SSC Maharashtra Board Maths 2 Question Paper July 2023 with Answers

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

SSC Maths 2 Question Paper July 2023 with Answers Pdf Download Maharashtra Board

Time: 2 Hours
Total Marks: 40

Note:

1. All questions are compulsory.
2. Use of calculator is not allowed.
3. The numbers to the right of the questions indicate full marks.
4. In case of MCQ’s [Q. No. 1(A)] only the first attempt will be evaluated and will be given credit.
5. For every MCQ, the correct alternative (A), (B), (C) or (D) with sub-question number is to be written as an answer.
6. Draw proper figures for answers wherever necessary.
7. The marks of construction should be clear. Do not erase them.
8. Diagram is essential for writing the proof of the theorem.

Question 1.
(A) For each of the following sub-question four alternative answers are given. Choose the correct alternative and write its alphabet: [4]
i. The volume of a cube of side 10 cm is ________. (D)
(A) 1 cm³
(B) 10 cm³
(C) 100 cm³
(D) 1000 cm³
Answer:
(D) 1000 cm³

ii. A line makes an angle of 30o with positive direction of X-axis, then the slope of the line is ________. (C)

Answer:
C

iii. ∠ACB is inscribed in arc ACB of a circle with centre O. If ∠ACB = 65°, find m(arc ACB) : (D)
(A) 65°
(B) 130°
(C) 295°
(D) 230°
Answer:
(D) 230°

iv. Find the perimeter of a square if its diagonal is 10√2 cm. (D)
(A) 10 cm
(B) 40√2 cm
(C) 20 cm
(D) 40 cm
Answer:
(D) 40 cm
Hints:
i. Volume of a cube = (side)³ = (10)³ = 1000 cm³

ii. Angle made with the positive direction of X-axis (θ) = 30°
Slope of line (m) = tan θ
∴ m = tan 30° = \(\frac{1}{\sqrt{3}}\)

iii. m∠ACB = \(\frac{1}{2}\) m(arc AB) …[Inscribed angle theorem]
∴ m(arc AB) = 2 m∠ACB
= 2 × 65 = 130°

m(arc ACB) = 360° – m(arc AB)
= 360° – 130° = 230° …[Measure of a circle Is 360°]

iv. In ∆ABC, ∠B = 90°, and ∠BAC = ∠BCA = 45°

∴ Perimeter of square = 4 (AB) = 4 x 10 = 40 cm

Question 1.
(B) Solve the following sub-questions: [4]
i. In the following figure, ∠ABC = ∠DCB = 90°, AB = 6, DC = 8, then \(\frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{DCB})}\)=?

ii. In the following figure, find the length of RP using the information given in ∆PSR.

iii. What is the distance between two parallel tangents of a circle having radius 4.5 cm?
iv. Find the co-ordinates of midpoint of the segment joining the points A(4, 6) and B(-2, 2).
Answer:
i. ∆ABC and ∆DCB have same base BC.

ii. In ∆PSR, ∠S = 90°, ∠P = 30° …[Given]
∴ ∠R = 60° …[Remaining angle of a triangle]
∴ ∆PSR Is a 30° – 60 – 90° triangle.
RS = \(\frac{1}{2}\)RP …[Side opposite to 30°]
∴ 6 = \(\frac{1}{2}\)RP
∴ RP = 6 × 2 = 12
∴ RP = 12 units
SMART TIP
To verify our answer, we use Pythagoras Theorem. If /(SR)² + /(SP)² = /(PR)², then our answer is correct.
/(SR)² + /(SP)² = (6)² + (6√3)² = 36 + 108 = 144
/(PR)² = (12)² = 144
∴ /(SR)² + /(SP)² = /(PR)²
Hence, our answer is correct.

iii. Distance between two parallel tangents of a circle
= diameter of the circle
= 2 × radius
= 2 × 4.5 cm
= 9 cm

iv. Let A(x₁, y₁) = A(4, 6)
B(x₂, y₂) = B(-2, 2)
Let the co-ordinates of the midpoint be P(x, y).
∴ By midpoint formula.

∴ The co-ordinates of the midpoint of the segment joining A(4, 6) and B(-2, 2) are (1, 4).

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

In the above figure, circle with centre D touches the sides of ∠ACB at A and B. If ∠ACB = 52°, complete the activity to find the measure of ∠ADB.
Activity:
In □ABCD,
∠CAD = ∠CBD = □° ….Tangent theorem
∴ ∠ACB + ∠CAD + ∠CBD + ∠ADB = □°
∴ 52° + 90° + 90° + ∠ADB = 360°
∴ ∠ADB + □° = 360°
∴ ∠ADB = 360° – 232°
∴ ∠ADB = □°
ii.

In the above figure, side of square ABCD is 7 cm with centre D and radius DA sector D-AXC is drawn.
Complete the following activity to find the area of square ABCD and sector D-AXC.
Activity:
Area of square = □ …..formula
= (7)²
= 49 cm²

iii. Complete the following activity to prove cot θ + tan θ = cosec θ × sec θ.
Activity:
L.H.S. = cot θ + tan θ


∴ L.H.S. = R.H.S.
∴ cot θ + tan θ = cosec θ × sec θ
(B) Solve the following sub-questions (Any four):
i. If cos θ = \(\frac{3}{5}\), then find sin θ.
ii. Find slope of line EF, where co-ordinates of E are (-4, -2) and co-ordinates of F are (6, 3).
iii.

In the above figure, ray PQ touches the circle at point Q.
If PQ = 12, PR = 8, find the length of seg PS.
iv.

In the above figure, ∠MNP = 90°, seg NQ ⊥ seg MP. MQ = 9, QP = 4. Find NQ.
v.

In the above figure, if AB ǁ CD ǁ EF, then find x and AE by using the information given in the figure.
Answer:
(A) i. In □ABCD,

ii.

iii. L.H.S. = cot θ + tan θ

∴ L.H.S. = R.H.S.
∴ cot θ + tan θ = cosec θ × sec θ

(B) i.

ii. E (x₁, y₁) = E (-4, -2) and F (x₂, y₂) = F (6, 3)
Here, x₁ = – 4, x₂ = 6, y₁ = – 2, y₂ = 3

iii. Ray PQ is a tangent to the circle at point Q and seg PS is the secant. …[Given]
∴ PR × PS = PQ² …[Tangent secant segments theorem]
∴ 8 × PS = 122
∴ 8 × PS = 144
∴ PS = \(\frac{144}{8}\)
∴ PS = 18 units

iv. In ∆MNP, ∠MNP = 90° and
seg NQ ⊥ seg MP …[Given]
∴ NQ² = MQ × QP …[Theorem of geometric mean]
∴ NQ = \(\sqrt{MQ\times QP}\) …[Taking square root of both sides]
= \(\sqrt{9\times4}\) = 3 × 2
∴ NQ = 6 units

v. line AB ǁ line CD ǁ line FE …[Given]

∴ x = 6 units
Now, AE = AC + CE
= 12 + x
= 12 + 6
= 18 units
∴ x = 6 units and AE = 18 units

Question 3.
(A) Complete the following activities and rewrite it (any one): [3]
i.

In the above figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle seg PQ ǁ seg DE, seg QR ǁ seg EF. Complete the following activity to prove seg PR ǁ seg DF.
Activity:
In ∆XDE, PQ ǁ DE

∴ seg PR ǁ seg DF …Converse of basic proportionality theorem
ii.

A, B, C are any points on the circle with centre O.
If m(arc BC) = 110° and m(arc AB) = 125°, complete the following activity to find m(arc ABC), m(arc AC), m(arc ACB) and m(arc BAC).
Activity :

(B) Solve the following sub-questions (any two): [6]
i. The radius of a circle is 6 cm, the area of a sector of this circle is 15 π sq.cm. Find the measure of the arc and the length of the arc corresponding to that sector.
ii. If A(3, 5) and B(7, 9), point Q divides seg AB in the ratio 2 : 3, find the co-ordinates of point Q.
iii. Prove that :
“In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides.”
iv. ∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR such that \(\frac{PQ}{LT}=\frac{3}{4}\).
Answer:
(A) i. In ∆XDE, PQ ǁ DE …(given)

seg PR ǁ seg DF ….Converse of basic proportionality theorem

ii.

(B) i. Given: Radius (r) = 6 cm, area of sector = 15π cm²
To find: i. Measure of the arc (θ),
ii. Length of the arc (/).

∴ The measure of the arc and the length of the arc are 150° and 5π cm respectively.

ii. Let the co-ordinates of point Q be (x, y) and A (x₁, y₁), B (x₂, y₂) be the given points.
Here, x₁ = 3, y₁ = 5, x₂ = 7, y₂ = 9, m = 2, n = 3
∴ By section formula,

iii. Given: In ∆ABC, ∠ABC = 90°.
To prove: AC² = AB² + BC²
Construction: Draw seg BD ⊥ hypotenuse AC, A-D-C.
Proof:

In ∆ABC, ∠ABC = 90° …[Given]
seg BD ⊥ hypotenuse AC …[Construction]
∴ ∆ABC ~ ∆ADB …[Similarity of right angled triangles]

iv.

Question 4.
Solve the following sub-questions (any two): [8]
i. A bucket is in the form of a frustum of a cone. It holds 28.490 litres of water. The radii of the top and the bottom are 28 cm and 21 cm respectively. Find the height of the bucket. (π = \(\frac{22}{7}\))
ii. Draw a circle with centre P and radius 3 cm. Draw a chord MN of length 4 cm. Draw tangents to the circle through points M and N which intersect in point Q. Measure the length of seg PQ.
iii. In ∆PQR, bisectors of ∠Q and ∠R intersect in point X. Line PX intersects side QR in point Y, then prove that: \(\frac{PQ+PR}{QR}=\frac{PX}{XY}\)
Answer:
i. Radius of the top of the frustum (r₁) = 28 cm
Radius of the bottom of the frustum (r₂) = 21 cm
Capacity of bucket = Volume of frustum

ii.

∴ l(PQ) = 4 cm.

iii. In ∆PQY, ray QX bisects ∠Q. …[Given]

Question 5.
Solve the following sub-questions (Any one): [3]
i. From top of the building, Ramesh is looking at a bicycle parked at some distance away from the building on the road.
If
AB → Height of building is 40 m
C → Position of bicycle
A → Position of Ramesh on top of the building
∠MAC is the angle of depression and m∠MAC = 30°, then:
(a) Draw a figure with the given information.
(b) Find the distance between building and the bicycle. (√3 = 1.73).
ii. □ABCD is a cyclic quadrilateral where side AB ≅ side BC, ∠ADC = 110°, AC is the diagonal, then:
(a) Draw the figure using given information
(b) Find measure of ∠ABC
(c) Find measure of ∠BAC
(d) Find measure of (arc ABC).
Answer:
i. Given: AB represents the height of building and point C represents the position of bicycle.

∴ The distance between building and the bicycle is 69.20 m.

ii. a.

b. □ABCD is a cyclic quadrilateral. …[Given]
∴ ∠ADC + ∠ABC = 180° …[Opposite angles of a cyclic quadrilateral are supplementary]
∴ 110° + ∠ABC = 180°
∴ ∠ABC = 180° – 110°
∴ m∠ABC = 70°

c. In ∆ABC,
side AB ≅ side BC …[Given]
∴ ∠ACB ≅ ∠BAC …[Isosceles triangle theorem]
Let ∠ACB = ∠BAC = x
Now, ∠ABC + ∠BAC + ∠ACB = 180° …[Sum of the measures of angles of a triangle is 180°]
∴ ∠ABC + x + x = 180°
∴ 70° + 2x = 180°
∴ 2x = 180° – 70°
∴ 2x = 110°
∴ x = \(\frac{110°}{2}\) = 55°
∴ m ∠BAC = 55° …(i)

d. ∠ADC = \(\frac{1}{2}\)m(arc ABC) …[Inscribed angle theorem]
∴ 110° = \(\frac{1}{2}\)m(arc ABC)
∴ m(arc ABC) = 220°

SSC Maharashtra Board Maths 2 Question Paper with Solutions