angle is equal to the sum of the interior opposite angles.
So, AOQ = APO + PAO .....(i)
Also, AO = PO (Radii)
So, APO = PAO (Angles opposite to equal sides are equal) .....(ii)
From (i) and (ii), we get
AOQ = APO + APO
AOQ = 2APO .....(iii)
Similalry,
In BOP, BOQ is an exterior angle
As, an exterior angle is equal to the sum of the interior opposite angles.
So, BOQ = BPO + PBO .....(iv)
Also, BO = PO (Radii)
So, BPO = PBO (Angles opposite to equal sides are equal) .....(v)
From (iv) and (v), we get
BOQ = BPO + BPO
BOQ = 2BPO .....(vi)
Adding (iii) and (vi), we get
AOQ + BOQ = 2APO + 2BPO
AOB = 2(APO + BPO)
So, AOB = 2APB
Note: In Case III, where AB is a major arc, AOB is replaced by reflex AOB.
So, reflex AOB = 2APB
Page No S.A.-2.5:
Question 26:
Prove that the opposite angle of an isosceles trapezium are supplementary.
Answer:
Given: In trapezium ABCD, AB = CD and AD || BC.
Construction: Draw AEBC and DFBC.
To prove:
Proof:
Hence, the opposite angle of an isosceles trapezium are supplementary.
Page No S.A.-2.5:
Question 27:
Sketch the graph of the equation 3x + 5y = 15. Find the area of the figure formed by this line and the two axes.
Answer:
The graph of the equation 3x + 5y = 15 is as follows:
The figure formed by the line and the two axes is a right angled AOB.
Page No S.A.-2.5:
Question 28:
In the following figures, ABCD is a parallelogram. A circle through A, B, C intersects CD or CD produced at E. Prove that AE = AD.
Answer:
Page No S.A.-2.6:
Question 23:
Construct a Δ ABC, in which base BC = 3 cm, ∠B = 30° and AB + AC = 5.2 cm.
Answer:
Given: In ΔABC, BC = 3 cm, ∠B = 30° and AB + AC = 5.2 cm
To construct: ΔABC
Steps of construction:
(1) Draw a line segment BC = 3 cm
(2) At point B, make an angle ∠PBC = 30°.
(3) With B as centre, draw an arc of radius 5.2 cm which cuts the ray PB at point D.
(4) Join DC.
(5) Draw perpendicular bisector of DC which cuts DB at point A.
(6) Join AC.
Thus, ΔABC is the required triangle.
Page No S.A.-2.6:
Question 24:
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, then how much soup the hospital has to prepare daily to serve 250 patients?
Answer:
So, the volume of the soup the hospital has to prepare daily to serve 250 patients is 38.5 litres.
Page No S.A.-2.6:
Question 25:
A heap of wheat is in the form of a cone, whose diameter is 10.5 m and height 7 m. Find the volume of wheat in the heap. The heap is to be covered by canvas to protect it from rain, find the area of the canvas required.
Answer:
Page No S.A.-2.6:
Question 26:
Find the mean of the following data by shortcut method.
Marks | 20 | 22 | 25 | 30 | 35 | 39 | 45 | 50 | Total |
Frequency | 4 | 6 | 8 | 10 | 8 | 7 | 5 | 2 | 50 |
Answer:
We have,
Let the marks be represented as xi and frequency be represented as fi.
Let a = 30
Hence, the mean of the given data is 31.8.
Page No S.A.-2.6:
Question 27:
On a page of a telephone directory, there are 200 telephone numbers. The frequency distribution of the digits at their units place is given below:
Unit digit | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Frequency | 22 | 26 | 22 | 22 | 20 | 10 | 14 | 28 | 16 | 20 |
Without looking at the page, a number is chosen at random from the page. What is the probability that the digit at the unit's place of the number chosen is greater than 6?
Answer:
We have, the frequency distribution of the digits at their units place:
Unit digit | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Frequency | 22 | 26 | 22 | 22 | 20 | 10 | 14 | 28 | 16 | 20 |
As, the frequency of getting unit digit geater than (i.e. 7, 8 and 9) = 28 + 16 + 20 = 64
Hence, the probability that the digit at the unit's place of the number chosen is greater than 6 is .
Page No S.A.-2.6:
Question 28:
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:
Outcome | 3 heads | 2 heads | 1 head | No Head |
Frequency | 23 | 72 | 77 | 28 |
Find the experimental probability of getting
(i) 2 Heads
(ii) at least 2 Heads
Answer:
We have,
Outcome | 3 heads | 2 heads | 1 head | No Head |
Frequency | 23 | 72 | 77 | 28 |
Page No S.A.-2.6:
Question 29:
In Fig. 7, a small indoor green house is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. What is the area of the glass? How much tape is required for all the 12 edges?
Answer:
Page No S.A.-2.6:
Question 30:
The daily earnings of 30 workers are given below:
Daily earning (in â¹) | No. of workers |
1-50 | 3 |
50-100 | 7 |
100-150 | 4 |
150-200 | 5 |
200-250 | 4 |
250-300 | 3 |
300-350 | 2 |
350-400 | 2 |
Draw a histogram and a frequency polygon to represent the above data.
Answer:
The frequency distribution table of the given data is as follows:
Daily earning (in â¹) | No. of workers |
1-50 | 3 |
50-100 | 7 |
100-150 | 4 |
150-200 | 5 |
200-250 | 4 |
250-300 | 3 |
300-350 | 2 |
350-400 | 2 |
The histogram and frequency polygon of the given data is as follows:
Page No S.A.-2.6:
Question 31:
The circumference of the base of a cone is cm and its slant height is 13 cm. Find the volume of the cone.
Answer:
Page No S.A.-2.7:
Question 30:
Prove that parallelogram on the same base and between the same parallels are equal in area.
Answer:
Given: The parallelograms ABCD and ABEF be on the same base AB and between same parallels AB and FC.
âTo prove: ar(ABCD) = ar(ABEF)
Proof:
Page No S.A.-2.7:
Question 31:
In Fig. 8, ABCD is a trapezium in which AB || DC. BD is a diagonal and E is the mid-point of AD. A line is drawn through E, parallel to AB, intersecting BC at F. Show that F is the mid-point of BC.
Answer:
Given: In trapezium ABCD, AB || DC, EF || AB and E is the mid-point of AD.
To prove: F is the mid-point of BC.
Proof:
In ΔABD,
As, EP || AB and E is the mid-point of AD (Given: EF || AB)
P is the mid-point of BD
Now, similarly in ΔBCD,
As, PF || DC and P is the mid-point of BD (Since, EF || AB || CD)
Therefore, F is the mid-point of BC.
Page No S.A.-2.7:
Question 32:
In Fig. 9, O is the centre of the circle. The distance between P and Q is 4 cm. Find the ∠ROQ.
Answer:
Given: A circle with centre O and radius OQ = 2 cm; and PQ = 4 cm.
Construction: Join OP.
As, OQ = 2 cm and PQ = 4 cm
So, PQ is the diameter of the circle.
In OPR,
Since, OP = OR (Radii)
So, ∠OPR = ∠ORP (Angles opposite to equal sides are equal)
∠OPR = 35
∠QPR = 35
Now,
As, the subtended by an arc at the centre is double the angle subtended by the same arc at the remaining part of the circle.
So, ∠ROQ = 2∠QPR = 2 35 = 70
Page No S.A.-2.8:
Question 33:
In Fig. 10, a right circular cone of diameter r cm and height 12 cm rests on the base of a right circular cylinder of radius r cm. Their bases are in the same plane and the cylinder is filled with water upto a height of 12 cm. If the cone is then removed, find the height to which water level will fall.
Answer:
So, the height to which the water level will fall is 1 cm.
Page No S.A.-2.8:
Question 34:
Draw a histogram for the following data:
Marks | 10-15 | 15-20 | 20-25 | 25-30 | 30-40 | 40-60 | 60-80 |
Number of candidates | 7 | 9 | 8 | 5 | 12 | 12 | 8 |
Answer:
We have,
As, in the above given table the classes are of unequal widths, so let us form the table with adjusted frequencies:
Now, the histogram for the given data is as follows:
Page No S.A.-2.8:
Question 10:
In Fig. 5, O is the centre of the circle. The angle by the arc BCD at the centre is 140°, BC is produced to P. Find ∠DCP.
Answer:
We have,
Using the property of circles, the angle subtended by a chord at the center is twice the angle subtended by it on the remaining part of the circle, we get
Also, in cyclic quadrilateral ABCD,
As, the opposite angles are supplementary.
So,
Now,
Page No S.A.-2.8:
Question 16:
In Fig. 6, ABCD is a square. If ∠PQR = 90° and PB = QC = DR, prove that ∠QPR = 45°.
Answer:
Page No S.A.-2.9:
Question 20:
The ratio of the curved surface area to the total surface area of a right circular cylinder is 1 : 3. Find the volume of the cylinder if its total surface area is 1848 cm2.
Answer:
So, the volume of the cylinder is 4312 cm3.
Page No SA. 1.6:
Question 4:
In the following figure, the measure of is
(a) 10°
(b) 40°
(c) 60°
(d) 30°
Answer:
Since ABC is a straight line, so
Now
Hence, the correct option is (b).
Page No SA. 1.6:
Question 5:
Find the value of k, if x − 1 is a factor of x2 + x + k.
Answer:
Let f(x) = x2 + x + k.
Since x − 1 is a factor of f(x), therefore
Hence, k = − 2.
Page No SA. 1.7:
Question 6:
Rationalize the denominator of
Answer:
The rationalising factor of is . So
Hence, .
Page No SA. 1.7:
Question 7:
Simplify :
Answer:
Using the identity (a + b)2 = a2 + 2ab + b2, we get
Hence, .
Page No SA. 1.7:
Question 8:
Show that 0.477777... can be expressed in the form , where p and q are integers and q ≠ 0.
Answer:
Let .
Multiplying it by 10, we get
Multiplying (i) by 10 again, we get
Subtracting (i) from (ii), we get
Hence, .
Page No SA. 1.7:
Question 9:
Evaluate (99)3 by using suitable identity.
Answer:
Here, (99)3 = (100 − 1)3.
So, using the identity (a − b)3 = a3 − b3 − 3ab(a − b), we get
(99)3 = (100)3 − 13 − 3(100)(1)(100 − 1)
= 1000000 − 1 − 300(99)
= 1000000 − 1 − 29700
= 1000000 − 29701
= 970299
Hence, (99)3 = 970299.
Page No SA. 1.7:
Question 10:
AD is a line segment. B and C are points in the interior if AC = BD, then prove that AB = CD.
Answer:
Given: AC = BD
AB + BC = BC + CD
AB = CD (Cancelling BC from both the sides)
Hence, AB = CD.
Page No SA. 1.7:
Question 11:
Find 6 rational numbers between 3 and 4.
Answer:
Let us write 3 and 4 as below:
Hence, 6 rational numbers between 3 and 4 are .
Page No SA. 1.7:
Question 12:
If a and b are rational numbers and , find the values of a and b.
Answer:
The rationalising factor for is . So
Hence, a = 7 and b = 4.
Page No SA. 1.7:
Question 13:
In the following figure, PQR = PRQ, then prove that PQS = PRT.
Answer:
Given: PQR = PRQ
Since ST is a straight line, so
Hence, .
Page No SA. 1.7:
Question 14:
In the following figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130° find ∠QRS.
Answer:
Extend line PQ to B intersecting RS at A.
AT and AB are parallel lines therefore
Now, in
Hence, .
Page No SA. 1.8:
Question 15:
In the following figure, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°, find SQT.
Answer:
In
are vertically opposite angles, so
Now, in
Hence, SQT = 60°.
Page No SA. 1.8:
Question 16:
Prove that the sum of all interior angles of a triangle is 180°.
Answer:
Let ABC be a triangle. Draw a line parallel to AC and passing hrough B as shown in the figure.
Since , therefore
Here, DBE is a straight line, so
Hence proved.
Page No SA. 1.8:
Question 17:
In the following figure, D and E are points on the base BC of a triangle ABC such that BD = CE and AD = AE. Prove that Δ ABE Δ ACD.
Answer:
Since AD = AE, so is a an isosceles triangle and therefore
Now, in , we have
In
AB = AC
AE = AD
Hence, Δ ABE Δ ACD.
Page No SA. 1.8:
Question 18:
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that and
. Show that
(i) Δ DAP Δ EBP
(ii) AD = BE
Answer:
(i)
In Δ DAP and Δ EBP, we have
Hence, .
(ii)
Page No SA. 1.8:
Question 19:
In which quadrant or on which axis do each of the points (−2, 4), (3, −1),(−1, 0), (−3, −5) and (0, −1) lie?
Answer:
The given points A(−2, 4), B(3, −1), C(−1, 0), D(−3, −5) and E(0, −1) are plotted as shown below:
From the figure, the points (−2, 4), (3, −1), and (−3, −5) lie in second, third and fourth quadrants respectively.
And the points (−1, 0) and (0, −1) lie on x-axis and y-axis respectively.
Page No SA. 1.9:
Question 20:
Plot the points E(4, 2), I (0, 2), L (−1, 3) and N (2, 0) on the Cartesian plane. Join these points in order and name the shape thus obtained.
Answer:
The points E(4, 2), I (0, 2), L (−1, 3) and N (2, 0) are plotted in the xy-plane and joined as shown below:
Hence, the shape obtained is a triangle.
Page No SA. 1.9:
Question 21:
Factorise : x3 + 13x2 + 32x + 20
Answer:
Let us split the middle terms as shown below:
x3 + 13x2 + 32x + 20 = x3 + 10x2 + 3x2 + 30x + 2x + 20
= x2(x + 10) + 3x(x + 10) + 2(x + 10)
= (x + 10) (x2 + 3x + 2)
= (x + 10) (x2 + 2x + x + 2)
= (x + 10) [x(x + 2) + (x + 2)]
= (x + 10)(x + 2) (x + 1)
Hence, x3 + 13x2 + 32x + 20 = (x + 1) (x + 2)(x + 10).
Page No SA. 1.9:
Question 22:
Without actually calculating the cubes, find the value of the following:
(−12)3 + (7)3 + (5)3
Answer:
Let a = −12, b = 7 and c = 5.
Here, a + b + c = −12 + 7 + 5 = 0.
If a + b + c = 0, then a3 + b3 + c3 = 3abc. Therefore
(−12)3 + (7)3 + (5)3 = 3(−12)(7)(5) = −1260
Hence, (−12)3 + (7)3 + (5)3 = −1260.
Page No SA. 1.9:
Question 23:
If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.
Answer:
Putting x + y + z = 0 in the identity x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2 − xy − yz − zx), we get
x3 + y3 + z3 − 3xyz = (0) (x2 + y2 + z2 − xy − yz − zx)
Thus, if x + y + z = 0, then x3 + y3 + z3 = 3 xyz.
Page No SA. 1.9:
Question 24:
Without actual division, prove that 2x4 − 6x3 + 3x2 + 3x − 2 is exactly divisible by x2 − 3x + 2.
Answer:
Let .
Thus, f(x) is divisible by g(x) if f(1) = 0 and f(2) = 0. Now
Hence, 2x4 − 6x3 + 3x2 + 3x − 2 is exactly divisible by x2 − 3x + 2.
Page No SA. 1.9:
Question 25:
Factorise: a7 − ab6
Answer:
Factorise the expression a7 − ab6 as shown below:
Now, using the identities , we get
Hence, .
Page No SA. 1.9:
Question 26:
In the following figure the side QR of Δ PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at a point T, then prove that ∠QTR = ∠QPR.
Answer:
From the figure,
Here, is an external angle for . So
is an external angle for . So
Substituting in (iii), we get
Hence, .
Page No SA. 1.9:
Question 27:
Three bus-stops situated at A, B and C in the following figure are operated by handicapped person. These three bus − stops are equidistant from each other. OB is the bisector of ∠ABC and OC is the bisector of ∠ACB.
(i) Find ∠BOC.
(ii) Do you think employment provided to handicapped persons is important for the development of the society? Express your views with relevant points.
Answer:
(i) In the figure, AB = BC = CA.
∴ âABC is an equilateral triangle.
⇒ ∠ABC = ∠ACB = ∠BAC = 60º
It is given that OB is the bisector of ∠ABC.
∴ ∠ABO = ∠OBC = .....(1)
Also, OC is the bisector of ∠ACB.
∴ ∠ACO = ∠OCB = .....(2)
In âOBC,
∠OBC + ∠OCB + ∠BOC = 180º (Angle sum property)
⇒ 30º + 30º + ∠BOC = 180º
⇒ ∠BOC = 180º − 60º = 120º
(ii) Yes, the employment provided to handicapped persons is important for the development of the society as employment to handicapped persons provide them with better opportunities of success and prosperity. It helps in their social and economic growth which in turns helps in the development of society.
Page No SA.1.10:
Question 28:
In the following figure, S is any point on the side QR of Δ PQR. Prove that PQ + QR + RP > 2 PS.
Answer:
The sum of any two sides of a triangle is grater than the third side.
In , we have
PQ + QS > PS ..... (i)
PR + SR > PS ..... (ii)
Adding (i) and (ii), we get
PQ + PR + (QS + SR) > 2PS
Since QS + SR = QR, therefore
PQ + PR + QR > 2PS
Hence, PQ + QR + RP > 2PS.
Page No SA.1.10:
Question 29:
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Answer:
Let a, b and c be the sides of the given triangle. Then
a = 18 cm, b = 10 cm and a + b + c = 42 cm
Substituting a = 18, b = 10 in a + b + c = 42, we get
18 + 10 + c = 42
Now
Using Heron's formula, we get
Hence, the area of the triangle is .
Page No SA.1.10:
Question 30:
Δ ABC is an isosceles triangle in which AB= AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.
Answer:
Since AB= AC and AD = AB, therefore AC = AD. Thus
Now, in
Hence, is a right angle.
Page No SA.1.10:
Question 31:
Simplify:
Answer:
The rationalising factor for are respectively. So
Hence, .
Page No SA1.2:
Question 1:
Decimal expansion of a rational number cannot be
(a) non-terminating
(b) non-terminating and recurring
(c) terminating
(d) non-terminating and non-recurring
Answer:
The decimal expansion of a rational number is either terminating or non-terminating recurring.
So, a rational number cannot be non-terminating and non-recurring.
Hence, the correct option is (d).
Page No SA1.2:
Question 2:
One of the factors of (9x2 − 1) − (1 + 3x)2, is
(a) 3 + x
(b) 3 − x
(c) 3x − 1
(d) 3x + 1
Answer:
To find all the factors of (9x2 − 1) − (1 + 3x)2, let us factorise it.
Thus, the factors of (9x2 − 1) − (1 + 3x)2 are –2 and 3x + 1.
Hence, the correct option is (d).
Page No SA1.2:
Question 3:
Simplify:
Answer:
Let us simplify the expression by simplifying each term separately.
Therefore
Hence, .
Page No SA1.2:
Question 4:
An exterior angle of a triangle is 110° and the two interior opposite angles are equal.
Each of these angles is
(a) 70°
(b) 55°
(c) 35°
(d) 110°
Answer:
Let be the triangle in which is the exterior triangle as shown in the figure.
Given:
An exterior angle of a triangle is equal to the sum of the opposite interior angles, therefore
Thus, the measure of the equal angles is .
Hence, the correct option is (b).
Page No SA1.2:
Question 5:
In ΔPQR, if , then
(a) QR > PR
(b) PQ > PR
(c) PQ < PR
(d) QR < PR
Answer:
In a given triangle, the side opposite to the greater angle is greater.
The side opposite to angles and are PQ and PR respectively.
In ΔPQR, , therefore
PQ > PR
Hence, the correct option is (b).
Page No SA1.2:
Question 6:
Evaluate the product without multiplying directly : 104×97.
Answer:
Let us find the product 104 × 97 using the identity .
Hence, 104 × 97 = 10088.
Page No SA1.3:
Question 7:
"Lines are parallel if they do not intersect," prove the above with suitable diagram.
Answer:
Let AB and CD be two parallel lines.
If possible, let AB and CD intersect at a point P.
But, two lines in the plane are intersecting lines if they have a common point.
This is a contradiction, because AB and CD are two parallel lines.
Hence, the lines are parallel if they do not intersect.
Page No SA1.3:
Question 8:
Prove that two distinct lines cannot have more than one point in common.
Answer:
Let AB and CD be two distinct lines. We need to prove that AB and CD cannot have more than one point in common.
Let us assume that AB and CD have two points, say P and Q, in common.
Then, AB contains both points P and Q.
Also, CD contains both points P and Q.
Now, incidence axiom states that there is one and only one line passing through two distinct points P and Q.
So, AB and CD are same lines.
But, this is a contradiction because AB and CD are two distinct lines.
Hence, two distinct lines cannot have more than one point in common.
Page No SA1.3:
Question 9:
Plot the points (3, 4), (−3,−3), (−7, 6) and (0, −6) on Cartesian plane and give their positions in quadrants / axes.
Answer:
Take 1 cm = 1 unit and draw the x-axis and y-axis.
Now plot the points (3, 4), (−3,−3), (−7, 6) and (0, −6) as shown below:
From the graph, the points (3, 4), (−7, 6) and (−3,−3) lie in the first, second and third quadrants respectively
and (0, −6) lies on the y-axis.
Page No SA1.3:
Question 10:
Find the area of the triangle with sides 35 cm, 54 cm and 61 cm.
Answer:
We will use Heron's formula to find the area of the triangle.
Here, a = 35 cm, b = 54 cm and c = 61 cm. Now
Therefore
Hence, the area of the required triangle is .
Page No SA1.3:
Question 11:
If x = 2 + , find .
Answer:
Given:
Now
Hence, .
Page No SA1.3:
Question 12:
Simplify
Answer:
Here, we need to simplify the inner most radicals first.
Hence, .
Page No SA1.3:
Question 13:
Expand
Answer:
Using the identity ( a + b + c )2 = a2 + b2 + c2 + 2(ab + bc + ac), we get
Hence, .
Page No SA1.3:
Question 14:
Evaluate the following :
Answer:
Using the identity (a + b)3 = a3 + b3 + 3ab(a + b), we get
Hence, .
Page No SA1.3:
Question 15:
In Δ ABC, if AB is the greatest side, then prove that .
Answer:
It is given that, AB is the longest side, so AB > BC and AB > AC.
Since, the angle opposite to the longer side is greater, therefore
Adding (i) and (ii), we get
Adding on both sides, we have
Hence, .
Page No SA1.4:
Question 16:
In the following figure, AE = AD, . Prove that AB = AC.
Answer:
Since AE = AD, so . Now
In
(Given)
(Given)
(Prove above)
(By ASA)
Thus, by corresponding parts of congruent triangles
AB = AC
Page No SA1.4:
Question 17:
Sides AB and CD of a quadrilateral ABCD are produced as shown in the following figure. Show that .
Answer:
In quadrilateral ABCD
Since, the sum of the angles of a quadrilateral is , therefore
Hence, .
Page No SA1.4:
Question 18:
AB is a line segment and line l is its perpendicular bisector. Show that every point on line l is equidistant from A and B.
Answer:
In the following figure, AB is the given line segment and l is its perpendicular bisector.
A any point on the given line l.
Now, in âAOP andâBOP
Hence, every point on line l is equidistant from A and B.
Page No SA1.4:
Question 19:
Find the area of a triangle whose sides are 5 cm, 12 cm and 13 cm. Also, find its shortest altitude.
Answer:
Here,
52 + 122
= 25 + 144
= 169
= 132.
Thus, the triangle is a right angled triangle.
So, the length of hypotenuse is 13 and the length of the perpendicular sides are 5 and 12.
Therefore, the length of its shortest altitude is 5.
Now
Hence, the area of the required triangle is 30 sq. units.
Page No SA1.4:
Question 20:
The perimeter of an isosceles triangle is 42 cm and its base is times each of the equal sides. Find the length of each side and area of the triangle.
Answer:
Let a, b and c be the sides of the triangle. Here, a = b and .
Since the perimeter of the triangle is 42 cm, so
Therefore, the lengths of the sides of the triangle are 12 cm, 12 cm and 18 cm.
Now
Hence, the area of the triangle is .
Page No SA1.4:
Question 21:
Varun was facing some difficulty in simplyfying His classmate Priya gave him a clue to rationalise the denominator for simplification. Varun Simplified the expression and thanked Priya for this good will. How Varun simplified ? What value does it indicate?
Answer:
The rationalising factor for is . So
Varun simplified by multiplying the numerator and denominator by .
It indicates the value of co-operation and help.
Page No SA1.5:
Question 22:
Prove that
Answer:
Using the laws of exponents , we get
Now
Hence, .
Page No SA1.5:
Question 23:
If the polynomials az3 + 4z2 + 3z − 4 and z3 − 4z + a leave the same remainder when divided by z − 3, find a.
Answer:
Let .
Since, p(z) and q(z) leave the same remainder when divided by z − 3, so
Hence, a = −1.
Page No SA1.5:
Question 24:
Factorize 2x2 − .
Answer:
Splitting middle term of , we get
Hence, the required factorisation is .
Page No SA1.5:
Question 25:
Find the quotient when f(x) = x3 + 3x2 + 3x + 5 is divided by g(x) = x + 2. Also, find the remainder.
Answer:
Here, f(x) = x3 + 3x2 + 3x + 5 and g(x) = x + 2.
Hence, the quotient is x2 + x + 1 and the remainder is 3.
Page No SA1.5:
Question 26:
Prove that:
(x + y)3 + (y + z)3 + (z + x)3 − 3(x + y) (y + z) (z + x) = (x3 + y3 + z3 − 3xyz).
Answer:
Using the identities , we get
Hence, (x + y)3 + (y + z)3 + (z + x)3 − 3(x + y) (y + z) (z + x) = 2(x3 + y3 + z3 − 3xyz).
Page No SA1.5:
Question 27:
In a ΔPQR, if PQ = QR and mid-points of three sides PQ, QR and RP are L, M and N respectively. Prove that LN = MN.
Answer:
Given: In and mid-points of three sides PQ, QR and RP are L, M and N respectively.
Now, in
(âµ PQ = QR)
PN = RN (N is mid-point of PR)
PL = RM (PQ = QR and L and M are mid-points of PQ and RQ respectively )
(By SAS)
Thus, by corresponding parts of congruent triangles, we have
LN = MN
Page No SA1.5:
Question 28:
In the following figure, if OA = OD and . Prove that ΔOCB is an isosceles triangle.
Answer:
Given: OA = OD and
In
OA = OD (Given)
(Vertically opposite angles)
(By ASA)
Thus, by corresponding parts of congruent triangles, we have
OC = OB
Hence, ΔOCB is an isosceles triangle.
Page No SA1.5:
Question 29:
In the following figure, ray OS stands on a line POQ, ray OR and ray OT are angle bisectors of POS and QOS respectively.
If POS = x, find ROT.
Answer:
Given: POQ is a straight line and ray OR and ray OT are angle bisectors of POS and QOS respectively.
Angles form a linear pair, so
Since OR and OT are angle bisectors of POS and QOS respectively, therefore
Hence, which is independent of x.
Page No SA1.6:
Question 30:
If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.
Answer:
Let m and l be two lines and p be the transversal line.
Let be the corresponding angles and CM and TN respectively be their angle bisectors. Therefore
Here, SM and TN are parallel. Therefore
Now
Thus, the sum of the interior angles on the same side of the line p and between the lines l and m is .
Therefore, l and m are parallel.
Hence, if a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then the two lines are parallel.
Page No SA1.6:
Question 31:
PQRS is a quadrilateral in which diagonals PR and QS intersect at O. Prove that PQ + QR + RS + SP < 2 (PR + QS).
Answer:
Let PQRS be the quadrilateral in which diagonals PR and QS intersect at O.
The sum of any two sides of a triangle is greater than the third side.
So, in , we have
OP + OQ > PQ .....(i)
OP + OS > PS .....(ii)
OS + OR > RS .....(iii)
OR + OQ > QR .....(iv)
Adding (i), (ii), (iii) and (iv), we get
2(OP + OQ + OS + OR) > PQ + PS + RS + QR
⇒ 2(OP + OR + OQ + OS ) > PQ + PS + RS + QR
⇒ 2(PR + QS ) > PQ + PS + RS + QR
⇒ PQ + PS + RS + QR < 2(PR + QS)
Page No SA1.6:
Question 32:
The value of 271/3. 41/2 is
(a) 6
(b) 12
(c) 18
(d) − 6
Answer:
Let us first simplify separately.
Therefore
Hence, the correct option is (a).
Page No SA1.6:
Question 33:
Which of the following is an irrational number?
(a)
(b) −
(c)
(d)
Answer:
Let us simplify each of the numbers , −, and separately.
Here, −, and are rational numbers while is an irrational number.
Hence, the correct option is (a).
Page No SA1.6:
Question 34:
If the perimeter of an equilateral triangle is 60 m, then its area is
(a)
(b)
(c)
(d)
Answer:
Let a be the side of the equilateral triangle. Then
Perimeter = 60 m
Now
Hence, the correct option is (d).
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