Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 2 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 2 with Solutions

Time: 3 hrs

Max. Marks:80

Instructions:

- This question paper has 5 Sections A-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).
- All questions are compulsory. However, an internal choice in 2 questions of 2 marks, 2 questions of 3 marks and 2 questions of 5 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.

Prime factorisation of 120 is

(a) 2^{4} × 3 × 5

(b) 2^{3} × 3^{2} × 4

(c) 2^{3} × 3 × 5

(d) 2^{2} × 3 × 5

Answer:

(c) 2^{3} × 3 × 5

We have, 120 = 2 × 2 × 2 × 3 × 5

= 2^{3} × 3 × 5

Question 2.

The distance between the points A(a + b, a – b) and B(a – b, – a – b) is

(a) 2\(\sqrt{b^2+a^2}\)

(b) \(\sqrt{b^2+a^2}\)

(c) 2\(\sqrt{b^2-a^2}\)

(d) \(\sqrt{b^2-a^2}\)

Answer:

(a) 2\(\sqrt{b^2+a^2}\)

The distance between the points A(a + b, a – b) and B(a – b, -a – b) is

Question 3.

The ratio in which the points P(\(\frac{3}{4}\), \(\frac{5}{12}\)) divides the line segments joining the points A(\(\frac{1}{2}\), \(\frac{3}{2}\)) and B (2, – 5) is

(a) 5 : 1

(b) 1 : 6

(c) 1 : 5

(d) 6 : 1

Answer:

(c) 1 : 5

Let the point P divides the line joining the points AB in the ratio k : 1.

On comparing the x and y-coordinates, we get

\(\frac{3}{4}\) = \(\frac{4 k+1}{2 k+2}\) and \(\frac{5}{12}\) = \(\frac{-10 k+3}{2 k+2}\)

⇒ 6k + 6 = 16k + 4

⇒ 10k = 2 ⇒ k = \(\frac{1}{5}\)

Hence, point P divides the Pine segment AB in the ratio 1 : 5.

Question 4.

If sin(A -B) = sin A cos B – cos Asin B, then the value of sin 15° is

(a) \(\frac{\sqrt{3}-1}{\sqrt{2}}\)

(b) \(\frac{\sqrt{2}-1}{2 \sqrt{2}}\)

(c) \(\frac{1-\sqrt{3}}{2 \sqrt{2}}\)

(d) \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

Answer:

(d) \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

We have, sin(A – B) = sinAcosB – cosAsinB

Put A = 45° and B = 30°, we get

sin(45° – 30°) = sin45°cos30° – cos45°sin30°

= \(\frac{1}{\sqrt{2}}\) × \(\frac{\sqrt{3}}{2}\) – \(\frac{1}{\sqrt{2}}\) × \(\frac{1}{2}\)

\(=\frac{\sqrt{3}}{2 \sqrt{2}}\) – \(\frac{1}{2 \sqrt{2}}\) = \(\frac{1}{2 \sqrt{2}}\)

Question 5.

The value of p in which the system of linear equations -x + py = 1 and px – y = 1 represent parallel lines is

(a) 0

(b) 1

(c) -1

(d) 2

Answer:

(b) 1

Given, pair of Linear equation is

-x + py – 1 = 0 and px – y – 1 = 0

On comparing the given equations with standard form, i.e. a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0, we get

a_{1} = -1, b_{1} = p, c_{1} = -1

and a_{2} = p, b_{2} = -1, c_{2} = -1

Condition for parallel lines,

\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)

⇒ \(\frac{-1}{p}\) = \(\frac{p}{-1}\) ⇒ p^{2} = 1 ⇒ p = ±1

But p = -1 does not satisfy the condition \(\frac{p}{-1}\) ≠ \(\frac{-1}{-1}\), so we neglect it.

Hence, for p = 1, the given system of equations will represent parallel lines.

Question 6.

A point P is 45 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 36 cm. The radius of the circle is

(a) 27 cm

(b) 17 cm

(c) 37 cm

(d) 12 cm

Answer:

(a) 27 cm

We know that, the radius drawn from centre O to the tangent line is perpendicular.

Question 7.

The H.C.F and L.C.M of two numbers are 12 and 5040 respectively. If one of the numbers is 144 find the other number

(a) 420

(b) 202

(c) 101

(d) 111

Answer:

(a) 420

Given, HCF =12, LCM = 5040

and also one number is 144.

Let the other number be x.

As we know that

HCF × LCM = 144 × x

12 × 5040 = 144 × x

⇒ x = \(\frac{12 \times 5040}{144}\)

⇒ x = 420

Question 8.

The 10th term from the end of the AP sequence 4, 9, 14, ……., 254 is

(a) 208

(b) 210

(c) 209

(d) None of these

Answer:

(c) 209

Given, AP sequence is 4, 9, 14, ………, 254.

Here, a = 4, d = 9 – 4 = 5 and l = 254

∴ 1oth term from the end = l – (10 – 1)d

= 254 – (9) × 5

= 254 – 45 = 209

Question 9.

is equal to

(a) \(\frac{3}{2 \sqrt{2}}\)

(b) 0

(c) \(\frac{1}{2 \sqrt{2}}\)

(d) \(\frac{1}{2}\)

Answer:

(d) \(\frac{1}{2}\)

We have,

Question 10.

The zeroes of the polynomial f(x) = 4x^{2} + 8x are

(a) 2, 0

(b) -2, 2

(c) 0, 1

(d) 0, -2

Answer:

(d) 0, -2

Given, f(x) = 4x^{2} + 8x

The zeroes of f(x) are given by f(x) = 0

∴ 4x^{2} + 8x = 0 ⇒ x(4x + 8) = 0

⇒ x = 0 or 4x + 8 = 0 ⇒ x = 0 or x = -2

Hence, zeroes of f(x) are 0 and -2,

Question 11.

In the given figure, if TP and TQ are the two tangents to a circle with centre O, so .that ∠POQ = 120°. The value of ∠PTQ is

(a) 30°

(b) 120°

(c) 60°

(d) None of these

Answer:

(a) 30°

Given, TP and TQ are two tangents to a circle and ∠POQ = 120°. Join PQ.

Question 12.

The coordinate of the point, in which the diagonals of the parallelogram formed by joining the points (-2, -1), (1,0), (4, 3) and (1, 2) intersect are

(a) (0, 1)

(b) (1, 1)

(c) (1, 2)

(d) None of these

Answer:

(b) (1, 1)

Given, vertices of a parallelogram are A(-2, -1) B(1, 0), C(4, 3) and D(1, 2).

We know that diagonals of a parallelogram intersect at mid-point.

∴ The coordinate of intersection point of diagonal = mid-point of AC

= \(\left(\frac{-2+4}{2}, \frac{-1+3}{2}\right)\) = \(\left(\frac{2}{2}, \frac{2}{2}\right)\) = (1, 1)

Question 13.

The exponent of 2 factorisation of 1440 is

(a) 5

(b) 6

(c) 3

(d) 4

Answer:

(a) 5

The prime factorisation of 1440 is

1440 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5

= 2^{5} × 3^{2} × 5^{1}

Hence, the exponent of 2 in the prime factorisation of given number is 5.

Question 14.

The probability that a number selected from the numbers 1, 2, 3, …, 15 is a multiple of 4 is

(a) 0

(b) 1

(c) \(\frac{1}{10}\)

(d) \(\frac{1}{5}\)

Answer:

(d) \(\frac{1}{5}\)

The possible outcomes = 15

The number of favourable multiple of 4 = {4, 8, 12}

∴ P(getting a multiple of 4) = \(\frac{3}{15}\) = \(\frac{1}{5}\)

Question 15.

The value of cos^{2}θ + \(\frac{1}{1+\cot ^2 \theta}\) is

(a) 0

(b) 1

(c) -1

(d) None of these

Answer:

(b) 1

[∵ cosec^{2}A – cot^{2}A = 1]

= cosec^{2}θ + sin^{2}θ = 1

Question 16.

The quadratic equation 9x^{2} + 6kx + 4 = 0 has equal roots. Then, the value of k is

(a) 2, -2

(b) 2, 2

(c) 0, 2

(d) – 2, 0

Answer:

(a) 2, -2

Given, quadratic equation is 9x^{2} + 6kx + 4 = 0.

On comparing with ax2 + bx + c = 0, we get

a = 9, b = 6k and c = 4

The condition for quadratic equation has equal roots i.e. b^{2} – 4ac = 0

i.e., b^{2} – 4ac = 0

∴ (6k)^{2} – 4 × 9 × 4 = 0

⇒ 36k^{2} – 144 = 0

⇒ k^{2} = \(\frac{144}{36}\)

⇒ k^{2} = 4

⇒ k = ±2 [taking square roots]

Question 17.

The median of the first 10 prime numbers is

(a) 12

(b) 10

(c) 24

(d) None of these

Answer:

(a) 12

First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Here, n = 10 which is even, therefore

Question 18.

A pair of dice is thrown once, then the probability of getting an even doublet is

(a) 0

(b) 1

(c) \(\frac{1}{12}\)

(d) None of these

Answer:

(c) \(\frac{1}{12}\)

When two dice are thrown, then possible number of outcomes = 6 × 6 = 36

The number of favourable even doublets = {(2, 2), (4, 4), (6, 6)}

∴ P(getting a doublet) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

Directions In the 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 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.

Question 19.

Assertion (A) If the outer and inner diameter of a circular path is 10 m and 6 m, then area of the path is 16 π m^{2}.

Reason (R) If R and r be the radius of outer and inner circular path respectively, then area of path = π(R^{2} – r^{2}).

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

∴ Area of the path = π\(\left[\left(\frac{10}{2}\right)^2-\left(\frac{6}{2}\right)^2\right]\)

= π(25 – 9)

= 16π m^{2}

Hence, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Question 20.

Assertion (A) If in a ∆ABC, a line DE || BC, intersects AB at D and AC at E, then \(\frac{A B}{A D}\) = \(\frac{A C}{A E}\)

Reason (R) If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Reason is true,

For Assertion (A),

Since, DE || BC

∴ By Thales Theorem

∴ Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).

Section B

(Section B consists of 5 questions of 2 marks each)

Question 21.

5 yr hence, the age of Shivangi small be 3 times the age of Anshika while 5 yr earlier the age of Shivangi was 7 times the age of Anshika. Find the present age of Shivangi.

Answer:

Let the present age of Shivangi = x

and present age of Anshika = y

After 5 yr, age of Shivangi = (x + 5)

After 5 yr, age of Anshika = (y + 5)

According to the question

x + 5 = 3(y + 5)

⇒ x + 5 = 3y + 15

⇒ x – 3y = 10

⇒ x = 10 + 3y …….. (i)

and 5 yr earlier

Age of Shivangi = x – 5

and age of Anshika = y – 5

According to the question,

x – 5 = 7(y – 5)

⇒ x – 7y = – 30 …… (ii)

Subtracting Eq. (i) from Eq. (ii), we get

– 4y = – 40

⇒ y = 10

On substituting the value of y in Eq. (i), we get

10 + 3 × 10 = 40

‘Hence, present age of Shivangi = 40 yr.

Question 22.

Without solving the following quadratic equation, find the value of m for which the given equation has real and equal roots.

x^{2} + 2 (m -1 )x + (m + 5) = 0

Answer:

Given, quadratic equation is

x^{2} + 2 (m – 1)x + (m + 5) = 0

On comparing it with ax^{2} + bx + c = 0, we get

a = 1, b = 2(m – 1) and c = m + 5

Since, the given equation has real and equal roots.

So, the discriminant will be zero

i.e. b^{2} – 4ac = 0

⇒ [2(m – 1)]^{2} – 4 × 1 × (m + 5) = 0

⇒ 4(m – 1)^{2} – 4(m + 5) = 0

⇒ 4[m^{2} + 1^{2} – 2m – m – 5] = 0

[∵ (a – b)^{2} = a^{2} + b^{2} – 2ab]

⇒ m^{2} – 3m – 4 = 0 [dividing by 4]

⇒ m^{2} – 4m + m – 4 = 0

[splitting the middle term]

⇒ m(m – 4) + 1(m – 4) = 0

⇒ (m + 1)(m -4) = 0

⇒ m + 1 = 0 or m – 4 = 0

m = -1 or m = 4

Hence, the values of m are -1 and 4.

Question 23.

In the figure, quadrilateral ABCD is circumscribing a circle with centre O and AD ⊥ AB. If radius of incirclte is 10 cm, then find the value of x.

Or

In the figure, two tangents TP and TQ are drawn to a circle with centre O from an external point T.

Prove that ∠PTQ = 2∠OPQ.

Answer:

Given, in radius of circle, r = 10 cm.

Draw a line perpendicular from centre to the tangent line AB.

∠A = ∠OPA = ∠OSA = 90°

[∵ all lines in quadrilateral APOS are perpendicular]

∠SOP = 90°

Also, AP = AS

[∵ pair of tangents drawn from outside point of a circle are equal in lengths]

Hence, OSAP is i i square.

AP = OS = 10 cm

CR = CQ = 27 cm

Now, BQ = BC – CQ

= 38 – 27 = 11 cm

Here, BP = BQ = 11 cm

∴ x = AB = AP + BP = 10 + 11 = 21 cm

Or

Given, TP and TO are two tangents of a circle with centre O and P and Q point of contact, let ∠PTQ = 0 As we know that the length of tangents drawn from an external point to a circle are equal.

So, ∆TPQ is an isosceles triangle

As we know that the tangents at any point of a circle is perpendicular to the radius through the point of contact.

∴ ∠OPT = 90°

Now, ∠OPQ = ∠OPT – ∠TPQ

= 90° – (90° – \(\frac{\theta}{2}\)) = \(\frac{\theta}{2}\) = \(\frac{\angle P T Q}{2}\)

⇒ ∠PTQ = 2∠OPQ

Hence proved.

Question 24.

In the given figure DE || BC find EC.

Also given AD = 1.5 cm, DB = 3 cm and AE = 1 cm.

Answer:

Since, DE || BC [given]

∴ Using the Basic Proportionality theorem.

We have, \(\frac{A D}{D B}\) = \(\frac{A E}{E C}\)

Since, AD = 1.5 DB = 3 cm and AE = 1 cm

Question 25.

If tan θ = \(\frac{7}{13}\), then find the value of \(\frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}\).

Or

If cos θ = \(\frac{b}{\sqrt{a^2+b^2}}\), 0 < θ < 90°, find the value of sin θ and tan θ.

Answer:

Section C

(Section C consists of 6 questions of 3 marks each)

Question 26.

Find a relation between x and y, such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).

Or

Find the ratio in which the point (-3, p) divides the line segment joining the points (-5, -4) and (-2, 3). Also, determine the value of p.

Answer:

Let the point A(x, y) be equidistant from the points B(3, 6) and C(-3, 4).

Or

Suppose the point P(-3, p) divides the line segment joining the point A(-5, -4)and B(-2, 3) in the ratio k : 1.

Then, the coordinates of P are \(\left(\frac{-2 k-5}{k+1}, \frac{3 k-4}{k+1}\right)\)

But, the coordinates of P are given as (-3, p).

∴ \(\frac{-2 k-5}{k+1}\) = -3 and \(\frac{3 k-4}{k+1}\) = p

⇒ -2k – 5 = -3k – 3

⇒ k = 2

Hence, the point P divides the line segment joining the points A and B in the ratio 2 : 1.

Now, consider p = \(\frac{3 k-4}{k+1}\)

p = \(\frac{3 \times 2-4}{2+1}\)

= \(\frac{6-4}{3}\) = \(\frac{2}{3}\)

Question 27.

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a queen of red colour.

(ii) a face card.

(iii) a black face card.

Answer:

Total number of cards in one deck of cards is 52.

∴ Total number of possible outcomes = 52

(i) Let E_{1} = Event of getting a queen of red colour

∴ Number of outcomes favourable to E_{1} = 2

[∵ there are four queen in a deck of playing cards out of which two are red and two are black]

Hence, probability of getting a queen of red colour,

P(E_{1}) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(ii) Let E_{2} = Event of getting a face card

∴ Number of outcomes favourable to E_{2} = 12

[∵ in a deck of cards, there are 12 face cards, namely 4 kings, 4 jacks, 4 queens]

Hence, probability of getting a face card,

P(E_{2}) = \(\frac{12}{52}\) = \(\frac{3}{13}\)

(iii) Let E_{3} = Event of getting a black face card

∴ Number of outcomes favourable to E_{3} = 6

[∵ in a deck of cards, there are 12 face cards out of which 6 are red and 6 are black]

Hence, probability of getting a black face card,

P(E_{3}) = \(\frac{6}{52}\) = \(\frac{3}{26}\)

Question 28.

Prove that 7 – 2\(\sqrt{5}\) is irrational.

Answer:

Let us assume to the contrary that 7 – 2\(\sqrt{5}\) is a rational number.

Then, it can be expressed in the form \(\frac{a}{b}\), where a and b are coprime integers and b ≠ 0.

Now, 7 – 2\(\sqrt{5}\) = \(\frac{a}{b}\), where a and b are integers and b ≠ 0.

⇒ 2\(\sqrt{5}\) = 7 – \(\frac{a}{b}\) ⇒ \(\sqrt{5}\) = \(\frac{7}{2}\) – \(\frac{a}{2b}\)

Since, a and b are integers b ≠ 0, therefore \(\frac{a}{2b}\) is rational number and so \(\frac{7}{2}\) – \(\frac{a}{2b}\) is a rational number as difference of two rational number is also a rational number,

⇒ \(\sqrt{5}\) is a rational number.

But this contradicts the fact that \(\sqrt{5}\) is an irrational number.

This shows that our assumption is incorrect.

So, 7 – 2\(\sqrt{5}\) is irrational. Hence proved.

Question 29.

In a class test, the sum of the marks obtained by Shyam in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks USS in Science, the product of marks obtained in two subjects would have been 180. Find the marks obtained by him in the two subjects separately.

Answer:

Suppose, Shyam gets x marks in Mathematics and y marks in Science. Then,

x + y = 28 ……… (i)

Again, (x + 3)(y – 4) = 180 ……… (ii)

On putting y = (28 – x) from Eq. (i) in Eq. (ii), we get

(x + 3)(28 – x – 4) = 180

⇒ (x + 3)(24 – x) = 180

⇒ -x^{2} + 21x + 72 = 180

⇒ x^{2} – 21x + 108 = 0

⇒ (x – 12) (x – 9) = 0

⇒ x -12 = 0 or x – 9 = 0

⇒ x = 12 or x = 9

Now, x = 12, then y = 28 -12 = 16

and when x = 9, then y = 28 – 9 = 19

∴ Either he got 12 marks in Mathematics and 16 marks in Science or he got 9 marks in Mathematics and 19 marks in Science.

Question 30.

From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB.

Answer:

Given P is an external point and two tangents PA and PB are drawn.

To prove OP is the perpendicular bisector of AB.

Proof Suppose OP intersects AB at C. In triangles ∆PAC and ∆PBC, we have PA = PB

[tangents drawn from an external point are equal]

∠APC = ∠BPC

[tangents PA and PB are equally inclined to OP]

and PC = PC [common side]

∴ ∆PAC\(\cong\) ∆PBC [by SAS congruency criterion]

⇒ AC = BC

and ∠ACP = ∠BCP [byCPCT]

But ∠ACP + ∠BCP = 180°

⇒ ∠ACP = ∠BCP = 90°

Hence, OP ⊥ AB

Hence proved.

Question 31.

Prove that

\(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) = cosec A + cot A

Using the identity cosec^{2} A – cot^{2} A = 1.

Or

Evaluate \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}-2}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}+1}\)

Answer:

Or

Section D

(Section D consists of 4 questions of 5 marks each)

Question 32.

ABCD is a trapezium in which AB||DC and its diagonal intersect each other at the point P. Show that \(\frac{A P}{B P}\) = \(\frac{C P}{D P}\).

Answer:

We have a trapezium ABCD such that AB || DC. The diagonals AC and BD intersect each other at P.

Let us draw PE parallel to either AB or DC.

In ∆ADC PE || DC [by Construction]

∴ Using the Basic proportionality theorem, we get

\(\frac{A E}{E D}\) = \(\frac{A P}{C P}\)

Question 33.

Find the missing frequencies f_{1} and f_{2} in the following frequency distribution table, if N = 100 and median is 32.

Or

The table below shows the salaries of 280 persons.

Calculate

(i) median of the data.

(ii) mode of the data.

Answer:

Given, median = 32 and N = Σf =100

Since, sum of frequencies =100

∴ 10 + f_{1} + 25 + 30 + f_{2} + 10 = 100

⇒ f_{1} + f_{2} = 100 – 75

⇒ f_{1} + f_{2} = 25

⇒ f_{2} = 25 – f_{1} ……. (i)

Now, the cumulative frequency table distribution is

Here, N = 100 ⇒ \(\frac{N}{2}\) = 50

Given, median = 32, which belongs to the class 30 – 40.

So, the median class is 30 – 40.

Then, l = 30, h = 10, f = 30 and cf = 35 + f_{1}

On putting the value of f_{1} in Eq. (i), we get

f_{2} = 25 – 9 = 16

Hence, the missing frequencies are

f_{1} = 9 and f_{2} = 16

Or

First, we construct a cumulative frequency table

∴ \(\frac{N}{2}\) = \(\frac{280}{2}\) = 140

(i) Here, median class is 10-15, because 140 lies in it.

∴ l = 10, f = 133, cf = 49 and h = 5

(ii) Here, the highest frequency is 133, which lies in the interval 10-15, called modal class.

∴ l = 10, h = 5, f_{1} = 133, f_{0} = 49 and f_{2} = 63

Hence, the median and modal salary are ₹ 13421 and ₹ 12727, respectively.

Question 34.

Find the area of sector OAYB and area of triangle AOB shown in figure, if radius of the circle is 56 cm and ∠AOB = 120°.

[Take, π = \(\frac{22}{7}\)]

Answer:

Given, r = 56cm and ∠AOB =120°

From Eq. (i),

AM = \(\frac{1}{2}\)AB

⇒ 2AM = AB

⇒ AB = 2AM ……….(iii)

From Eqs. (ii) and (iii), we get

AB = 2 × \(\frac{\sqrt{3}}{2}\) × 56

⇒ AB = 56\(\sqrt{3}\)

Now, area (∆MOB) = \(\frac{1}{2}\) × AB × OM

= \(\frac{1}{2}\) × 56\(\sqrt{3}\) × \(\frac{56}{2}\)

= 784\(\sqrt{3}\) cm^{2}.

Question 35.

The sum of n terms of three arithmetic progressions are S_{1}, S_{2} and S_{3}. The first term of each is unity and the common difference are 1, 2, 3 respectively. Prove that S_{1} + S_{3} = 2 S_{2}

Or

The ratio of the sums of m and n terms of an AP is m^{2} : n^{2}. Show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1).

Answer:

Given, S_{1} = Sum of n terms of an AP with first term 1 and common difference 1.

Or

Let a and d be the first term and common difference of an AP. Then, the sums of m and n terms are

Replacing on both sides m by 2m – 1 and n by 2n – 1, we get

Section E

(Case study based questions are compulsory)

Question 36.

Applications of Parabolas-Highway Overpasses/Underpasses

A highway underpass is parabolic in shape.

A parabola is the graph that result from p(x) = ax^{2} + bx + c. Parabolas are symmetric about a vertical line known as the Axis of Symmetry.

The axis of symmetry runs through the maximum or minimum point of the parabola which is called the vertex.

On the basis of above information, answer the following questions.

(i) If the highway overpass is represented by x^{2} – 2x – 8. Then, find its zeroes. (2)

Or

What is the representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0? (2)

(ii) Graph of a quadratic polynomial is a ……….. (1)

(iii) Find the number of zeroes of the polynomial f(x) = (x – 2)^{2} +2. (1)

Answer:

(i) Let p(x) = x^{2} – 2x – 8

= x^{2} – (4 – 2)x – 8

[splitting middle term]

= x^{2} – 4x + 2x – 8

= x(x – 4) + 2(x – 4)

= (x + 2)(x – 4)

For finding the zeroes, put p(x) = 0

⇒ (x + 2)(x – 4) = 0

⇒ x + 2 = 0 or x – 4 = 0

⇒ x = -2 or x = 4

Or

Given, one of the zeroes of a polynomial is α = 6

and sum of zeroes, α + β = 0

⇒ 6 + β = 0

⇒ β = -6

Now product of zeroes,

αβ = 6 × (-6) = -36

Now, polynomial is

x^{2} – (sum of zeroes)x + product of zeroes

∴ x^{2} – (0)x + (-36) or x^{2} – 36

Hence, the representation of highway underpass is x^{2} – 36.

(ii) The graph of a quadratic polynomial is the shape of parabola.

(iii) We have, f(x) = (x – 2)^{2} + 2

⇒ f(x) = x^{2} + 4 – 4x + 2

⇒ f(x) = x^{2} – 4x + 6

On comparing ax^{2} + bx + c = 0, we get

a =1, b = -4 and c = 6

Now, b^{2} – 4ac = (-4)^{2} – 4 × 1 × 6

= 16 – 24

= -8 < 0

Hence, number of zeroes of given polynomial is zero.

Question 37.

School Trophy

A school decide to give a trophy of the best student in the class, which is the form of cylinder mounted on a solid hemisphere with the same radius and is made from some metal. This trophy is mounted on a wooden cuboid as shown in the figure.

Suppose the diameter of the hemisphere is 24 cm and total height of the trophy is 28 cm.

On the basis of above information, answer the following questions.

(i) Find the curve surface area of cylinder. (1)

(ii) Find the volume of cylinder. (1)

(iii) Find the curved surface area of the trophy. (2)

Or

Find the volume of the trophy. (2)

Answer:

Given, diameter of hemisphere,

d = 24 cm

Radius,

r = \(\frac{d}{2}\) = \(\frac{24}{2}\) = 12 cm

Now, height of cylinder,

h = 28 – 12

= 16 cm

(iii) Curved surface area of trophy = Curved surface area of cylinder + Curved surface area of hemisphere

= 2πrh + 2πr^{2} = 2πr(h+ r)

= 2 × \(\frac{22}{7}\) × 12(16+ 12)

= \(\frac{528 \times 28}{7}\) = 2112 cm^{2}

Or

Question 38.

A satellite flying at height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka, them being Nanda Devi (height 7,816 m) and Mullayanagiri (height 1,930 m). The angles of depression from the satellite, to the top of Nanda Devi and Mullayanagiri are 30° and 60° respectively. If the distance between the peaks of two mountains is 1937 km and the satellite is vertically above the midpoint of the distance between the two mountains.

(i) Find the distance of the satellite from the top of Nanda Devi.

(ii) Find the distance of the satellite from the top of Mullayanagiri.

(iii) Find the distance of the satellite from the ground.

Or

What is the angle of elevation if a man is standing at a distance of 7816 m from Nanda Devi?

Answer:

(ii) We have, HP = 1S = 968500 m

Now, in ∆FHP,

cos 60° = \(\frac{H P}{F P}\)

\(\frac{1}{2}\) = \(\frac{968500}{F P}\)

FP = 968500 × 2 = 1937000m

= 1937 km

(iii) In ∆FAG, tan 30° = \(\frac{F G}{A G}\)

⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{F G}{968500}\)

FG = \(\frac{968500}{\sqrt{3}}\)

= 559180.138 m

= 559.18 km

Height of satellite from ground

= FI = FG + GI

= 559.18 + 7.816 [∵ GH = AD = 7816 m]

= 566.996 km

Or

Let E be the position of man.