SSC Maharashtra Board Maths 2 Question Paper Dec 2020 with Answers

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

SSC Maths 2 Question Paper Dec 2020 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.
For each of the following sub-question four alternative answers are given. Choose the correct alternative and write its alphabet: [4]
i. ∆ABC ~ ∆PQR and ∠A = 45°, ∠Q = 87°, then ∠C = ________. (C)
(A) 45°
(B) 87°
(C) 48°
(D) 90°
Answer:
(C) 48°

ii. ∠PRQ is inscribed in the arc PRQ of a circle with centre ‘O’. (D)
If ∠PRQ = 75°, then m(arc PRQ) = ________.
(A) 75°
(B) 150°
(C) 285°
(D) 210°
Answer:
(D) 210°

iii. A line makes an angle of 60° with the positive direction of X-axis, so the slope of a line is ________. (C)
Answer:
(C) √3

iv. Radius of a sector of a circle is 5 cm and length of arc is 10 cm, then the area of a sector is ________. (B)
(A) 50 cm²
(B) 25 cm²
(C) 25 m²
(D) 10 cm²
Answer:
(B) 25 cm²
Hints:
i. ∆ABC ~ ∆PQR
∴ ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R …[Corresponding angle of similar triangle]
In ∆ABC, ∠A + ∠B + ∠C = 180°
∴ 45° + 87° + ∠C = 180°
∴ ∠C = 48°

ii. m(arc PRQ) = 360° – 2(m ∠PRQ)
= 360° – 2 × 75°
= 360° – 150°
= 210°

iii. Angle made w1th the pos1t1ve d1rect1on of X-ax1s (θ) = 60°
Slope of the l1ne (m) = tan θ
∴ m = tan 60° = √3
∴ The slope of the l1ne 1s √3.

iv.

Question 1.
(B) Solve the following sub-questions: [4]
i.

In the above figure, seg AB ⊥ seg BC and seg DC ⊥ seg BC.
If AB = 3 cm and CD = 4 cm, then find \(\frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{DCB})}\)
ii. In cyclic □ABCD, ∠B = 75°, then find ∠D.
iii. Point A, B, C are collinear. If slope of line AB is –\(\frac{1}{2}\), then find the slope of line BC.
iv. If 3 sin θ = 4 cos θ, then find the value of tan θ.
Answer:
i. ∆ABC and ∆DCB have same base BC.

ii. As oppos1te angles of a cycl1c quadrilateral are supplementary,
∠B + ∠D = 180°
∴ 75° + ∠D = 180°
∴ ∠D = 180° – 75°
∴ ∠D = 105°

iii. Points A, B, C are coll1near.
∴ Slope of l1ne AB = Slope of line BC
∴ Slope of line BC = –\(\frac{1}{2}\)

iv. 3 sin θ = 4 cos θ
∴ \(\frac{\sin \theta}{\cos \theta}=\frac{4}{3}\)
∴ tan θ = \(\frac{4}{3}\)

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

In ∆ABC, seg DE ǁ side BC. If AD = 6 cm, DB = 9 cm, EC = 7.5 cm, then complete the following activity to find AE.
Activity:
In ∆ABC, seg DEǁside BC  …(given)

ii.

In the above figure, chord AB and chord CD intersect each other at point E. If AE = 15, EB = 6, CE = 12, then complete the activity to find ED.
Activity:
Chord AB and chord CD intersect each other at point E …(given)

iii. If C(3, 5) and D(-2, -3), then complete the following activity to find the distance between points C and D.
Activity:

(B) Solve the following sub-questions (Any four):
i. ∆ABC ~ ∆PQR, A(∆ABC) = 81 cm², A(∆PQR) = 121 cm².
If BC = 6.3 cm, then find QR.
ii. In ∆PQR, ∠P = 60°, ∠Q = 90° and QR = 6√3 cm, then find the values of PR and PQ.
iii. Find the slope of a line passing through the points A(2, 5) and B(4, -1).
iv. Draw a circle with centre ‘O’ and radius 3.2 cm. Draw a tangent to the circle at any point P on it.
v. Find the surface area of a sphere of radius 7 cm.
Answer:
(A) i. In ∆ABC, seg DE ǁ side BC …(given)

ii. Chord AB and chord CD intersect each other at point E …(given)
∴ CE × ED = AE × EB …[Theorem of internal division of chords]

iii.

(B) i. ∆ABC ~ ∆PQR …[Given]

ii. In ∆PQR, ∠P = 60°, ∠Q = 90° …[Given]
∴ ∠R = 30° …[Remaining angle of a triangle]
∴ ∆PQR is a 30° – 60° – 90° triangle.

Also, PQ = \(\frac{1}{2}\)PR …[Side opposite to 30°]
∴ PQ = \(\frac{1}{2}\) × 12 …[From(i)]
∴ PQ = 6cm
∴ PR = 12 cm and PQ = 6 cm

iii. Let A(x₁, y₁) = A(2, 5) and B(x₂, y₂) = B(4, -1)
Here, x₁ = 2, x₂ = 4, y₁ = 5, y₂ = -1

∴ The slope of line AB is – 3.

iv.

v. Given: For the sphere, radius (r) = 7 cm
To find: Surface area of the sphere.
Solution:
Surface area of sphere = 4πr²
= 4 × \(\frac{22}{7}\) ×(7)²
= 88 × 7
= 616 cm²
∴ The surface area of the sphere is 616 cm².

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

In ∆PQR, seg PS ⊥ side QR, then complete the activity to prove PQ² + RS² = PR² + QS².
Activity:
In ∆PSQ, ∠PSQ = 90°
∴ PS² + QS² = PQ² …(Pythagoras theorem)
∴ PS² = PQ² – □ …(i)
Similarly,
In ∆PSR, ∠PSR = 90°
∴ PS² + □ = PR² …(Pythagoras theorem)
∴ PS² = PR² – □ …(ii)
∴ PQ² – □ = □ – RS² …from (i) and (ii)
∴ PQ² + □ = PR² + QS²
ii. Measure of arc of a circle is 36° and its length is 176 cm. Then complete the following activity to find the radius of circle.
Activity:
Here, measure of arc = θ = 36°
Length of arc = l = 176 cm


(B) Solve the following sub-questions (Any two): [6]
i. Prove that, “The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.”
ii. Draw a circle with centre ‘O’ and radius 3.4 cm. Draw a chord MN of length 5.7 cm in it. Construct tangents at points M and N to the circle.
iii. Prove that:

iv. Radii of the top and base of frustum are 14 cm and 8 cm respectively. Its height is 8 cm. Find its curved surface area.(π = 3.14)
Answer:
i. In ∆PSQ, ∠PSQ = 90°
∴ PS² + QS² = PQ² …(Pythagoras theorem)

ii. Here, measure of arc = θ = 36°
Length of arc = l = 176 cm

(B) i. Given: line lǁ line mǁ line n
Transversals t₁ and t₂ intersect the parallel lines at points A, B, C and P, Q, R respectively.

ii.

iii.

iv. Given: Radii (r₁) = 14 cm, (r₂) = 8 cm and height (h) = 8 cm
To find: Curved surface area of the frustum

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

In ∆ABC, ∠BAC = 90°, seg AP ⊥ side BC, B-P-C. Point D is the mid-point of side BC, then prove that 2AD² = BD² + CD².
ii.

In the above figure, chord AB = chord AD. Chord AC and chord BD intersect each other at point E. Then prove that:
AB² = AE × AC.
iii. A straight road leads to the foot of the tower of height 48 m. From the top of the tower the angles of depression of two cars standing on the road are 30° and 60° respectively. Find the distance between the two cars.(√3 =1.73)
Answer:
i. Given: ∠BAC = 90°,
seg AP ⊥ side BC, B-P-C.
Point D is the mid-point of side BC.
To prove: 2AD² = BD² + CD²
Proof: In ∆ABC, ∠BAC = 90° …[Given]
∴ BC² = AB² + AC² …(i)[Pythagoras theorem]
D is the mud-point of hypotenuse BC.
∴ AD = \(\frac{1}{2}\)BC …(ii)[In right angled triangle, median to hypotenuse is half the hypotenuse.]
BD = CD = \(\frac{1}{2}\)BC …(iii)[D is the mud-pomt of side BC]
In ∆ABD, ∠ADB 1s an acute angle and AP ⊥ BC
∴ AB² = BD² + AD² – 2BD × PD …(iv)[Application of Pythagoras theorem]
In ∆ADC, ∠ADC is an obtuse angle and AP ⊥ BC
∴ AC² = CD² + AD² + 2CD × PD …(v)[Application of Pythagoras theorem]
∴ AB² + AC² = BD² + AD² – 2BD × PD + CD² + AD² + 2CD × PD …[Adding (iv) and (v)]
∴ AB² + AC² = BD² + CD² + 2AD² – 2BD × PD + 2BD × PD …[From (iii)]
∴ BC² = BD² + CD² + 2AD² …[From (i)]
∴ (2AD)² = BD² + CD² + 2AD² …[From (ii)]
∴ 4AD² = BD² + CD² + 2AD²
∴ 2AD² = BD² + CD²

ii. Given : chord AB ≅ chord AD
chord AC and chord BD intersect at point E.
To prove : AB² = AE × AC
Construction: Draw seg BC.
Proof:

chord AB ≅ chord AD …[Given]
∴ ∠ADB ≅ ∠ABD …(i)[Isosceles triangle theorem]
∴ ∠ADB ≅ ∠ACB …(ii)[Angles Inscribed in the same arc]
∴ ∠ABD ≅ ∠ACB …[From (i) and (ii)]
∴ ∠ABE ≅ ∠ACB …(iii) [B-E-D]
In ∆ABE and ∆ACB,
∠EAB ≅ ∠CAB …[Common angle]
∴ ∠ABE ≅ ∠ACB …[From (iii)]
∴ ∆ABE ~ ∆ACB …[AA test of similarity]
∴ \(\frac{AB}{AC}=\frac{AE}{AB}\) …[Correspond1ng sides of similar triangles]
∴ AB² = AE × AC

iii. Let AB be the tower of height 48m.

Let C and D be the position of two cars at angles of depression 60° and 30° respectively.
AM is the horizontal line and ∠MAD and ∠MAC are angles of depression of measures 30° and 60° respectively. ∠MAD and ∠ADB, ∠MAC and ∠ACB are pairs of alternate angles.
∴ m∠ADB = 30° and m∠ACB = 60°
To find: Distance between C and D.
i.e., l(seg CD)
In ∆ADB, ∠B = 90°

∴ The distance between the two cars is 55.36 m.

Question 5.
Solve the following sub-questions (Any one): [3]
i. Let M be a point of contact of two internally touching circles. Let line AMB be their common tangent. The chord CD of the bigger circle touches the smaller circle at point N. The chord CM and chord DM of bigger circle intersect the smaller circle at point P and R respectively.
a. From the above information draw the suitable figure.
b. Draw seg NR and seg NM and write the two pairs of congruent angles in smaller circle considering tangent and chord.
c. By using the property which is used in (b) write the two pairs of congruent angles in the bigger circle.
ii. Draw a circle with centre ‘O’ and radius 3 cm. Draw a tangent segment PA having length √40 cm from an exterior point P.
Answer:
i.

ii. In ∆OAP,
seg OA ⊥ tangent AP …[Tangent is perpendicular to radius]
∴ ∠OAP = 90°
∴ OP² = OA² + AP² … [Pythagoras theorem]
∴ OP² = (3)² + (√40)²
∴ OP² = 9 + 40
∴ OP² = 49 OP = 7 cm
∴ Point P is at a distance 7cm from the centre.

SSC Maharashtra Board Maths 2 Question Paper with Solutions