SSC Maharashtra Board Maths 2 Question Paper 2022 with Solutions

Maharashtra Board SSC Class 10 Maths 2 Question Paper 2022 with Solutions Answers Pdf Download.

SSC Maths 2 Question Paper 2022 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) For each of the following sub-questions four alternatives answers are given. Choose the correct alternative and write its alphabet: [4]

(i) If ∆ABC ~ ∆DEF and ∠A = 48°, then ∠D =
(A) 48°
(B) 83°
(C) 49°
(D) 132°
Solution: (A) 48°.

(ii) AP is a tangent at A drawn to the circle with centre O from an external point P.
OP = 12 cm and ∠OPA = 30°, then the radius of a circle is …………..
(A) 12 cm
(B) 6√3 cm
(C) 6 cm
(D) 12√3 cm
Solution : (C) 6 cm.

(iii) Seg AB is parallel to X-axis and coordinates of the point A are (1, 3), then the coordinates of the point B can be
(A) (-3,1)
(B) (5,1)
(C) (3,0)
(D) (-5,3)
Solution : (D) (- 5, 3)

(iv) The value of 2 tan 45° – 2 sin 30° is …………….
(A) 2
(B) 1
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{4}\)
Solution: (B) 1

(B) Solve the following subquestions : [4]

In ∆ABC, ∠ABC = 90°, ∠BAC = ∠BCA = 45°. If AC = 9√2, then find the value of AB.
Solution:
Given, in ∆ABC
∠ABC = 90°,
∠BAC = ∠BCA = 45°
AC = 9√2
AB = \(\frac{1}{\sqrt{2}}\) × AC [Property of 45° – 45° – 90°]
∴ AB = \(\frac{1}{\sqrt{2}}\) × 9√2
∴ AB = 9

(ii) Chord AB and chord CD of a circle with centre O are congruent. If tn(arc AB) = 120°, then find the m(arc CD).
Solution:
Given, Chord AB = Chord CD
m(arc AB) = 120°
We know that,
arc AB = arc CD [Corresponding arcs of congruent chord of a circle are congruent]
∴ m(arc AB) = m(arc CD)
∴ 120° = m(arc CD)
∴ m(arc CD) = 120°

(iii) Find the Y-coordinate of the centroid of a triangle whose vertices are (4, -3), (7, 5) and (-2,1).
Solution:
Vertices of a triangle. (4, – 3), (7, 5) and (-2,1) [Given]
x₁ = 4, x₂ = 7, x₃ = – 2
y₁ = -3, x₂ = 5, x₃ = 1
By using centroid formula,
[\(\frac{x_1+x_2+x_3}{3}\), \(\frac{y_1+y_2+y_3}{3}\)] = coordinate of centroid
Now, Y-coordinate of centroid = [lattex]\left[\frac{y_1+y_2+y_3}{3}\right][/latex]
= \(\frac{-3+5+1}{3}\)
= \(\frac{3}{3}\)
= 1
∴ Y-coordinate of centroid = 1

(iv) If sin θ = cos θ, then what will be the measure of angle θ ?
Solution:
Given, sin θ = cos θ
We know that,
sin θ = cos (90 – θ)
∴ cos θ = cos (90 – θ)
∴ θ = 90 – θ
∴ θ + θ = 90°
∴ 2θ = 90°
θ = \(\frac{90 \Upsilon}{2}\)
∴ θ = 45°

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

(i)

In the above figure, seg AC and seg BD intersect each other in point P. If \(\frac{A P}{C P}=\frac{B P}{D P}\), then complete the following activity to prove ∆ABP ~ ∆CDP.
Activity : In ∆APB and ∆CDP

Solution:
Activity : In ∆APB and ∆CDP
\(\frac{A P}{C P}=\frac{B P}{D P}\) ………….
∴ ∠APB = ∠CPD ………….. vertically opposite angles
\(\text { ØABP }\) ~ ∆CDP …………..By SAS of similarity

(ii)

In the above figure, ▭ABCD is a rectangle. If AB = 5, AC = 13, then complete the following activity to find BC.
Activity:
∆ABQ is triangle.
∴ By Pythagoras theorem
AB² + BC² = AC²
∴ 25 + BC² =
BC² =
BC =
Solution:
Activity:
∆ABC is right angle triangle.
∴ By Pythagoras theorem
AB² + BC² = AC²
∴ 25 + BC² = 169
BC² = 144
BC = 12

(iii) Complete the following activity to prove:
cot θ + tan θ = cosec θ × sec θ
Activity:

Solution:

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

(i) If ∆ABC ~ ∆PQR, AB : PQ = 4: 5 and A(∆PQR) = 125 cm², then find A(∆ABC).
Solution:
Given:
∆ABC ~ ∆PQR
We know that,

(ii)

In the above figure, m(arc DXE) = 105°, m(arc AYC) = 47°, then find the measure of ∠DBE.
Solution: From fig.

Chord AD and CE intersect externally at point B.
∴ m∠DBE = \(\frac{1}{2}\) [m(arc DXE) – m( arc AYC)] [Theorem]
∴ m∠DBE = \(\frac{1}{2}\)[105°-47°]
∴ m∠DBE = \(\frac{1}{2}\)[58°]
∴ m(∠DBE) = 29°

(iii) Draw a circle of radius 3.2 cm and centre ‘O’. Take any point P on it. Draw tangent to the circle through point P using the centre of the circle.
Solution:
Step 1: Draw a circle of radius 3.2 cm with centre O.
Step 2: Take any point P on the circle.
Step 3: Draw a ray OP.
Step 4: Draw a line perpendicular to OM through point P.

(iv) If sin θ = \(\frac{11}{61}\), then find the value of cos θ using trigonometric identity.
Solution:
Given: sin θ = \(\frac{11}{61}\)
We know that,
sin² θ + cos² θ = 1
∴ \(\frac{11}{61}\)² + cos² θ = 1
∴ \(\frac{121}{3721}\) + cos² θ = 1
∴ cos² θ = 1 – \(\frac{121}{3721}\)
∴ cos² θ = 1 – \(\frac{3721-121}{3721}\)
∴ cos² θ = \(\frac{3600}{3721}\)
∴ cos θ = \(\frac{60}{61}\) [Taking square root of the both sides]

(v) In AABC, AB = 9 cm, BC = 40 cm, AC = 41 cm. State whether ∆ABC is a rightangled triangle or not ? Write reason.
Solution:
Sides of ∆ABC are AB = 9 cm, BC = 40 cm, AC = 41 cm
The longest side of triangle is 41 cm.
∴ (41)² = 1681 ……(i)
Now, sum of remaining sides square is,
(9)2 + (40)2 = 81 + 1600 = 1681 ………(ii)
From equations (i) and (ii), given sides form a right angle triangle.
Because, square of longest side is equal to sum of square of remaining two sides.
[Converse of Pythagoras theorem]

Question 3.
(A) Complete the following activity and rewrite it (any one). [4]

(i)

In the above figure, chord PQ and chord RS intersect each other at point T. If ∠STQ = 58° and ∠PSR = 24°, then complete the following activity to verify:
∠STQ = \(\frac{1}{2}\)[m(arc PR) + m(arc SQ)]
Activity:
In ∆PTS,

Solution:
Activity: In ∆PTS,

(ii) Complete the following activity to find the coordinates of point P which divides seg AB in the ratio 3 :1 where A(4, -3) and B(8,5).
Activity:

Solution:
Activity:

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

(i) In AABC, seg XY || side AC. If 2AX = 3BX and XY = 9, then find the value of AC.
Solution:
Given, 2AX = 3BX

AC = 15 units

(ii) Prove that “Opposite angles of cyclic quadrilateral are supplementary”.
Proof:
Given : ▭ABCD is a cyclic quadrilateral.

To prove : ∠BAD + ∠BCD = 180° and ∠ABC + ∠ADC = 180°
Proof : Arc BCD subtend angle BAD at the circumference.
∴ ∠BAD = \(\frac{1}{2}\) m(arc BCD)
[Inscribed angle theorem] …(i)
Arc BAD subtends by ∠BCD at the circumference.
∴ ∠BCD = \(\frac{1}{2}\) m(arc DAB) [Inscribed angle theorem] …(ii)
From equations (i) and (ii), we get
∠BAD + ∠BCD = \(\frac{1}{2}\) [m(arc BCD) + m(arc DAB)]
∴ (∠BAD + ∠BCD) = \(\frac{1}{2}\) × 360° [Complete circle]
= 180°
As sum of all angles of quadrilateral is 360°
∠ADC + ∠ABC = 360° – [∠BAD + ∠BCD]
= 360° -180°
= 180°
Hence, opposite angles of cyclic quadrilateral are supplementary.

(iii) ∆ABC ~ ∆PQR. In ∆ABC, AB = 5.4 cm, BC = 4.2 cm, AC = 6.0 cm, AB : PQ = 3:2, then construct ∆ABC and ∆PQR.
Solution:
∆ABC ~ ∆PQR [Given]
We knoW that corresponding sides of triangle which are similar are in proportion
∴ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\) = \(\frac{3}{2}\)
∴ \(\frac{A B}{P Q}\) = \(\frac{3}{2}\)
∴ \(\frac{5.4}{P Q}\) = \(\frac{3}{2}\)
∴ PQ = \(\frac{5.4×2}{3}\)
∴ PQ = 3.6 cm
Also,
∴ \(\frac{B C}{Q R}\) = \(\frac{3}{2}\)
∴ \(\frac{4.2}{Q R}\) = \(\frac{3}{2}\)
∴ QR = \(\frac{4.2×2}{3}\)
∴ QR = 2.8 cm
Also,
∴ \(\frac{A C}{P R}\) = \(\frac{3}{2}\)
∴ \(\frac{6}{P R}\) = \(\frac{3}{2}\)
∴ PR = \(\frac{6×2}{3}\)
∴ PR = 4 cm
Now, draw angle AABC with sides AB = 5.4 cm, BC = 4.2 cm and AC = 6 cm.
Also draw triangle APQR with sides PQ = 3.6 cm, QR = 2.8 cm and PR = 4 cm.
∆ABC,

∆PQR

(iv) Show that:

SSC Maharashtra Board Maths 2 Question Paper with Solutions