Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 7 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Standard Set 7 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question Paper has 5 Sections A, B, C, D and E.
- Section A has 20 MCQs carrying 1 mark each.
- Section B has 5 questions carrying 2 marks each.
- Section C has 6 questions carrying 3 marks each.
- Section D has 4 questions carrying 5 marks each.
- Section E has 3 case based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
- All questions are compulsory. However, an internal choice in 2 question of 5 marks, 2question of 3 marks and 2 questions of 2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.
- Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.
Section – A
Section A consists of 20 questions of 1 mark each.
Question 1.
In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm, then the length of AP (in cm) is:
(a) 7.5
(b) 15
(c) 10
(d) 9
Answer:
(a) 7.5
Explanation: Given, AB = 5 cm, AC = 6 cm and BC = 4 cm.
Let PB = x cm, CQ = y cm and D be the point where the tangent BC touches the circle.
So, AP = (5 + x) cm and AQ = (6 + y) cm
Also PB = BD = x cm [tangents from B]
and CQ = CD = y cm [tangents from C]
Now, 5 + x = 6 + y [∵ AP = AQ]
x – y = 1 …… (i)
and x + y = 4 [∵ BC = BD + CD = 4] …….. (ii)
Adding (i) and (ii), we get
2x = 5
⇒ x = 2.5
Thus, AP = (5 + x) cm = 7.5 cm
Question 2.
If tan2 45° – cos2 30° = x sin 30°cos 60°, then the value of x is (1)
(a) 0
(b) 2
(c) 1
(d) -1
Answer:
(c) 1
We have, tan2 45° – cos2 30° = x sin 30°cos 60°
⇒ (1)2 – \(\left(\frac{\sqrt{3}}{2}\right)^2\) = x × \(\frac{1}{2}\) × \(\frac{1}{2}\)
⇒ 1 – \(\frac{3}{4}\) = \(\frac{x}{4}\) ⇒ \(\frac{1}{4}\) = \(\frac{x}{4}\)
⇒ x = 1
Question 3.
In the figure, P and Q are points on the sides AB and AC of the triangle ABC such that AP = 3.5 cm, PB = 7 cm, AQ = 3 cm and QC = 6 cm. If PQ = 4.5 cm, what is the length of BC?
(a) 13.7 cm
(b) 12.7 cm
(c) 13 cm
(d) 13.5 cm
Answer:
(d) 13.5 cm
Explanation: Given, AP = 3.5 cm, PB = 7 cm, AQ = 3 cm and QC = 6 cm.
We get, \(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{AP}}{\mathrm{AP}+\mathrm{PB}}=\frac{3.5}{10.5}=\frac{1}{3}, \frac{\mathrm{AQ}}{\mathrm{AC}}\) = \(\frac{\mathrm{AQ}}{\mathrm{AQ}+\mathrm{QC}}=\frac{3}{6+3}=\frac{3}{9}=\frac{1}{3}\)
⇒ \(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{AQ}}{\mathrm{AC}}=\frac{1}{3}\)
∴ ΔABC ~ ΔAQP [By SS criterion]
∴ \(\frac{\mathrm{QP}}{\mathrm{BC}}=\frac{\mathrm{AQ}}{\mathrm{AC}}=\frac{\mathrm{AP}}{\mathrm{AB}}\)
⇒ \(\frac{4.5}{B C}=\frac{3}{9}\) ⇒ [∵ Given PQ = 4.5 cm]
⇒ BC = 4.5 × 3 = 13.5
⇒ BC = 13.5 cm
Question 4.
If x – 2 and x = 0 are the zeroes of the polynomial f(x) = 5x2 + ax + b. Then, the values of a and b are (1)
(a) a = 2, b = 1
(b) a = 10, b = 1
(c) a = 1, b = 2
(d) a = -10, b = 0
Answer:
(d) a = -10, b = 0
Given, f(x) = 5x2 + ax + b
Since x = 2 and x = 0 are the zeroes of f(x).
f(2) = 0 and f(0) = 0 ⇒ 5(2)2 + 2a + b = 0
and 5(0) + a(0) + b = 0
⇒ 20 + 2a + b = 0 and b = 0
⇒ 20 + 2a + 0 = 0 and b = 0
⇒ 2a = -20 and b = 0
a = -10 and b = 0
Question 5.
An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. What is the width of the river?
(a) 317.5 m
(b) 315.5 m
(c) 314.5 m
(d) 314.4
Answer:
(b) 315.5 m
Explanation: Let, the width of river, AB = x m
Height of aeroplane, DC = 200 m
Question 6.
The ratio in which the X-axis divides the line segment joining A(3, 6) and B(12, -3) is (1)
(a) 1 : 2
(b) 2 : 1
(c) -2 : 1
(d) -1 : 2
Answer:
(b) 2 : 1
Let P(x, 0) be the point of intersection of X-axis with the line segment joining A(3, 6) and B(12, -3) which divides the line segment AB in the ratio λ : 1.
Then, by using distance formula,
(x, 0) = \(\left(\frac{12 \lambda+3}{\lambda+1}, \frac{-3 \lambda+6}{\lambda+1}\right)\)
Now, equating the y component of both sides, we get
\(\frac{-3 \lambda+6}{\lambda+1}\) = 0
⇒ -3λ + 6 = 0 ⇒ λ = \(\frac{2}{1}\)
So, X-axis divides AB in the ratio 2: 1.
Question 7.
If the mean of the following distribution is 2.6, then the value of y is:
Variable (x) | Frequency |
1 | 4 |
2 | 5 |
3 | y |
4 | 1 |
5 | 2 |
(a) 3
(b) 8
(c) 13
(d) 24
Answer:
(b) 8
Explanation: Mean = 2.6
Mean = \(\frac{\Sigma f x}{\Sigma f}=\frac{28+3 y}{12+y}\) 2.6 = \(\frac{28+3 y}{12+y}\)
⇒ 2.6(12 + y) = 28 + 3y ⇒ 31.2 + 2.6y = 28 + 3y
⇒ o.4y = 3.2 ⇒ y = 8.
Question 8.
The smallest number by which \(\sqrt{27}\) should be multiplied, to get a rational number is (1)
(a) \(\sqrt{27}\)
(b) 3\(\sqrt{3}\)
(c) \(\sqrt{3}\)
(d) 3
Answer:
(c) \(\sqrt{3}\)
As \(\sqrt{27}\) = \(\sqrt{3 \times 3 \times 3}\) = 3\(\sqrt{3}\)
If we multiply it by \(\sqrt{3}\), then it will become 3\(\sqrt{3}\) × \(\sqrt{3}\) = 3 × 3 = 9
i.e. a rational number.
Question 9.
An urn contains 100 lottery tickets numbered from 1 to 100. If a ticket is selected at random, then what is the probability that it is a perfect square?
(a) 0.01
(b) 0.1
(c) 0.11
(d) 0.001
Answer:
(b) 0.1
Explanation: Perfect square = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Number of perfect square = 10
Total number of lottery tickets = 100
P(Number of perfect square) = \(\frac{\text { Number of favourable event }}{\text { Total number of event }}\)
= \(\frac{10}{100}\)
= 0.1
Question 10.
The tangents drawn at the extremities of the diameter of a circle are (1)
(a) parallel
(b) perpendicular
(c) neither parallel nor perpendicular
(d) None of the above
Answer:
(a) parallel
Since, OP ⊥ AB and OQ ⊥ CD
∴ ∠1 = 90° and ∠2 = 90°
⇒ ∠1 = ∠2, which are alternate angles.
∴ AB || CD
Question 11.
The value of cot 45° + (\(\frac{\sec 45^{\circ}}{\sin 45^{\circ}}\)):
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3
Explanation: We know, cot 45° = 1, sec 45° = √2, sin45°= \(\frac{1}{\sqrt{2}}\)
On putting the values, cot 45° + \(\frac{\sec 45^{\circ}}{\sin 45^{\circ}}\) = 1 + \(\frac{\frac{\sqrt{2}}{1}}{\sqrt{2}}\)
⇒ = 1 + 2
= 3.
Question 12.
A toy is the shape of cone over a hemisphere and the radius of the hemisphere is 3.5 cm. If the height of the toy is 15.5 cm, then the total area of the toy is (1)
(a) 214.5 cm2
(b) 215.4 cm2
(c) 216.5 cm2
(d) 210 cm2
Answer:
(a) 214.5 cm2
Height of the cone = (15.5— 3.5)cm = 12 cm
Slant height, l = \(\sqrt{h^2+r^2}\) = \(\sqrt{12^2+(3.5)^2}\)
= \(\sqrt{144+12.25}\) = 12.5 cm
Curved surface area of the cone
= πrl = \(\frac{22}{7}\) × 3.5 × 12.5 = 137.5 cm2
Curved surface area of the hemisphere = 2πr2
= \(\frac{2 \times 22}{7}\) × 3.5 × 3.5 cm2 = 77 cm2
∴ Total area of the toy = (137.5 + 77) cm2 = 214.5 cm2
Question 13.
If the root of the equation Ax2 – Bx + C = 0 are -1 and 1, then which one of the following is correct?
(a) A and C are both zero
(b) A and B are both positive
(c) A and C are both negative
(d) A and C are of opposite sign
Answer:
(d) A and C are of opposite sign
Explanation: Ax2 – Bx + C = 0
Since, the given roots are – 1 and 1
Sum of roots = \(-\frac{(-B)}{A}=\frac{B}{A}\) = – 1 + 1 = 0
Product of roots = \(\frac{\mathrm{C}}{\mathrm{A}}\) = 1 × (-1) = – 1
But \(\frac{\mathrm{C}}{\mathrm{A}}\) = -1
⇒ C = -A
Question 14.
In an isosceles right angled triangle, if the hypotenuse is 5\(\sqrt{2}\) cm, then the length of the sides of triangle is (1)
(a) 4 cm
(b) 6 cm
(c) 5 cm
(d) None of these
Answer:
(c) 5 cm
Let ABC is an isosceles right angled triangle, right angled at B with AB = BC and AC = 5\(\sqrt{2}\)cm.
By pythagoras theorem,
(AC)2 = (AB)2 + (BC)2
⇒ (AC)2 = 2(AB)2
⇒ (5\(\sqrt{2}\))2 = 2(AB)2
⇒ \(\frac{50}{2}\) = (AB)2
⇒ (AB)2 = 25
⇒ AB = 5 cm
Hence, the length of the equal sides of a triangle is 5 cm.
Question 15.
If one root of the quadratic equation 6x2 – x- k – 0 is \(\frac{2}{3}\), what is the value of k?
(a) 1
(b) 3
(c) 2
(d) 5
Answer:
(c) 2
Explanation: Given quadratic equation is 6x2 – x – k = 0.
Put x = \(\frac{2}{3}\) in the given equation
⇒ 6(\(\frac{2}{3}\))2 – (\(\frac{2}{3}\)) – k = 0
⇒ 6 × \(\frac{4}{9}\) – \(\frac{2}{3}\) – k = 0
⇒ \(\frac{8}{3}\) – \(\frac{2}{3}\) – k = 0
⇒ 2 – k = 0
⇒ k = 2
Question 16.
The mean and median of a distribution are 16 and 17, respectively. Then, the value of mode is (1)
(a) 17
(b) 19
(c) 16
(d) 15
Answer:
(b) 19
Given, mean =16 and median = 17
By empirical relationship, we have
Mode = 3 median – 2 mean
= 3 × 17 – 2 × 16
= 51 – 32 = 19
Question 17.
If a chord of a circle of radius 14 cm makes an angle of 90° at the centre, then the area of major segment is:
(a) 560 cm2
(b) 300 cm2
(c) 160 cm2
(d) None of these
Answer:
(a) 560 cm2
Explanation:
Area of sector OABO = \(\frac{\theta}{360^{\circ}}\) × πr2
= \(\frac{90^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 14 × 14
= 154 cm2
Area of circle = πr2
\(\frac{22}{7}\) × 14 × 14
= 616 cm2
Area of right angled ΔABO = \(\frac{1}{2}\) × 14 × 14 = 98 cm2
Area of the minor segment of the circle = Area of sector OABO – Area of right angled ΔABO
(154 – 98) = 56 cm2
Area of major segment = Area of circle – Area of minor segment
= 616 – 56 = 560 cm2
Question 18.
The difference between the circumference and the radius of a circle is 37 cm. The area of the circle is (1)
(a) 149 cm2
(b) 154 cm2
(c) 121 cm2
(d) 169 cm2
Answer:
(b) 154 cm2
Let r be the radius of the circle.
Given, the circumference of a circle – radius of the circle = 37
⇒ 2πr – r = 37
⇒ r(2π -1) = 37
⇒ r = \(\frac{37}{2 \pi-1}\)
= \(\frac{37}{2\left(\frac{22}{7}\right)-1}\) = \(\frac{37 \times 7}{37}\) = 7 cm
∴ Area of circle = πr2 = \(\frac{22}{7}\) × (7)2 = 154 cm2
DIRECTION: In question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true, but reason (R) is false.
(d) Assertion (A) is false, but reason (R) is true.
Question 19.
Statement A (Assertion): The area enclosed by a chord and the major arc is major segment. Statement R (Reason): If a circle is divided into three equal arcs, then each is a major arc.
Answer:
(c) Assertion (A) is true, but reason (R) is false.
Explanation: For assertion, the area enclosed by a sector is proportional to the arc length of the sector.
So, A = \(\frac{\mathrm{RL}}{2}\), A = area, R = radius and L = arc length
Hence, the resulting enclosed area is a major segment if it is a major arc.
So the assertion is true.
For reason,
We denote an arc as a major arc when it is greater than the semicircle and if we divide a circle into 3 arc each of it will be less than a semicircle so it is a minor arc.
Question 20.
Assertion (A) 2 is a rational number.
Reason (R) The square roots of all positive integers are irrational. (1)
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer:
(c) Assertion (A) is true but Reason (R) is false.
∵ \(\sqrt{4}\) = ±2, which is not an irrational number.
∴ Assertion (A) is true but Reason (R) is false.
Section – B
Section B consists of 5 questions of 2 marks each.
Question 21.
Two numbers are in the ratio 2 : 3 and their LCM is 180. What is the HCF of these numbers?
Solution:
Let first number be 2x and second be 3x
∴ their LCM = 2 × 3 × x
but given LCM = 180
∴ 2 × 3 × x = 180
⇒ x = 30
First number = 2x = 2 × 30 = 60 = 2 × 2 × 3 × 5
Second number = 3x = 3 × 30 = 90 = 3 × 3 × 2 × 5
Now, HCF of 60 and 90 = 2 × 3 × 5 = 30
Question 22.
In a right-angled ΔABC, right-angled at B, if the ratio of AB to AC is 1 : \(\sqrt{2}\), then find the values of
(i) \(\frac{2 \tan A}{1-\tan ^2 A}\)
(ii) \(\frac{2 \tan A}{1+\tan ^2 A}\) (2)
Answer:
We have, AB : AC = 1 : \(\sqrt{2}\) ⇒ \(\frac{A B}{A C}\) = \(\frac{1}{\sqrt{2}}\)
Let AB = x and AC = \(\sqrt{2}\)x, for some x
By Pythagoras theorem, we have
(AC)2 = (AB)2 + (BC)2
⇒ \((\sqrt{2} x)^2\) = x2 + (BC)2
⇒ (BC)2 = 2x2 – x2 = x2
⇒ BC = x [taking positive square roots]
∴ tan A = \(\frac{B C}{A B}\) = \(\frac{x}{x}\) = 1 [∵ tanθ = \(\frac{\text { perpendicular }}{\text { base }}\)]
(i) We have,
\(\frac{2 \tan A}{1-\tan ^2 A}\) = \(\frac{2 \times 1}{1-1}\) = \(\frac{2}{0}\), which is not defined
(ii) We have,
\(\frac{2 \tan A}{1+\tan ^2 A}\) = \(\frac{2 \times 1}{1+1^2}\) = \(\frac{2}{2}\) = 1
Question 23.
Find the values of x for which the distance between the points A(2, – 3) and B(x, 5) is 10 units.
OR
If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k? Solution:
Given points are A(2, – 3), B(x, 5) and AB = 10 units
Now, AB = 10 units
By distance formula, AB = \(\left|\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right|\)
10 = \(\left|\sqrt{(x-2)^2+(5+3)^2}\right|\)
On squaring both sides, we get 100 = (x – 2)2 + 64
⇒ (x – 2)2 = 100 – 64 = 36
Taking square root on both sides, we get
x – 2 = ± 6
⇒ x = 2 ± 6
∴ x = 2 + 6 = 8
and x = 2 – 6 = – 4
Hence, x = 8 and – 4.
OR
Given points are (4, k) and (1, 0), the distance is 5 units.
By distance formula,
Distance = \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\(\)
5 = \(\sqrt{(1-4)^2+(0-k)^2}\)
On squaring both sides,
⇒ 25 = (- 3)2 + k2 = 9 + k2
⇒ k2 = 16
⇒ k = ± \(\) = ± 4
Hence, k = ± 4
Question 24.
Find the value of K for given equation has real and equal roots.
Kx2 – 5x + K = 0 (2)
Answer:
Given, quadratic equation is
kx2 – 5x + k = 0
On comparing with ax2 + bx + c = 0, we get
Here, a = K . b = – 5 and C = K
∴ D = b2 – 4ac = (-5)2 – 4 × K × (K)
= 25 – 4K2
∴ The given equation have real and equal roots,
So, D = 0
⇒ 25 – 4K2 = 0 ⇒ 4K2 = 25
K = ±\(\frac{5}{2}\)
Question 25.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Solution:
We know that, Mean, \(\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{499}{40}\)
\(\bar{x}\) = 12.475
∴ Mean number of days a student was absent is 12.475.
Section – C
Section C consists of 6 questions of 3 marks each.
Question 26.
A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel. (3)
Answer:
Let the radius of the wheel be r cm.
Distance covered by the wheel in one revolution
\(=\frac{\text { Distance moved }}{\text { Number of revolutions }}\) = \(\frac{11}{5000}\)km
= \(\frac{11}{5000}\) × 1000 × 100 cm = 220 cm
∴ Circumference of the wheel = 220 cm
⇒ 2πr = 220 cm
⇒ 2 × \(\frac{22}{7}\) × r = 220 ⇒ r = 35 cm
∴ Distance = 2r = (2 × 35) = 70 cm
Hence, the diameter of the wheel is 70 cm.
Or
If the perimeter of a semi-circular protractor is 108 cm. Find the diameter of the protractor. [take π = \(\frac{22}{7}\)] (3)
Answer:
Let the radius of the protractor be r cm.
Then, perimeter = 108 cm
⇒ \(\frac{1}{2}\)(2πr) + 2r = 108
[∴ perimeter of a semi-circle = \(\frac{1}{2}\)(2πr)]
⇒ πr + 2r = 108
⇒ \(\frac{22}{7}\) × r + 2r = 108
⇒ 36r = 108 × 7
⇒ r = 3 × 7 = 21 cm
∴ Distance of the protractor = 2r = (2 × 21) cm = 42 cm
Question 27.
Find the zeroes of the quadratic polynomial 5x2 – 4 – 8x and verify the relationship between the zeroes and the coefficients of the polynomial.
Solution:
Given, the polynomial is 5x2 – 8x -4
Here, a = 5, b = – 8 and c = – 4
5x2 – 8x – 4 = 5x2 – 10x + 2x – 4 = 5x(x – 2) + 2(x – 2)
= (5x + 2) (x – 2)
∴ x = 2, \(\frac{-2}{5}\)
Therefore, the zero of 5x2 – 8x – 4 are 2 and \(\frac{-2}{5}\)
Now, Sum of zeroes = 2 + (\(\frac{-2}{5}\)) = \(\frac{8}{5}\) = \(\frac{-b}{a}\)
Product of zeroes 2 × (\(\frac{-2}{5}\)) = \(\frac{-4}{5}\) = \(\frac{c}{a}\)
Question 28.
The polynomial x2 – (k + 6)x + 2(2k – 1) has sum of its zeroes equal to half of their product. Find the value of k. (3)
Answer:
Let α and β are the roots of given quadratic equation
x2 – (k + 6)x + 2(2k – 1) = 0
Now, sum of roots = α + β
= –\(\left\{\frac{-(k+6)}{1}\right\}\)
= k + 6
Product of roots = αβ = \(\frac{2(2 k-1)}{1}\)
= 2(2k – 1)
According to the question,
Sum of roots (zeroes)
= \(\frac{1}{2}\) × products of roots (zeroes)
⇒ k + 6 = \(\frac{1}{2}\) × 2(2k – 1)
⇒ k + 6 = 2k – 1
⇒ 6 + 1 = 2k – k
⇒ k = 7
Question 29.
ABCD is a trapezium with AB || DC. E and Fare points on non-parallel sides AD and BC respectively such that EF is parallel to AB. Show that AE × FC = BF × ED.
OR
In the given figure, PA, QB, and RC are perpendicular to AC such that PA = x, RC = y, QB = z, AB = a, and BC = b. Prove that:
\(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\)
Solution:
Given: ABCD is a trapezium in which AB || DC || EF
To Prove : AE × FC = BF × ED.
Construction : Join AC which intersect EF at the point G.
Proof: In ΔADC,
EG || DC
∴ \(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AG}}{\mathrm{GC}}\) [By basic proportionality theorem] …..(i)
In ΔABC, GF || AB
∴ \(\frac{C F}{B F}=\frac{C G}{A G}\) [By B.P.T.]
or \(\frac{B F}{C F}=\frac{A G}{C G}\) ……. (ii)
From equations (i) and (ii),
\(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{CF}}\)
AE × CF = BF × ED. Hence proved.
OR
Given : PA, QB and RC are perpendicular to AC.
∴ PA || QB || RC
Thus, in ΔPAC and ΔQBC,
QB || PA and ∠PAC = ∠QBC = 90°
So, ΔQBC ~ ΔPAC
∴ \(\frac{\mathrm{QB}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{AC}}\)
or \(\frac{z}{x}=\frac{b}{a+b}\) …….. (i)
Similarly, in ΔRAC and ΔQAB, QB || RC and ∠A is common
ΔQAB ~ ΔRAC
∴ \(\frac{\mathrm{QB}}{\mathrm{RC}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
or \(\frac{z}{y}=\frac{a}{a+b}\) ………… (ii)
Adding equations (i) and (ii), we get
\(\frac{z}{x}+\frac{z}{y}=\frac{b}{a+b}+\frac{a}{a+b}\)
or z(\(\frac{1}{x}+\frac{1}{y}\)) = \(\frac{a+b}{a+b}\)
or \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\) Hence Proved.
Question 30.
Find the value of
\(\frac{5 \sin ^2 30^{\circ}+\cos ^2 45^{\circ}-4 \tan ^2 30^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\tan 45^{\circ}}\) (3)
Answer:
Or
Given, cos θ = \(\frac{21}{29}\), determine the value of \(\frac{\sec \theta}{\tan \theta-\sin \theta}\). (3)
Answer:
Given, cos θ = \(\frac{21}{29}\)
Thus, base and hypotenuse are in the ratio 21:29.
Let us take a right ΔABC in which ∠ACB = θ.
Let BC = 21 k and AC = 29k, where k is any positive integers. Using Pythagoras theorem, we have
Question 31.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. Based on the following information, answer the following questions.
(i) Find the total length of the silver wire required.
(ii) Find the area of each sector of the brooch.
(iii) Find the total cost of silver wire, if cost of silver wire is ₹100 per mm.
OR
Find the area of the segment shown in Fig. if radius of the circle is 21 cm and ∠AOB = 120°. (Use π = \(\frac{22}{7}\))
Solution:
Given: Diameter = 35 mm
∴ Radius (r) = \(\frac{35}{2}\) rnm
(i) Silver wire used to make a brooch = Circumference of circle
= 2πr
= 2 × \(\frac{22}{7}\) × \(\frac{35}{2}\)
= 110 mm.
Wire used in 5 diameters = 5 × 35 mm
= 175 mm
Total wire used = 110 mm + 175 mm
= 285 mm
(ii) Area of circle = πr2
= \(\frac{22}{7}\) × \(\frac{35}{2}\) × \(\frac{35}{2}\) mm2
= \(\frac{11 \times 5 \times 35}{2}\) mm2
Area of each sector of brooch = \(\frac{11 \times 5 \times 35}{2} \times \frac{1}{10}\)
= \(\frac{385}{4}\) mm2
(iii) Cost of silver wire = ₹100 × 285
= ₹28500.
OR
Given, radius of the circle = 21 cm and ∠AOB = 120°
Area of the segment AYB = Area of sector AOB – Area of ΔAOB
Area of sector AOB = \(\frac{120^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 21 × 21
= 462 cm2
To find the area of ΔOAB, draw OM ⊥ AB
ΔAMO ≅ ΔBMO (by R.H.S)
∠AOM = ∠BOM = \(\frac{1}{2}\) × 120°
= 60°
In ΔOMA, \(\frac{\mathrm{OM}}{\mathrm{OA}}\) = cos 60°
⇒ \(\frac{\mathrm{OM}}{21}=\frac{1}{2}\)
⇒ OM = \(\frac{21}{2}\) cm
Also, \(\frac{\mathrm{AM}}{\mathrm{OA}}\) = sin 60°
⇒ AM = 21 × \(\frac{\sqrt{3}}{2}\)
∴ AB = 2 × AM = 21√3
since perpendicular from the centre bisects its chord.
So, Area of ΔOAB = \(\frac{1}{2}\) × AB × OM
= \(\frac{1}{2}\) × 21√3 × \(\frac{21}{2}\)
= \(\frac{441}{4}\)√3 cm2
∴ Area of segment = (462 – \(\frac{441}{4}\)√3) cm2
\(\frac{21}{4}\) (88 – 21√3) cm2
= 271.04 cm2
Section – D
Section D consists of 4 questions of 5 marks each.
Question 32.
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, then find the value of length of her shadow after 4s. (5)
Answer:
Let AB be the lamp-post, CD be the girl and D be the position of girl after 4s.
Again, let DE = x m be the length of shadow of the girl.
Given, CD = 90 cm = 0.9 m, AB = 3.6 m
and speed of the girl = 1.2 m/s
∴ Distance of the girl from lamp-post after 4s,
BD = 1.2 × 4 = 4.8 m
[∵ distance = speed × time]
In ΔABE and ΔCDE,
∠B = ∠D [Each 90°]
∠E = ∠E [common angle]
∴ ΔABE ~ ΔCDE [by AA similarity criterion]
⇒ \(\frac{B E}{D E}\) = \(\frac{A B}{C D}\) ……… (i)
[since, corresponding sides of similar triangles are pcoportional]
On substituting all the values in Eq. (i), we get
\(\frac{4.8+x}{x}\) = \(\frac{3.6}{0.9}\) [∵ BE = BD + DE = 4.8 + x]
⇒ \(\frac{4.8}{3}\) = 1.6 m
⇒ 4.8 + x = 4x
⇒ 3x = 4.8
⇒ x = \(\frac{4.8}{3}\)
Hence, the length of her shadow after 4s is 1.6 m.
Question 33.
Prove that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.
OR
Prove that tangents to a circle from an external point are equal in length.
Solution:
Given : AABC and a line DE intersecting AB at D and AC at E, such that
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
To prove : DE || BC
Construction : Draw DE || BC.
Pj-oof : ∵ DE’ || BC
If a line is drawn parallel to one side of a triangle to intersect other two sides on distinct points, the other two sides are divided in the same ratio.
OR
Given : PA and PB are tangents to a circle having centre O from an external point P.
To prove : PA = PB.
Construction : Join OA, OB and OP.
Proof : In ΔOAP and ΔOBP,
∠OAP = ∠OBP = 90° [Radius tangent angle]
OA = OB [Radius of the circle]
OP = OP [Common]
∴ ΔOAP ≅ ΔOBP [R.H.S. congruency rule]
⇒ AP = BP [C.P.C.T.] Hence proved.
Question 34.
Find the sum of first 51 terms of an AP -Those second and third terms are 14 and 18, respectively. (5)
Answer:
Here, n = 51, T2 = 14 and T3 = 18
Let the first term of The AP be a and the common difference is d.
We have, T2 ⇒ a + d = a + d = 14 ……(i)
T3 = a + 2d ⇒ a + 2d = 18 ……..(ii)
Subtracting Eq. (i) from Eq. (ii), we get
a + 2d – a – d = 18 – 14 ⇒ d = 4 (2)
From Eq. (i), we get a + d = 14
⇒ a + 4 = 14 ⇒ a = 14 – 4 = 10
Now, Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
S51 = \(\frac{51}{2}\)[(2 × 10) + (51 – 1) × 4]
= \(\frac{51}{2}\)[20 + 200] = \(\frac{51}{2}\) [220]
= 51 × 110 = 5610
Thus, the sum of 51 terms is 5610.
Or
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289. Find the sum of first n terms. (5)
Answer:
Here, we have S7 = 49 and S17 = 289
Let the first term of the AP be ‘a’ and ‘d’ be the common difference, then
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
⇒ S7 = \(\frac{7}{2}\)[2a + (7 – 1)d] = 49
⇒ 7(2a + 6d) = 2 × 49 = 98
⇒ 2a + 6d = \(\frac{98}{7}\) = 14 ⇒ 2[a + 3d] = 14
⇒ a + 3d = \(\frac{14}{2}\) = 7 ⇒ a + 3d = 7 …… (i)
Also, S17 = \(\frac{17}{2}\)[2a + (17 – 1)d] = 289
⇒ \(\frac{17}{2}\)(2a + 16d) = 289
⇒ a + 8d = \(\frac{298}{17}\) = 17 ⇒ a + 8d = 17 …….. (ii)
Subtracting Eq. (i) a from Eq. (ii), we have
a + 8d – a – 3d = 17 – 7
⇒ 5d = 10 ⇒ d = 2
Now, from Eq. (i), we have
a + 3(2) = 7
⇒ a = 7 – 6 = 1
Now, Sn = \(\frac{n}{2}\)[2a + (n – 1)d] = \(\frac{n}{2}\)[2 × 1 + (n – 1) × 2]
= \(\frac{n}{2}\)[2 + 2n – 2] = \(\frac{n}{2}\)[2n] = n × n = n2
Thus, the required sum of n terms = n2
Question 35.
Prove that:
\(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cos θ.
OR
Amit, standing on a horizontal plane, finds a bird flying at a distance of 200 m from him at an
elevation of 30°. Deepak standing on the roof of a 50 m high building, finds the angle of elevation
of the same bird to be 45°. Amit and Deepak are on opposite sides of the bird. Find the distance of
the bird from Deepak.
Solution:
OR
Let Amit be at C point and bird is at A point, such that ∠ACB = 30°. AB is the height of bird from point B on ground and Deepak is at D point, DE is the building of height 50m.
Now, in right ΔABC, we have
sin 30° = \(\frac{P}{H}=\frac{A B}{A C}\)
⇒ \(\frac{1}{2}=\frac{\mathrm{AB}}{200}\)
⇒ AB = 100 m
In right ΔAFD, we have
sin45° = \(\frac{\mathrm{AF}}{\mathrm{AD}}\)
⇒ \(\frac{1}{\sqrt{2}}=\frac{50}{\mathrm{AD}}\) (∵ AB = AF + BF, 100 = AF + 50, AF = 50)
⇒ AD = 50√2 m
Hence, the distance of bird from Deepak is 50√2 m .
Section – E
Question 36.
Morning Walk
In a morning walk, Pankaj, Saksham and Mayank step off together, their steps measuring 240 cm, 90 cm and 120 cm
respectively. They want to go for a juice shop for a health issue, which is situated near by them.
Based on the above information, answer the following questions.
(i) If a and b are two numbers, then find the relation between LCM and HCF. (1)
(ii) Find the minimum distance of the shop from where they start to walk together so that one can cover the distance in complete steps. (2)
Answer:
(i) The relation between LCM and HCF is LCM(a, b) × HCF (a, d) = a × b
(ii) Minimum required distance to reach the juice shop = LCM (240, 90, 120)
∴ 240 = 2 × 2 × 2 × 2 × 3 × 5
90 = 2 × 3 × 3 × 5
and 120 = 2 × 2 × 2 × 3 × 5
Now, LCM = 24 × 32 × 5
= 16 × 9 × 5 = 720
Hence, the required distance is 720 cm.
Or
Find the number of common steps cover by all of them to reach the juice shop. (2)
(iii) A largest positive integer that divides given two positive integers is called (1)
Answer:
The number of common steps cover by all of them = HCF (240, 90, 120)
Now, 240 = 2 × 2 × 2 × 2 × 3 × 5
90 = 2 × 3 × 3 × 5 and 120 = 2 × 2 × 2 × 3 × 5
So, HCF = 2 × 3 × 5 = 30
(iii) The largest positive integer that divides given two positive integers is called HCF.
Question 37.
A practical application of the trigonometry functions is to find the length that you cannot measure. Very frequently, angles of depression and elevation are used in these type of problems. Angle of depression is the angle measured from the horizon or horizontal line down.
A boy is riding in a hot air bolloon. When he is at the point A, he spotted a car parked in the ground at an angle of depression of 45°. After some time, his balloon rises to some of height and reached at A’. Now for the some point, the angle of depression changes to 60°.
Now, answer the following questions :
(i) What is the value of ∠AQP?
(ii) If the height of air-balloon at point A is ‘h’m from the ground. Find the distance of the point ‘A’ from the parked car? (in terms of h)
(iii) If the balloon rises to a height of 12 m, find the height of the balloon at the initial point i.e., height ‘h’ ?
OR
What is the final height of balloon after a lapse of time at point A?
Solution:
(i) Since, line AX || QP and QA is a transversal.
⇒∠XAQ = ∠AQP = 45° (Alternate pair of angles)
(ii) In ΔAPQ, sin 45° = \(\frac{\mathrm{AP}}{\mathrm{AQ}}\)
⇒ \(\frac{1}{\sqrt{2}}=\frac{h}{\mathrm{AQ}}\)
⇒ AQ = √2h
(iii) In ΔAPQ,
tan 45° = \(\frac{\mathrm{AP}}{\mathrm{PQ}}\)
⇒ AP = PQ = h ……. (i)
Now, in ΔA’PQ, tan 60° = \(\frac{A^{\prime} P}{P Q}\)
⇒ √3 = \(\frac{(h+12)}{h}\)
⇒ √3h – h = 12
⇒ h = \(\frac{12}{0.73}\) = 16.43 m
OR
As, h = 16.43
Then, A’P = 12 + 16.43
= 28.43 m
Question 38.
Telecasting Tower
A straight highway leads to the foot of a national communication and telecasting tower. A watchman standing at the top of the tower observes a car at an angle of depression of 300 which is approaching the foot of the tower with a uniform speed.
Two minutes later, the angle of depression was found to be 60°. The watchman suspects that some terrorists are approaching the tower. It needs half a minute for the watchman to inform the security staff so that it may alert them.
Based on the above information, answer the following questions.
(i) Find the angle of depression from the object to the point on the ground and the angle of elevation of the same point on the ground to the same object. (1)
(ii) The angle of an object viewed, is the angle formed by the line of sight with the horizontal. (1)
(iii) How much time the car will take to reach the foot of the tower? (2)
Answer:
(i) The angle of depression from the object to the point ‘ on the ground and the angle of elevation of the same point on the ground to the same object is equal.
(ii) The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal.
(iii) Let AB be the tower of height h m, C and D be the position of the car at an angle of depression of 30° and 60°, respectively.
Also, let the speed of the car be x m/min and y min will be taken by the car to cover distance BD.
Then, CD = 2x m
[∵ time taken to cover distance CD is 2 min and distance = speed × time]
and BD = xy m
⇒ 2 + y = 3y ⇒ 2 = 2y
∴ y = 1 min
Hence, the car will take 1 minute to reach the foot of the tower.
Or
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60°, respectively. Find the height of the tower. (2)
Answer:
Let h be the height of the transmission tower and x be the distance between the point and foot of the building.