SSC Maharashtra Board Maths 2 Question Paper 2023 with Solutions

Maharashtra Board SSC Class 10 Maths 2 Question Paper 2023 with Solutions Answers Pdf Download.

SSC Maths 2 Question Paper 2023 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 answrs are given for every subquestion. Select the correct alternative and write the alphabet of that answer: [4]
(1) If a, b, c are sides of a triangle and a²+ b² = c², name the type of triangle:
(A) Obtuse angled triangle
(C) Right angled triangle
(B) Acute angled triangle
(D) Equilateral triangle
Solution : (c) Right angled triangle

(2) Chrods AB and CD of a circle intersect inside the circle at point E. If AE = 4, EB = 10, CE = 8/then find ED:
(A) 7
(B) 5
(C) 8
(D) 9
Solution: (B) 5

(3) Co-ordinates of origin are .
(A) (0,0)
(B) (0,1)
(C) (1,0)
(D) (1,1)
Solution : (A) (0, 0)

(4) If radius of the base of cone is 7 cm and height is 24 cm, then find its slant height:
(A) 23 cm
(B) 26 cm
(C) 31 cm
(D) 25 cm
Solution : (D) 25 cm

(B) Solve the following sub-questions : [4]
(1) If ∆ABC ~ ∆PQR and \(\frac{\mathrm{A}(\Delta \mathrm{ABC})}{\mathrm{A}(\Delta \mathrm{PQR})}\) = \(\frac{16}{25}\), then find AB : PQ.
Solution :
∆ABC ~ ∆PQR
\(\frac{\mathrm{A}(\Delta \mathrm{ABC})}{\mathrm{A}(\Delta \mathrm{PQR})}\) = \(\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}[\latex]= [latex]\frac{16}{25}\)
\(\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}[\latex]= [latex]\frac{(4)^2}{(5)^2}\)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}[\latex]= [latex]\frac{4}{5}\)

(2) In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm, then find RS.
Solution :
In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm
∴ By using angle sum property of triangle.
∠R = 60°
∴ By 30° – 60°- 90° theorem,
∴ l(RS) = \(\frac{1}{2}\) × RT
∴ = \(\frac{1}{2}\) × 12
l(RS) = 6 cm

(3) If radius of a circle is 5 cm, then find the length of longest chord of a circle.
Solution:
Radius = 5 cm.
As, diameter is the longest chord of the circle.
∴ Diameter = 2 × Radius
= 2 × 5
= 10 cm.
The longest chord is 10 cm.

(4) Find the distance between the points O(0,0) and P(3,4).
Solution:
O(0, 0), P(3, 4)
∴ (x₁, y₁) = (0, o)
(x₂, y₂) = (3, 4).
∴ x₁ = o, y₁ = 0
x₂ = 3, y₂ = 4
d(OP) = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(3-0)^2+(4-0)^2}\)
= \(\sqrt{3^2+4^2}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
d(OP) = 5 cm.

Question 2.
(A) Complete the following activities (any two): [4]

(1)

In the above figure, ∠L = 35°, find:
(i) m(arc MN)
(ii) m(arc MLN)
Solution:

Solution:
(i) ∠L = \(\frac{1}{2}\)m(arc MN) ……….(By inscribed angle theorem)
35° = \(\frac{1}{2}\)m(arc MN)
2 × 35° = m(arc MN)
m(arc MN) = 70°

(ii) m(arc MLN) = 360° – m(arc MN) ……..(Definition of measure of arc)
= 360° – 70°
m(arc MLN) = 270°

(2) Show1 that, cot θ + tan θ = cosec θ × sec θ.
Solution:

Solution:

(3) Find the surface area of a sphere of radius 7 cm.
Solution:

Solution:

(B) Solve the following sub-questions (any four): [8]
(1)

In trapezium ABCD side AB || side PQ || side DC. AP = 15, PD = 12, QC = 14, find BQ.
Solution :
In a trapezium ABCD.
Side AB || side PQ || side DC.
By using property of three paraller lines and their transversals.
\(\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{BQ}}{\mathrm{QC}}\)
\(\frac{15}{12}=\frac{B Q}{14}\)
15 × 14 = BQ × 12
\(\frac{15×14}{12}\) = BQ
\(\frac{15×7}{6}\) = BQ
BQ = 17.5 units

(2) Find the length of the diagonal of a rectangle whose length is 35 cm and bfeadth is 12 cm.
Solution:
Let, ABCD is a rectangle
l(AB) = 35 cm
l(BC) = 12 cm

Let AC be the diagonal of rectangle
as ∠A = ∠B = ∠C = ∠D = 90°
In AABC, as ∠B = 90°
By using Pythagoras theorem.
AC² = AB² + BC²
AC² = 35² + 12²
AC² = 1225 + 144
AC² = 1369
AC = 37 cm
∴ The diagonal of the rectangle is 37 cm.

(3)

In the given figure points G, D, E, F are points of a circle with centre C, ECF = 70°, m(arc DGF) = 200°
Find:
(i) m(arc DE)
(ii) m(arc DEF)
Solution:
∠ECF = 70°
m(arc DGF) = 200°
(i) as ∠ECF = m(arc EF) …(Definition of measure of minor arc)
= 70°
∴ m(arcEF) =70°
as, measure of a circle is 360°.
∴ m(arc DGF) + m(arc EF) + m(arc DE) = 360°
200° + 70° + m(arc DE) = 360°
m(arc DE) = 360° – 270°
m(arcDE) =90°
∴ m(arcDE) =90°

(ii) m(arc DEF) = m(arc DE) + m(arc EF) …(arc addition propert)
= 90° + 70° …from (1)
= 160°
∴ m( arc DJEF) = 160°

(4) Show that points A(-1, -1), B(0,1), C(l, 3) are collinear.
Solution:
Let A(xi, yi) = (-1, -1)
B(*2/ Vi) = (0,1)
C(x3/ y3) = (1, 3)
Slope of line = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{1-(-1)}{0-(-1)}\)
= \(\frac{1+1}{1}\) = 2

Slope of line = \(\frac{y_3-y_2}{x_3-x_2}\)
= \(\frac{3-1}{1-0}\)
= \(\frac{2}{1}\) = 2
As, slope of line AB = slope of line BC
Also AB and BC lines contain common point B

∴ Points A, B, C are collinear.
Hence proved.

(5) A person is standing at a distance of 50 m from a temple looking at its top. The angle of elevation is 45°. Find the height of the temple.
Solution:
Let PQ be the Temple with height h and R be the point from where the person is looking at its top.
The angle of elevation is 45°.

And the distance of a person from temple is 50 m.
∴ l(QR) = 50 m
As PQ⊥ QR
∴ ∠PQR = 90° and ∠PRQ = 45°
∴ By using tan 45° formula.
tan θ = \(\frac{1}{2}\)
tan 45° = \(\frac{1}{2}\)
1 = \(\frac{1}{2}\)
∴ 1 × 50 = h
h = 50 m
∴ The height of the temple is 50 m.

Question 3.
(A) Complete the following activities (any one): [3]

(1)

In ∆PQR, seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.
Complete the proof by filling in the boxes.
Solution:
In ∆PMQ,

Solution:
In ∆PMQ,
Ray MX is the bisector of ZPMQ

(2) Find the co-ordinates of point P where P is the midpoint of a line segment AB with A(-4,2) and B(6,2).
Solution:

Suppose, (- 4,2) = (x₁, y₁) and (6,2) = (x₂, y₂), and co-ordinates of P are (x, y)
∴ According to midpoint theorem

Solution:
Suppose, (- 4,2) = (x₁, y₁) and (6,2) = (x₂, y₂), and co-ordinates of P are (x, y)
∴ According to midpoint theorem

(B) Solve the following subquestions (any two): [4]

(1) In AABC, seg AP is a median. If BC = 18, AB² + AC² = 260, find AP.
Solution :
ABC is given triangle with segment AP is medium.

BC =18
AB² + AC² = 260
As P is the midpoint
PC = \(\frac{1}{2}\) × 18
PC = 9 cm
By using Apollunius theorem.
AB² + AC² = 2AP² + 2PC²
260 = 2AP² + 2(9)²
260 = 2AP² + 2(81)
260 = 2AP² + 162
130 = AP² + 81
130 – 81 = AP²
49 = AP² ……..Taking square root on both sides
7 = AP
AP = 7 unit.

(2) Prove that, “Angles inscribed in the same arc are congruent”.
Solution :
In a circle
∠ABC and ∠ADC are inscribed in the same arc i.e., arc AMC.
To prove : ∠ABC ≅ ∠ADC

Proof: mZABC = \(\frac{1}{2}\)m(arcAMC) ……..(i)
…by inscribed angle theorem
∠ADC = \(\frac{1}{2}\) m(arcAMC) ………(ii)
∴ from (i) and (ii)
∠ABC ≅ ∠ADC …(by Angle with equal measurement)
Hence proved

(3) Draw a circle of radius 3.3 cm. Draw a chord PQ of length 6.6 cm. Draw tangents to the circle at points P and Q.
Solution:
Steps of construction:
(1) Draw a circle with centre O with radius 3.3 cm, mark any point P on it.
(2) Draw chord PQ = 6.6 cm (PQ is the diameter of the circle).
(3) Draw rays OX and OY.
(4) Draw line l perpendicular to ray OX through point P.
(5) Draw line m perpendicular to ray OY through point Q.

(4) The radii of circular ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its curved surface area. (π = 3.14)
Solution:
Let r₁ = 14 cm
r₂ = 6 cm
height = h = 6 cm.
∴ Slant height of frustum (l) = \(\sqrt{h^2+\left(r_1-r_2\right)^2}\)
= \(\sqrt{6^2+(14-6)^2}\)
= \(\sqrt{36+64}\)
= \(\sqrt{100}\)
l = 10 cm …(1)
∴ Curved surface area of frustum
= πl (r₁ + r₂)
= 3.14 × 10(14 + 6)….from (1)
= 3.14 × 10×20
= 3.14 × 200
= 628 cm²
∴ The curved surface area of the frustum is 628 cm².

Question 4.
Solve the following sub-questions (any two): [8]

(1) In ∆ABC, seg DE || side BC. If 2A(∆ADE) = A(DBCE), find AB : AD and show that BC = √3DE.
Solution:
Given: In ∆ABC, DE || side BC.
∴ 2A(∆ADE) = A(DBCE)
To prove: \(\frac{AB}{AD}[/lattex]
Proof: BC = √3DE.

In ∆ABC
DE || BC
ZADE = ZABC
In ∆ADE and ∆ABC
∠BAC = ∠DAE
∠ADE = ∠ABC
ByA-Atest,
∆ADE = AABC
[latex]\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{AE}}\)
Also, 2A(∆ADE) = A(DBCE)
as A(∆ABC) = A(∆ADE) + A (DBCE)
A(A∆ABC = A(∆ADE) + 2A(∆ADE)
A(∆ABC) = 3A(∆ADE)
\(\frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{ADE})}=\frac{3}{1}\) ……(2)
\(\frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{ADE})}=\frac{\mathrm{BC}^2}{\mathrm{DE}^2}\) …(Theorem of Area of similar triangle)
\(\frac{3}{1}\) = \(\frac{\mathrm{BC}^2}{\mathrm{DE}^2}\)
3DE² = BC² …(By taking square root on both sides)
BC = √3DE.
Hence proved

SSC Maharashtra Board Maths 2 Question Paper with Solutions