Maharashtra Board Class 10 Maths 2 Sample Paper Set 4 with Answers

Maharashtra Board SSC Class 10 Maths 2 Sample Paper Set 4 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 2 Model Paper Set 4 with Answers

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-questions four alternative answers are given. Choose the correct alternative and write its alphabet: [4]
i. Distance of point (-3, 4) from the origin is ________. (C)
(A) 7
(B) 1
(C) 5
(D) -5
Answer:
(C) 5

ii. If in ADEF and ∆PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E, then which of the following statements is false? (B)

Answer:
B

iii. The maximum number of tangents that can be drawn to a circle from a point outside it is ________. (A)
(A) 2
(B) 1
(C) one and only one
(D) 0
Answer:
(A) 2

iv. If tan θ = \(\frac{12}{5}\) , then 5 sin θ – 12 cos θ = (B)
(A) \(\frac{119}{13}\)
(B) 0
(C) 1
(D) \(\frac{1}{13}\)
Answer:
(B) 0

Question 1.(B)
(B) Solve the following sub-questions: [4]
i. In ∆ABC, ∠B = 90°, ∠A = 30°, ∠C = 60°, AC = 14 cm, find AB.

ii. If the area of a circle is 314 cm2 and area of minor sector is 100 cm2, find the area of its , corresponding major sector.
iii. ∆DEF ~ ∆MNK. If DE = 2, MN = 5, then find the value of \(\frac{\mathrm{A}(\triangle \mathrm{DEF})}{\mathrm{A}(\triangle \mathrm{MNK})}\).
iv. If sec θ =\(\frac{2}{\sqrt{3}}\)= , then find the value of acute angle θ.
Answer:
i. By 30° – 60° – 90° theorem,

ii. area of major sector = area of circle – area of minor sector
= 314 – 100
= 214 cm²

iii. ∆DEF ~ ∆MNK …[Given]

iv. Given, sec θ = \(\frac{2}{\sqrt{3}}\)
But, sec 30° = \(\frac{2}{\sqrt{3}}\)
∴ sec θ = sec 30°
∴ θ = 30°

Question 2.
(A) Complete the following activities and rewrite it (Any two): [4]
i. The radii of two circular ends of frustum shaped bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. Complete the activity to find how many litres of water it can hold?
Capacity of bucket = Volume of frustum

ii. In the given figure, O is the centre of the circle.

∠AOB = 120°, m(arc AC) = 55°.
Use the given information and fill the boxes.
a. m(arc AXB) = □
b. m(arc CAB) = □
c. ∠COB = □
d. m(arc AYB) = □
iii. For finding AB and BC with the help of information given in the adjoining figure, complete the following activity.

(B) Solve the following sub-questions (Any four): [8]
i. Are the triangles in the given figure similar? If yes, by which test?

ii. Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (π = 3.14 )
iii. Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.
iv. If cot θ = \(\frac{40}{9}\), find the value of cosec θ.
v. Find the volume of a sphere of diameter 6 cm. [π = 3.14]
Answer:
(A) i. Capacity of bucket = Volume of frustum

ii. m(arc AXB) = m∠AOB …[Definition of measure of minor arc]
∴ m(arc AXB) = [120°]
m(arc CAB) = m(arc AC) + m(arc AXB) …[Arc addition property]
∴ m(arc CAB) = 55° + 120° = [175°]
m∠COB = m(arc CAB) …[Definition of measure of minor arc]
∴ ∠COB = [175°]
m(arc AYB) + m(arc AXB) = 360° …[Measure of a circle is 360°]
∴ m(arc AYB) + 120° = 360°
∴ m(are AYB) = [240°]

ii. AB = BC …[Given]
∴ ∠BAC = ∠BCA …[Isosceles triangle therem]

(B) i. In ∆PQR and ∆LMN,

∆PQR ~ ∆LMN …[SSS test of similarity]

ii. Given : Radius (r) = 10 cm,
Measure of the arc (θ) = 54°
To find : Area of the sector.
Solution:

∴ The area of the sector is 47.1 cm²

iii.

Steps of construction:
a. With centre P, draw a circle of radius 3.2 cm.
b. Take any point M on the circle and draw ray PM.
c. Draw line l ⊥ ray PM at point M. Line l is the required tangent to the circle at point M.

iv.

v. Given: For the sphere, diameter (d) = 6 cm
To find: Volume of the sphere.
Solution:

∴ The volume of the sphere is 113.04 cm³.

Question 3.
(A) Complete the following activity and rewrite it (Any one): [3]
i. In the given 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. Fill in the blanks to prove that, seg PR ǁ seg DF.

In ∆XDE, PQ ǁ DE … [Given]

ii. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram. Complete the following activity.

(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. Two buildings are facing each other on a road of width 12 metres. From the top of the first building, which is 10 metres high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?
iii. In the adjoining figure, two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and P, Q respectively. Prove that: seg SQ ǁ seg RP.

iv. Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Answer:
(A) i. In ∆XDE, PQ ǁ DE …[Given]

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

ii.

∴ Slope of side AB = Slope of side [CD], and
Slope of side BC = Slope of side [AD]
∴ side AB ǁ side CD, and side BC ǁ side AD
i.e., both the pairs of opposite sides of □ABCD are parallel.
∴ Points A, B, C and D are the vertices of a [parallelogram].

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

To prove: \(\frac{AB}{BC}=\frac{PQ}{QR}\)
Construction: Draw seg PC which intersects line m at point D.
Proof:

ii. Let AB and CD represent the heights of the two buildings, and BD represent the width of the road.
AB = 10 m
BD = 12 m
Draw seg AM ⊥ seg CD.

Angle of elevation = ∠CAM = 60°
In □ABDM,
∠B = ∠D = 90°
∠M = 90° …[seg AM ⊥ seg CD]
∴ ∠A = 90° …[Remaining angle of □ABDM]
∴ □ABDM is a rectangle. …[Each angle is 90°]
∴ AM = BD = 12 m …[Opposite sides of a rectangle]
DM = AB = 10 m
In right angled ∆AMC,

Now, CD = DM + CM …[C-M-D]
∴ CD = (10 + 12√3)m
∴ The height of the second building is (10 + 12√3)m.

iii. Given: Two circles intersect each other at points M and N.

To prove: seg SQ ǁ seg RP
Construction: Join seg MN.
Proof:
□RMNP is a cyclic quadrilateral.
∴ ∠MRP = ∠MNQ …(i) [Corollary of cyclic quadrilateral theorem]
Also, □MNQS is a cyclic quadrilateral.
∴ ∠MNQ + ∠MSQ = 180° …[Theorem of cyclic quadrilateral]
∴ ∠MRP + ∠MSQ = 180° …[From (i)]
But, they are a pair of interior angles on the same side of transversal RS on lines SQ and RP.
∴ seg SQ ǁ seg RP …[Interior angles test]

iv.

∴ PQ = RS …[From (i) and (iii)]
QR = PS …[From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
∴ □ PQRS is a parallelogram.
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 4.
Solve the following sub-questions (Any two): [8]
i. In ∆ABC, ∠ACB = 90°, seg CD ⊥ seg AB, seg DE ⊥ seg CB.
Show that CD² × AC = AD × AB × DE.

ii. Draw a triangle ABC, right angled at B such that AB = 3 cm, BC = 4 cm. Now, construct a triangle PBQ similar to ∆ABC each of whose sides is \(\frac{7}{4}\) times the corresponding sides of ∆ABC.
iii. In the given figure, line AB is tangent to both the circles touching at A and B.
OA = 29, BP = 18, OP = 61, then find AB.

Answer:
i. Given: In ∆ABC, ∠ACB = 90°, seg CD ⊥ seg AB, seg DE ⊥ seg CB.
To prove: CD² × AC = AD × AB × DE.
In ∆ACB,
∠ACB = 90° …[Given]
seg CD ⊥ hypotenuse AB
∴ CD² = AD × DB …(i)[Theorem of geometric mean]
In ∆DEB and ∆ACB,
∠DEB ≅ ∠ACB …[Each is of measure 90°]
∠DBE ≅ ∠ABC …[Common angle]
∆DEB ~ ∆ACB …[By AA test of similarity]

ii.

Steps of construction:
a. Draw AABC with AB = 3 cm, BC = 4 cm and ∠B = 90°.
Draw ray BX making an acute angle with side BQ.
b. Taking convenient distance on the compass, mark 7 points B₁, B₂, B₃, B₄, B₅, B₆ and B₇ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
c. Join B₄C. Draw a line parallel to B₄C through B₇. The line through B₇ intersects ray BC at Q.
d. Draw a line parallel to side AC through Q. Name the point of intersection of this line and ray BA as P.
∆PBQ is the required triangle similar to ∆ABC.

iii. Construction: Draw seg PM ⊥ seg OA.

In □MABP,
∠A = ∠B = 90° …[Tangent theorem]
∠M = 90° …[By construction]
∴ ∠P = 90° …[Remaining angle of a □MABP]
∴ □MABP is a rectangle. …[Each angle is 90°]
∴ AB = PM …(i)[Opposite sides of a rectangle]
AM = BP = 18 …(ii) [Opposite sides of a rectangle]
AM + OM = OA …[A-M-O]
∴ 18 + OM = 29 …[Substituting the given value and from (ii)]
∴ OM = 29 – 18
∴ OM = 11 units …(iii)
In ∆OMP, ∠OMP = 90° …[By construction]
∴ OM² + PM² = OP² …[By Pythagoras theorem]
∴ (11)² + PM² = (61)² …[Substituting the given value and from (iii)]
∴ PM² = (61)² – (11)² = 3721 – 121
∴ PM² = 3600
∴ PM = √36×100 …[Taking square root of both sides]
∴ PM = 6 × 10 = 60 units
AB = PM = 60 units …[From (i)]
∴ AB = 60 units

Question 5.
Solve the following sub-questions (Any one):
i. □ABCD is cyclic quadrilateral, lines AB and DC intersect in the point F and lines AD and BC intersect in the point E. Show that the circumcircles of ∆BCF and ∆CDE intersect in a point G on the line EF.

ii. There are three grassfields – one of the shape of an equilateral triangle, the other square and the third one hexagonal. A cow is to be tied to a peg by means of a rope 6 m long. The peg is fixed at any one vertex of the field. In which field should the cow be tied so that it has maximum area to graze?
Answer:
i. Given: □ABCD is a cyclic quadrilateral.
To prove: Circumcircles of ∆BCF and ∆CDE intersect in a point G on line EF
Construction: Since circumcircles of ∆BCF and ∆CDE are intersecting circles, let them intersect at the point G. Join seg CG.
Proof: □BFGC is a cyclic quadrilateral. …[By definition]

∴ ∠ABC = ∠CGE ..(i)[Exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle]
□DCGE is a cyclic quadrilateral. …[By definition]
∴ ∠ADC = ∠CGE …(ii)[Exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle]
□ABCD is a cyclic quadrilateral. …[Given]
∠ABC + ∠ADC = 180° …(iii)[Opposite angles of a cyclic quadrilateral are supplementary]
∴ ∠CGF + ∠CGE = 180° …[From (i), (ii) and (iii)]
∴ ∠CGF and ∠CGE form a linear pair of angles.
By definition of linear pair of angles, seg GF and seg GE are on the same line.
∴ G is a point on line EF.
∴ Circumcircles of ABCF and ACDE intersect in a point G on the line EF.

ii. Here, radius (r) = 6m
For equilateral triangle, θ₁ = 60°
For square, θ₂ = 90°
For hexagon, θ₃ = 120°
Area available for grazing = area of sector

∴ The cow must be tied in the hexagonal field so that it has maximum area to graze.

SSC Maharashtra Board Maths 2 Question Paper with Solutions