SSC Maharashtra Board Maths 2 Question Paper 2024 with Solutions

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
r₁ = 5 cm and r₂ = 3 cm
Then, distance between their centres = r₁ + r₂
= 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)
AC² = AD² + DC²
⇒ (10\(\sqrt{2}\))² = x² + x²
⇒ 100 × 2 = 2x²
⇒ 200 = 2x²
⇒ x² = \(\frac{200}{2}\)
⇒ x² = 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°
AB² + BC² = AC² ……(Pythagoras theorem)
Divide both sides by AC²
\(\frac{\mathrm{AB}^2}{\mathrm{AC}^2}\) + \(\frac{B C^2}{A C^2}\) = \(\frac{\mathrm{AC}^2}{\mathrm{AC}^2}\)
\(\left(\frac{\mathrm{AB}}{\mathrm{AC}}\right)^2\) + \(\left(\frac{\mathrm{BC}}{\mathrm{AC}}\right)^2\) = 1
But \(\frac{AB}{AC} = sin θ and [latex]\frac{BC}{AC}\) = cos θ
∴ sin² θ + cos² 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)
= 14²
= 196 cm²

Area of circle = πr² ….(Formula)
= \(\frac{22}{7}\) × 7 × 7
= 154 cm²
(Area of shaded portion) = (Area of square) – (Area of circle)
= 196 – 154
= 42 cm²

(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,
AC² = BC² + AB²
AC² = (9)² + (12)²
AC² = 81 + 144
AC² = 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(x₁, y₁)
B(4, 7) = B(x₂, y₂)
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(x₁, y₁)
B(2, – 5) = B(x<su\(\frac{1}{2}\)b>2, y₂)
C(- 4, 7) = C(x₃, y₃)
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 PQ² + PR² = 290, then find QR.
Solution:
In ∆PQR, M is the mid-point of line QR.
∴ QM =MR …(i)

By Apollonius theorem,
PQ² + PR² = 2PM² + 2QM²
⇒ 290 = 2(9)² + 2QM²
⇒ 290 = 2(81) + 2QM²
⇒ 290 = 162 + 2QM²
⇒ 290 – 162 = 2QM²
⇒ 128 = 2QM²
⇒ \(\frac{128}{2}\) = QM²
⇒ QM² = 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 = r₁
Water level in cylinder = 20 cm
On dropping sphere ball,
rise in height = 6.75 cm = h
Radius of sphere = r₂
Now, Volume of water raised in cylinder = Volume of the sphere
πr₁²h = \(\frac{4}{3}\)πr₂³
(12 × 12 × 6.75) = \(\frac{4}{3}\)r₂³
\(\frac{12 \times 12 \times 6.75 \times 3}{4}\) = r₂³
r₂³ = 729
Taking the cube roots on both sides,
r₂ = 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.]

SSC Maharashtra Board Maths 2 Question Paper with Solutions