Maharashtra Board SSC Class 10 Maths 2 Question Paper 2024 with Solutions Answers Pdf Download.

## SSC Maths 2 Question Paper 2024 with Solutions Pdf Download Maharashtra Board

Time: 2 Hours

Max. Marks: 40

General Instructions:

- All questions are compulsory.
- Use of a calculator is not allowed.
- The numbers to the right of the questions indicate full marks.
- In case of MCQs [Q. No. 1(A)] only the first attempt will beevaluated and will be given credit.
- Draw proper figures wherever necessary.
- The marks of construction should be clear. Do not erase them.
- Diagram is essential for writing the proof of the theorem.

Question 1.

(A) Four alternative answers for each of the following sub-questions are given. Choose the correct alternative and write its alphabet: [4]

(1) Out of the dates given below which date constitutes a Pythagorean triplet ?

(A) 15/8/17

(B) 16/8/16

(C) 3/5/17

(D) 4/9/15

Answer:

(A) 15/8/17

Explanation: Since, 172 = 152 + 8

2. sin 0 x cosec 0 = ?

(A) 1

(B) 0

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

(D) \(\sqrt{2}\)

Answer: (A) 1

Explanation : Since, by reciprocal relation sin 0 = \(\frac{1}{cosec θ}\)

(3) Slope of X-axis is ……………….

(A) 1

(B) -1

(C) 0

Answer:

(C) 0

Explanation: Since, X-axis is inclined at 0°.

(4) A circle having radius 3 cm, then the length of its largest chord is …………..

(A) 1.5 cm

(B) 3 cm

(C) 6 cm

(D) 9 cm

Answer: (C) 6 cm

Explanation : The largest chord of a circle is its diameter.

(B) Solve the following sub-questions:

(1) If AABC ~ APQR and AB : PQ = 2:3, then find the value of \(\frac{\mathbf{A}(\triangle \mathrm{ABC})}{\mathbf{A}(\triangle \mathrm{PQR})}\)

Solution:

Given ∆ABC ~ ∆PQR

and AB : PQ = 2 : 3

Since, \(\frac{\mathbf{A}(\triangle \mathrm{ABC})}{\mathbf{A}(\triangle \mathrm{PQR})}\) = \(\frac{(\mathrm{AB})^2}{(\mathrm{PQ})^2}\) [Area of similar triangles]

\(\frac{\mathbf{A}(\triangle \mathrm{ABC})}{\mathbf{A}(\triangle \mathrm{PQR})}\) = \(\frac{(\mathrm{2})^2}{(\mathrm{3})^2}\)

\(\frac{\mathbf{A}(\triangle \mathrm{ABC})}{\mathbf{A}(\triangle \mathrm{PQR})}\) = \(\frac{4}{9}\)

(2) Two circles of radii 5 cm and 3 cm touch each other externally. Find the distance between their centres.

Solution:

Let, the radii of two circles be

r_{1} = 5 cm and r_{2} = 3 cm

Then, distance between their centres = r_{1} + r_{2}

= 5 + 3 = 8 cm

Hence, distance between their centres is 8 cm.

(3) Find the side of a square whose diagonal is 10\(\sqrt{2}\) cm.

Solution:

Let ⇄ABCD be the square and AC be the diagonal of length 10\(\sqrt{2}\) cm.

Let side of squqre be x.

In ADC

∠ADC = 90° (By Pythagoras theorem)

AC^{2} = AD^{2} + DC^{2}

⇒ (10\(\sqrt{2}\))^{2} = x^{2} + x^{2}

⇒ 100 × 2 = 2x^{2}

⇒ 200 = 2x^{2}

⇒ x^{2} = \(\frac{200}{2}\)

⇒ x^{2} = 100

Taking square root on both sides

x = 10 cm

Hence, side of the square = 10 cm.

(4) Angle made by the line with the positive direction of X-axis is 45°. Find the slope of that line.

Solution:

Since, Slope = tan θ

Here, θ = 45°

∴ Slope = tan 45°

Hence, Slope = 1

Question 2.

(A) Complete any two activities and rewrite it: [4]

In the above figure, ∠ABC is inscribed in arc ABC.

If ∠ABC = 60°, find m∠AOC.

Img 3

Solution:

∠ABC = \(\frac{1}{2}\)m(arc AXC)

60° = \(\frac{1}{2}\)m(arc AXC)

__120°__ = m(arc AXC)

But m∠AOC = m(arc __AXC__)

∴ m∠AOC = __120°__

(2) Find the value of sin2 0 + cos2 0.

Solution:

In ∆ABC,

∠ABC = 90°, ∠C = 0°

AB^{2} + BC^{2} = __AC ^{2}__ ……(Pythagoras theorem)

Divide both sides by AC

^{2}

\(\frac{\mathrm{AB}^2}{\mathrm{AC}^2}\) + \(\frac{B C^2}{A C^2}\) = \(\frac{\mathrm{AC}^2}{\mathrm{AC}^2}\)

\(\left(\frac{\mathrm{AB}}{\mathrm{AC}}\right)^2\) + \(\left(\frac{\mathrm{BC}}{\mathrm{AC}}\right)^2\) = 1

But \(\frac{AB}{AC} =

__sin θ__and [latex]\frac{BC}{AC}\) =

__cos θ__

∴ sin

^{2}θ + cos

^{2}θ = 1

Question 3.

In the figure given above, ABCD is a square and a circle is inscribed in it. All sides of a square touch the circle. If AB = 14 cm, find the area of shaded region.

Solution:

Area of square = (__Side__)^{2} ….(Formula)

= 14^{2}

= __196__ cm^{2}

Area of circle = __πr²__ ….(Formula)

= \(\frac{22}{7}\) × 7 × 7

= 154 cm^{2}

(Area of shaded portion) = (Area of square) – (Area of circle)

= 196 – 154

= __42__ cm^{2}

(B) Solve any four of the following sub-questions: [8]

(1) Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.

Solution:

Radius of circle, r = 3.5 cm

Length of its arc, l = 2.2 cm

Then, Area of sector = \(\frac{l×r}{2}\)

= \(\frac{2.2×3.5}{2}\)

= 1.1 × 3.5

= 3.85 cm²

area of sector = 385 cm²

(2) Find the length of the hypotenuse of a right-angled triangle if remaining sides are 9 cm and 12 cm.

Solution:

Let ∆ ABC be the right angled triangle, right angled at B.

Then, BC = 9 cm, AB = 12 cm

In ∆ ABC, ∠ABC = 90°

By Pythagoras theorem,

AC^{2} = BC^{2} + AB^{2}

AC^{2} = (9)^{2} + (12)^{2}

AC^{2} = 81 + 144

AC^{2} = 225

Taking square root on both sides,

AC = 15 cm

Length of the hypotenuse = 15 cm

Question (3).

In the above figure, m(arc NS) = 125°, m(arc EF) = 37°. Find the measure of ∠NMS.

Solution :

Given, m(arc NS) = 125°

m(arc EF) = 37°

m∠NMS = \(\frac{1}{2}\) [m(arc NS) – m(arc EF)]

= \(\frac{1}{2}\) [125°-37°]

= \(\frac{1}{2}\) × 88°

= 44°

∴ m∠NMS = 44°.

Question (4)

Find the slope of the line passing through the points A(2,3), B(4, 7). Solution: For the given points A(2,3), B(4, 7).

A(2,3) = A(x_{1}, y_{1})

B(4, 7) = B(x_{2}, y_{2})

Slope of line AB = \(\frac{y_2-y_1}{x_2-x_1}\)

= \(\frac{7-3}{4-2}\)

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

Slope of line AB = 2.

(5) Find the surface area of a sphere of radius 7 cm.

Solution:

Given : radius of sphere, r = 7 cm

Then, surface area of sphere = 4πr²

= 4 × \(\frac{22}{7}\) × 7 × 7

= 88 × 7 = 616 cm²

surface area of sphere = 616 cm².

Question 3.

(A) Complete any one activity of the following and rewrite it: [3]

(1)

In ∆ABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing \(\frac{AB}{BC}=\frac{AE}{EB}\), complete the following activity:

Proof : In AABC, ray BD bisects ∠B

Solution:

Proof : In AABC, ray BD bisects ∠B

\(\frac{AB}{BC}=\frac{AE}{EB}\) …………..(I) [__Angle Bisector theorem__]

In ∆ABC

DE || BC

∴ \(\frac{AE}{EB}=\frac{AD}{DC}\) …………(II) [__Basic proportionality theorem__]

∴ \(\frac{AB}{BC}=\frac{AE}{EB}\) ….[from (I) and (II)]

(2)

Given : Chords AB and CD of a circle with centre P intersect at point E.

To prove: AE × EB = CE × ED

Construction:

Draw seg AC and seg BD.

Fill in the blanks and complete the proof.

Proof: In ∆CAE and ∆BDE,

Solution:

Proof: In ∆CAE and ∆BDE,

∠AEC ≅ ∠DEB …__(vertically opposite angle)__

∠CAE ≅ ∠BDE ……..(angles inscribed in the same arc)

∆CAE ~ ∆BDE … __[By AA test of similarity]__

\(\frac{AE}{DE}=\frac{CE}{EB}\) ………__(Corresponding sides of similar triangle)__

∴ AE x EB = CE x ED.

(B) Solve any two of the following sub-questions : [6]

(1) Determine whether the points are collinear.

A(1, – 3), B(2, – 5), C(- 4, 7)

Solution :

Let A(1, – 3) = A(x_{1}, y_{1})

B(2, – 5) = B(x<su\(\frac{1}{2}\)b>2, y_{2})

C(- 4, 7) = C(x_{3}, y_{3})

d(A, B) = \(\sqrt{(2-1)^2+(-5+3)^2}\)

= \(\sqrt{(1)^2+(-2)^2}\)

= \(\sqrt{1+4}\)

= \(\sqrt{5}\)

d (B, C) = \(\sqrt{(-4-2)^2+(7+5)^2}\)

= \(\sqrt{(-6)^2+(12)^2}\)

= \(\sqrt{36+144}\)

= \(\sqrt{180}\)

= \(6 \sqrt{5}\)

and d (A, C) = \(\sqrt{(-4-1)^2+(7+3)^2}\)

= \(\sqrt{(-5)^2+(10)^2}\)

= \(\sqrt{25+100}\)

= \(\sqrt{125}\)

= \(5 \sqrt{5}\)

Therefore, d(A, B) + d(A, C) = d(B, C)

⇒ Points A, B and C are collinear.

2) ∆ABC ~ ∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ∆ABC and BC

∆LMN such that \(\frac{BC}{MN}=\frac{4}{5}\)

Solution:

Given, ∆ABC ~ ∆LMN

\(\frac{AB}{LM}=\frac{BC}{MN}=\frac{AC}{LN}\)

\(\frac{AB}{LM}=\frac{BC}{MN}=\frac{AC}{LN}\) = \(\frac{4}{5}

AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm AB 5

Also, Given [latex]\frac{AB}{LM}=\frac{5}{4}\)

\(\frac{5.5}{LM}=\frac{5}{4}\)

LM = \(\frac{5.5×4}{5}\)

LM = 1.1 × 4

LM = 4.4 cm

\(\frac{BC}{MN}=\frac{5}{4}\)

\(\frac{5}{MN}=\frac{5}{4}\)

MN = \(\frac{6×4}{5}\)

MN = \(\frac{24}{5}\)

MN = 4.8 cm

\(\frac{AC}{LN}=\frac{5}{4}\)

\(\frac{4.5}{LN}=\frac{5}{4}\)

LN = \(\frac{4.5×4}{5}\)

LN = 0.9 × 4

LN = 3.6 cm

In ∆ABC AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm

Then, in ∆LMN LM = 4.4 cm, MN = 4.8 cm, LN = 3.6 cm.

Constrution:

(3) Seg PM is a median of ∆PQR, PM = 9 and PQ^{2} + PR^{2} = 290, then find QR.

Solution:

In ∆PQR, M is the mid-point of line QR.

∴ QM =MR …(i)

By Apollonius theorem,

PQ^{2} + PR^{2} = 2PM^{2} + 2QM^{2}

⇒ 290 = 2(9)^{2} + 2QM^{2}

⇒ 290 = 2(81) + 2QM^{2}

⇒ 290 = 162 + 2QM^{2}

⇒ 290 – 162 = 2QM^{2}

⇒ 128 = 2QM^{2}

⇒ \(\frac{128}{2}\) = QM^{2}

⇒ QM^{2} = 64

Taking square root on both sides

QM = 8 units

QM = MR = 8 units [from (i)]

QR = QM + MR [∵ QNR is a straight line]

QR = 8 + 8

QR = 16 units

QR = 16 units.

(4) Prove that, ‘If a line parallel“to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the side in the same proportion.

Solution:

Given : In ∆ABC, line l || line BC and line l intersects AB and AC at point P and Q respectively.

To Prove: \(\frac{\mathrm{AP}}{\mathrm{~PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}\)

Construction : Draw seg PC and seg BQ.

Proof: AAPQ and APQB have equal heights.

\(\frac{\mathrm{A}(\Delta \mathrm{APQ})}{\mathrm{A}(\Delta \mathrm{PQB})}\) = \(\frac{\mathrm{AP}}{\mathrm{~PB}}\) ………(areas proportionate to bases) …(i)

Similarly, \(\frac{\mathrm{A}(\Delta \mathrm{APQ})}{\mathrm{A}(\Delta \mathrm{PQB})}\) = \(\frac{\mathrm{AQ}}{\mathrm{QC}}\) ……….(areas proportionate to bases) …(ii)

Seg PQ is common base of ∆PQB and ∆PQC, seg PQ 11 seg BC.

Hence, ∆PQB and ∆PQC have equal areas.

A(∆PQB) = A(∆PQC) ……(iii)

\(\frac{\mathrm{A}(\Delta \mathrm{APQ})}{\mathrm{A}(\Delta \mathrm{PQB})}\) = \(\frac{\mathrm{A}(\Delta \mathrm{APQ})}{\mathrm{A}(\Delta \mathrm{PQC})}\) [From (i), (ii) and (iii)]

\(\frac{\mathrm{AP}}{\mathrm{~PB}}\) = \(\frac{\mathrm{AQ}}{\mathrm{QC}}\) [From (i) and (ii)]

Hence, Proved.

Question 4.

Solve any two of the following sub-questions : [8]

2tan² θ = -3 + 1 – 2

tan² θ = -2

tan² θ = 1

tan θ = 1 (Taking square root on both sides)

tan θ = tan 45°

θ =45°.

(2) A cylinder of radius 12 cm contains water up to the height 20 cm. A spherical iron ball is dropped into the cylinder and thus water level raised by 6.75 cm. What is the radius of iron ball ?

Solution:

Given: Radius of cylinder = 12 cm = r_{1}

Water level in cylinder = 20 cm

On dropping sphere ball,

rise in height = 6.75 cm = h

Radius of sphere = r_{2}

Now, Volume of water raised in cylinder = Volume of the sphere

πr_{1}²h = \(\frac{4}{3}\)πr_{2}³

(12 × 12 × 6.75) = \(\frac{4}{3}\)r_{2}³

\(\frac{12 \times 12 \times 6.75 \times 3}{4}\) = r_{2}³

r_{2}³ = 729

Taking the cube roots on both sides,

r_{2} = 9

∴ The radius of the iron ball is 9 cm.

(3) Draw a circle with centre O having radius 3 cm. Draw tangent segments PA and PB through the point P outside the circle such that ∠APB = 70°.

Solution:

Construction:

(i) Construct a circle of radius 3 cm with centre O.

(ii) Join OA, OB and increase it.

(iii) Now draw the perpendicular bisectors of these two lines.

(iv) Join A to P and B to P. They meet at point P which is of 70°.

Question 5.

Solve any one of the following sub-questions : [3]

(1) OABCD is trapezium, AB || CD diagonals of trapezium intersects in point P.

Write the answers of the following questions:

(a) Draw the figure using given information.

(b) Write any one pair of alternate angles and opposite angles.

(c) Write the names of similar triangles with test of similarity.

Solution:

(a)

(b) Alternate angles, ∠BAP = ∠PCD (•.• AB || DC and BD is their transversal.)

Opposite angles, ∠APB = ∠CPD (Vertically opposite angles)

(c) ∆APB ~ ∆CPD (By AA test of similarity)

(2) AB is a chord of a circle with centre O. AOC is diameter of circle, AT is a tangent at A.

Write answers of the following questions :

(a) Draw the figure using given information.

(b) Find the measures of ∠CAT and ∠ABC with reasons.

(c) Whether ∠CAT and ∠ABC are congruent ? Justify your answer.

Solution:

(a)

(b) ∠CAT = ∠OAT = ∠90° (By tangent theorem)

and ∠ABC = 90° (Angle in a semi-circle is a right angle.)

(c) ∠CAT = ∠ABC = 90°

∠CAT ≅ ∠ABC

Therefore,

[∵ The angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle.]