Maharashtra Board SSC Class 10 Maths 2 Question Paper 2020 with Solutions Answers Pdf Download.

## SSC Maths 2 Question Paper 2020 with Solutions Pdf Download Maharashtra Board

Time allowed : 2 hours

Maximum marks : 40

General Instructions:

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

Question 1.

(A) Four alternative answers are given for every sub-question. Select the correct alternative and write the alphabet of that answer : [4]

(i) Out of the following which is the Pythagorean triplet?

(A) (1, 5, 10)

(B) (3, 4, 5)

(C) (2, 2, 2)

(D) (5, 5, 2)

Solution:

(B) (3, 4, 5)

(ii) Two circles of radii 5.5 cm and 3.3 cm respectively touch each other externally. What is the distance between their centres?

(A) 4.4 cm

(B) 2.2 cm

(C) 8.8 cm

(D) 8.9 cm

Solution:

(C) 8.8 cm

(iii) Distance of point (-3, 4) from the origin is ____.

(A) 7

(B) 1

(C) -5

(D) 5

Solution:

(D) 5

(iv) Find the volume of a cube of side 3 cm :

(A) 27 cm^{3}

(B) 9 cm^{3}

(C) 81 cm^{3}

(D) 3 cm^{3}

Solution:

(A) 27 cm^{3}

(B) Solve the following questions : [4]

(i) The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.

Solution:

Ratio of areas of similar triangles = (Ratio of corresponding sides of similar triangles)^{2} = \(\left(\frac{3}{5}\right)^2\)

Ratio of their areas = \(\frac{9}{25}\)

(ii) Find the diagonal of a square whose side is 10 cm.

Solution:

Let □ABCD is a square

l(AB) = l(BC) = l(CD) = l(AD) = 10 cm (Given)

In ∆ABC,

AC^{2} = AB^{2} + BC^{2} (Pythagoras theorem)

∴ AC^{2} = AB^{2} + AB^{2} (∵ AB = BC)

∴ AC^{2} = 2AB^{2}

∴ AC = \(\sqrt{2}\)AB

= \(\sqrt{2}\)(10)cm (AB = 10cm)

∴ AC = 10 × 1.414 = 14.14 cm

∴ Diagonal of the square AC = 14.14 cm

(iii) □ABCD is cyclic. If ∠B = 110, then find measure of ∠D.

Solution:

□ABCD is cyclic

∴ m∠B + m∠D = 180°

∴ 110° + m∠D = 180° (Given, m∠B = 110°)

∴ m∠D = 180° – 110°

m∠D = 70°

(iv) Find the slope of the line passing through the points A(2, 3) and B(4, 7).

Solution:

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

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

∴ Slope of line AB = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{7-3}{4-2}\)

= \(\frac{4}{2}\)

∴ Slope of line AB = 2 unit

Question 2.

(A) Complete and write the following activities (Any two): [4]

(i) In the figure given, ‘O’ is the centre of the circle, seg PS is a tangent segment and S is the point of contact. Line PR is a secant. If PQ = 3.6, QR = 6.4, find PS.

Solution:

Refer figure given in the question

(ii) If sec θ = \(\frac{25}{7}\), find the value of tan θ.

Solution:

(iii) In the figure given, O is the centre of the circle. Using given information complete the following table :

Solution:

(B) Solve the following sub-questions (Any four) :

(i) In ∆PQR, NM || RQ. If PM = 15, MQ = 10, NR = 8, then find PN. [8] Solution:

Given NM || RQ

∴ \(\frac{\mathrm{PN}}{\mathrm{NR}}\) = \(\frac{P M}{M Q}\) (Basic proportionality theorem) …(i)

But PM = 15, MQ = 10, NR = 8 (Given)

∴ Equation (i) becomes,

\(\frac{\mathrm{PN}}{8}\) = \(\frac{15}{10}\)

∴ PN = \(\frac{15 \times 8}{10}\)

= \(\frac{15 \times 4}{5}\) = 3 × 4

PN = 12 unit

(ii) In ∆MNP, ∠MNP = 90, seg NQ ⊥ seg MP. If MQ = 9, QP = 4, then find NQ.

Answer:

In ∆MNP, ∠MNP = 90°, seg NQ ⊥ seg MP

∴ According to right-angled triangle geometric mean subtheorem

NQ^{2} = MQ × QP

= 9 × 4 = 36

NQ = \(\sqrt{36}\)

= 6 unit

(iii) In the figure given above, M is the centre of the circle and seg KL is a tangent segment. L is a point of contact. If MK = 12, KL = 6\(\sqrt{3}\) then find the radius of the circle.

Answer:

In given figure, radius ML ⊥ tangent segment KL … (Tangent theorem)

∴ m∠MLK = 90°

In right-angled ∆MLK

MK^{2} = ML^{2} + LK^{2} (According to Pythagoras theorem)

∴ (12)^{2} = ML^{2} + (6\(\sqrt{3}\))^{2}

∴ 144 = ML^{2} + 108

∴ ML^{2} = 144 – 108 = 36

∴ ML = 6

∴ Radius ML = 6 unit

(iv) Find the co-ordinates of midpoint of the segment joining the points (22, 20) and (0, 16).

Answer:

Given points are (22, 20) and (0, 16)

Let, x_{1} = 22, x_{2} = 0, y_{1} = 20, y_{2} = 16

We Know, Midpoint = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

= \(\left(\frac{22+0}{2}, \frac{20+16}{2}\right)\)

= \(\left(\frac{22}{2}, \frac{36}{2}\right)\)

= (11, 8)

(v) A person is standing at a distance of 80 metres from a Church and looking at its top. The angle of elevation is of 45. Find the height of the Church.

Solution:

Let the height of church be BC

In ∆ABC,

∴ tan 45° = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)

1 = \(\frac{B C}{80}\) {∵ tan 45° = 1}

BC = 80 m

∴ The height of the church BC = 80 m.

Question 3.

(A) Complete and write the following activities (Any one):

(i) In the given figure, X is any point in the interior of the triangle. Point X is joined to the vertices of triangle, seg PQ || seg DE, seg QR || seg EF. Complete the activity and prove that seg PR || seg DF.

Solution:

(ii) If A(6,1), B(8,2), C(9,4) and D(7,3) are the vertices of ▭ABCD, show that ▭ABCD is a parallelogram.

Solution:

Solution:

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

(i) If ∆PQR, point S is the mid-point of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

Solution:

In ∆PQR, point S is the mid-point of side QR.

∴ Segment PS is median of ∆PQR

According to Apollonius’s theorem

PQ^{2} + PR^{2} = 2PS^{2} + 2QS^{2}

As per given values,

∴ (11)^{2} + (17)^{2} = 2(13)^{2} + 2QS^{2}

∴ 121 + 289 = 2(169) + 2QS^{2}

∴ 410 = 338 + 2QS^{2}

∴ 2QS^{2} = 410 – 338 = 72

∴ QS^{2} = \(\frac{72}{2}\) = 6 unit

We know, point S is the mid-point of side QR

∴ 2QS = QR (∵ QS = SR)

∴ QR = 2 (6) [From equation (i)]

QR = 12 unit

∴ Length of side QR is 12 unit.

(ii) Prove that, tangent segments drawn from an external point to the circle are congruent.

Solution:

Point O is the centre of the circle and point P is external to the circle. Segment PA and segment PB are tangent segments to the circle. Point A and point B are touch points of the tangent segments.

Prove : PA ≅ PB

Construction : Draw OA, OB and OP.

Proof : ∵ Each tangent of a circle is perpendicular to the radius drawn through the point of contact …… (Theorem)

∴ Radius OA ⊥ AP and, Radius OB ⊥ BP ……..(i)

∴ m∠PAO = 90° and m∠PBO = 90°

∴ ∆PAO and ∆PBO are right-angled triangles.

Now in ∆PAO and ∆PBO,

OA = OB (∵ Radius of same circle)

∠PAO = ∠PBO [using (i)]

Hypotenuse OP = Hypotenuse OP (∵ common side)

∴ ∆PAO ≅ ∆PBO (RHS conguruency criterion)

∴ line PA ≅ line PB (∵ corresponding sides of congruent triangles)

Line PA and line PB are tangent.

Hence proved.

(iii) Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.

Solution:

Steps of construction :

Step 1 : Draw a circle of radius 4.1 cm with centre O.

Step 2 : Take a point P in the exterior of the circle such that OP = 7.3 cm.

Step 3 : Draw segment OP. Draw perpendicular bisector of segment OP to get its midpoint M.

Step 4 : Draw a circle with radius OM and centre M.

Step 5 : Name the point of intersection of the two circles as A and B.

Step 6 : Join PA and PB.

Thus PA and PB are required tangents.

(iv) A metal cuboid of measures 16 cm 11 cm 10 cm was melted to make coins. How many coins were made, if the thickness and diameter of each coin was 2 mm and 2 cm respectively? (π = 3.14)

Solution:

We know,volume of cuboid = l × b × h

= 16 × 11 × 10

= 1760 cm^{3}

Now, Volume of coin = (Area of coin) × (thickness)

Area of coin = \(\frac{\pi d^2}{4}\)

= \(\frac{3.14 \times(2)^2}{4}\)

= \(\frac{3.14 \times 4}{4}\)

= 3.14 cm^{2}

∴ Volume of coin = 3.14 × t

= 3.14 × (0.2) ……(t = 2 mm = 0.2 cm)

= 0.628 cm^{2}

Now, let N number of coins are made from melted cuboid.

∴ Volume of cuboid = N × volume of coin

∴ 1760 = N × 0.628

∴ N = \(\frac{1760}{0.628}\) = 2802.547

∴ N ≅ 2802

∴ 2802 coins were made.

Question 4.

Solve the following sub-questions (Any two): [8]

(i) In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that seg PQ || seg BC. If PQ divides ∆ABC into two equal parts having equal areas, find \(\frac{B P}{\mathbf{A B}}\).

Answer:

(ii) Draw a circle of radius 2.7 cm and draw a chord PQ of length 4.5 cm. Draw tangents at points P and Q without using centre.

Solution:

Steps of construction :

Step 1 : Draw a circle of with centre O and radius 2.7 cm.

Step 2 : Draw a chord PQ of length 4.5 cm

Step 3 : Taking a point R on the major arc QP, join PR and QR.

Step 4 : Make ∠QPT = ∠PRQ and ∠PQS = ∠PRQ.

Step 5 : Produce TP to L and SQ to M.

Hence, TPL and SQM are the required tangents.

(iii) In the figure given □ABCD is a square of side 50 m. Points Q, R, S are midpoints of side AB, side BC, side CD, side AD respectively. Find area of shaded region.

Solution:

Area of □ABCD = side^{2} ……(∵ Area of square)

= 50^{2} = 2500 m^{2}

There are 4 sectors drawn within square.

In □ABCD with its points A, B, C, D as vertices and radius of each equal to half of side of square i.e., 25m

∴ Area of any one sector = \(\frac{\theta}{360}\) × πr^{2}

= \(\frac{90}{360}\) × 3.14 × 25 × 25 ……(For square θ = 90°)

= \(\frac{1}{4}\) × 3.14 × 625

= 0.785 × 625

= 490.625 m^{2}

Area of all three sectors = 4 × (Area of one sector)

= 4 × 490.625

= 1962.5 m^{2}

Now, Area of shaded portion = (Area of square) – (Area of four sectors)

= 2500 – 1962.5

∴ Area of shaded portion = 537.5 m^{2}

Question 5.

Solve the following sub-questions (Any one): [3]

(i) Circles with centres A, B and C touch each other externally. If AB = 3 cm, BC = 3 cm, CA = 4 cm, then find the radii of each circle.

Solution:

Suppose radius of circle with centre A is x cm

∴ Radius of circle with centre B = (3 – x) cm (∵ AB = 3 cm) and radius of circle with centre C = (4 – x) cm (∵ CA = 4 cm)

∴ (3 – x) + (4 – x) = BC = 3

∴ 3 – x + 4 – x = 3

∴ 7 – 2x =3

∴ 2x = 7 – 3

∴ 2x = 4

∴ x = 2

∴ Radius of circle with centre A = 2 cm

∴ Radius of circle with centre B = (3 – x)

= (3 – 2)

= 1 cm

∴ Radius of circle with centre C = (4 – x)

= (4 – 2)

= 2 cm

(ii) If sin θ + sin^{2}θ = 1

show that : cos^{2} θ + cos^{2} θ = 1

Solution:

sin θ + sin^{2} θ = 1 …(Given)

But sin^{2}θ + cos^{2}θ = 1 …….(Standard result)

∴ Putting the value 1 in given relation we get, comma

sin θ + sin^{2} θ = sin^{2} θ + cos^{2} θ

∴ sin θ = cos^{2}θ …… (i)

Now as per given relation

sin θ + sin^{2}θ = 1

∴ cos^{2}θ + (cos^{2}θ)^{2} = 1 […From equation (i)]

∴ cos^{2}θ + cos^{4}θ = 1

Hence proved.