Maharashtra Board Class 10 Maths 2 Sample Paper Set 3 with Answers

Maharashtra Board SSC Class 10 Maths 2 Sample Paper Set 3 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 2 Model Paper Set 3 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. □PQRS is a trapezium, and AB ǁ PS ǁ QR. If PA = 3 cm, (D)
AQ = 1.4 cm, BR = 2.1 cm, then SB =

(A) 2 cm
(B) 2.5 cm
(C) 4 cm
(D) 4.5 cm
Answer:
(D) 4.5 cm

ii. If ∆ABC ~ ∆PQR and \(\frac{AB}{PQ}=\frac{7}{5}\), then ________ (A)
(A) ∆ABC is bigger.
(B) ∆PQR is bigger.
(C) both triangles will be equal.
(D) can not be decided.
Answer:
(A) ∆ABC is bigger.

iii. In a right angled triangle, if sum of the squares of the sides making right angle is 169, then what is the length of the hypotenuse? (B)
(A) 15
(B) 13
(C) 5
(D) 12
Answer:
(B) 13

iv. If the area of a pircle is 200 cm² and that of the minor sector is 50 cm², then the area of the corresponding major sector is (C)
(A) 50 cm²
(B) 100 cm²
(C) 150 cm²
(D) 250 cm²
Answer:
(C) 150 cm²
[Note: For this question, student should write the correct option along with its contents in the answer sheet. (See above)
Here hints provided for understanding purpose of the students.]
Hints:
Since AB ǁ PS ǁ QR,
∴ \(\frac{PA}{AQ}=\frac{SB}{BR}\) …[Property of three parallel lines and their transversals]
∴ \(\frac{3}{1.4}=\frac{SB}{2.1}\)
∴ SB = 4.5 cm,

iii. In ∆PQR, ∠Q = 90°

∴ PR² = PQ² + QR² …[Pythagoras theorem]
∴ PR² = 169
∴ PR = √l69 = 13

iv. Area of major sector = Area of circle – area of minor sector
= 200 – 50
= 150 cm²

(B) Solve the following sub-questions: [4]
i. If the sides of a triangle are 3 cm, 4 cm and 5 cm respectively, determine whether the triangle is a right angled triangle or not?
ii. In the given figure,
∠L = 35°. Find m(arc MAN).

iii. What is the value of sin²θ + \(\frac{1}{1+\tan ^2 \theta}\)?
iv. If the radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm, find its slant height.
Answer:
i. Since, 3² + 4² = 5²
∴ By converse of pythagoras theorem the triangle is a right angled triangle.

ii. ∠L = \(\frac{1}{2}\)m (arc MAN) …[Inscribed angled theorem]
∴ 35°= \(\frac{1}{2}\)m(arc MAN)
∴ m(arc MAN) = 2 × 35° = 70°

Question 2.
(A) Complete the following activities and rewrite it (Any two): [4]
i. In the figure, DABCD is a cyclic quadrilateral, seg AB is a diameter. If Z ADC = 120°, complete the following activity to find the measure of Z BAC.

□ABCD is a cyclic quadrilateral.
∴ ∠ADC + ∠ABC = 180° …[Opposite angles of a cyclic quadrilateral are supplementary]
∴ 120° + ∠ABC = 180°

ii. Complete the table below the graph with the help of the following graph.

Write your observation from the table.
iii. In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC, then prove that ED ǁ BC. Complete the proof by filling the boxes.
In ∆ABC, ray BD bisects ∠ABC. …[Given]

But, seg AB = seg AC …(iii)[Given]

(B) Solve the following sub-questions (Any four): [8]
i. Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.

ii. Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.
iii. Find the co-ordinates of a point on Y-axis which is equidistant from M (-5, -2) and N (3, 2).
iv. Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.
v. In the given figure, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following.
a. ∠AOB
b. ∠ACB

Answer:
(A) i. □ABCD is a cyclic quadrilateral.
∠ADC + ∠ABC = 180° …[Opposite angles of a cyclic quadrilateral are supplementary]
120° + ∠ABC = 180°

ii.

∴ For any two points (x₁, y₁) and (x₂, y₂) on a line, the ratio \(\frac{y_2-y_1}{x_2-x_1}\) is always constant.

iii. In ∆ABC, ray BD bisects ∠ABC. …[Given]

(B) i. Let □ABCD be the given rectangle.
AB = 12 cm, BC = 35 cm
In ∆ABC, ∠B = 90° …[Angle of a rectangle]
∴ AC² = AB² + BC² …[Pythagoras theorem]
= 12² + 35²
= 144 + 1225 = 1369
∴ AC = √l369 …[Taking square root of both sides]
= 37 cm
∴ The diagonal of the rectangle is 37 cm.

ii. Given: For the cone,
radius (r) = 1.5 cm,
perpendicular height (h) = 5 cm
To find: Volume of the cone.
Solution:

∴ The volume of the cone is 11.79 cm³.

iii. Suppose point P(0 , y) on the Y-axis is equidistant from points M and N.
∴ PM = PN
∴ PM² = PN² …[Squaring both sides]
∴ [0 – (-5)]² + [y – (-2)]² = (0 – 3)² + (y – 2)²
∴ (5)² + (y + 2)² = (-3)² + (y – 2)²
∴ 25 + y² + 4y + 4 = 9 + y² – 4y + 4
∴ 8y = -16
∴ y = -2
∴ The point on Y-axis which is equidistant from points M and N is (0. -2).

iv. Analysis:
seg PM ⊥ line l …[Tangent is perpendicular to radius]
The perpendicular to seg PM at point M will give the required tangent at M.

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.

v. a. seg OA = seg OB = radius …[Radii of the same circle]
seg AB = radius …[Given]
seg OA = seg OB = seg AB
∴ ∆OAB is an equilateral triangle.
∴ m∠AOB = 60° …[Angle of an equilateral triangle]

b. m∠ACB = \(\frac{1}{2}\) m∠AOB …[Inscribed angle theorem]
= \(\frac{1}{2}\) × 60°
∴ m∠ACB = 30°

Question 3.
(A) Complete the following activity and rewrite it (Any one): [3]
In the adjoining figure, circles with centres X and Y touch each other at point Z. A secant passing through Z intersects the circles at points A and B respectively. Prove that, radius XA ǁ radius YB.

Fill in the blanks and complete the proof.
Construction: Draw segments XZ and YZ.
Proof:
By theorem of touching circles, points X, Z, Y are collinear.

ii. In the adjoining figure, seg PQ ǁ seg DE, A(∆PQF) = 20 sq. units,
PF = 2 DP, then find A (□ DPQE) by completing the following activity.
A(∆PQF) = 20 sq.units, PF = 2 DP, …[Given]

Let us assume DP = x.
∴ PF = 2x
DF = DP + PF = x + 2.r = 3x …[D – P – F]
In ∆FDE and ∆FPQ,

(B) Solve the following sub-questions (Any two):
i. Prove that in a right angled triangle, if an altitude is drawn to the hypotenuse, then the two triangles formed will be similar to the original triangle and to each other.
ii. A tree was broken due to storm. Its broken upper part was so inclined that its top touched the ground making an angle of 30° with the ground. The distance from the foot of the tree and the point where the top touched the ground was 10 metres. What was the height of the tree.
iii. A regular hexagon is inscribed in a circle of radius 14 cm. Find the area of the region between the circle and the hexagon, (π = \(\frac{22}{7}\), √3 = 1.732)
iv. Find the co-ordinates of the points of trisection of the line segment AB with A(2, 7) and B(-4, -8)
Answer:
(A) i. By theorem of touching circles, points X, Z, Y are collinear.

ii. A(∆PQF) = 20 sq.units, PF = 2 DP, …[Given]
Let us assume DP = x.
∴ PF = 2x
DF = DP + PF = x + 2x = 3x …[D – P – F]
In ∆FDE and ∆FPQ,

(B) i. Given: In ∆ABC, ∠ABC = 90°, seg BD ⊥ hypotenuse AC, A-D-C.
To prove: ∆ABC ~ ∆ADB
∆ABC ~ ∆BDC
∆ADB ~ ∆BDC

Proof:
In ∆ABC and ∆ADB,
∠ABC ≅ ∠ADB …[Each angle is of measure 90°]
∠BAC ≅ ∠DAB …[Common angle]
∆ABC ~ ∆ADB …(i)[AA test of similarity]
In ∆ABC and ∆BDC,
∠ABC ≅ ∠BDC …[Each angle is of measure 90°]
∠ACB ≅ ∠BCD …[Common angle]
∆ABC ~ ∆BDC …(ii)[AA test of similarity]
∆ADB ~ ∆BDC …(iii)[From (i) and (ii)]
∆ABC ~ ∆ADB ~ ∆BDC …[From (i), (ii) and (iii)]

ii. Let AB Represent the height of the tree.
Suppose the tree broke at point C and its top touches the ground at D.
BD = 10 m

∴ The height of the tree was 10√3 m.

iii. Side of the hexagon = 14 cm …[Given]

Radius of circle (r) = side of hexagon = 14 cm
A(circle) = πr²
= \(\frac{22}{7}\) × 14 × 14
= 616 cm²
Area of the region between the circle and the hexagon= A(circle) – A(hexagon)
= 616 – 509.208
= 106.792 cm²
∴ The area of the region between the circle and hexagon is 106.792 cm².

iv. A (2, 7), B (-4, -8)
Suppose the points P and Q trisect seg AB.
∴ AP = PQ = QB

Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,

Co-ordinates of Q are (-2, -3).
∴ The co-ordinates of the points of trisection seg AB are (0, 2) and (-2, -3).

Question 4.
Solve the following sub-questions (Any two): [8]
i. G is the centroid of ∆ABC. GE and GF are drawn parallel to AB and AC respectively.
Find A(∆GEF) : A(∆ABC).

ii. Draw a ∆ABC with side BC = 6 cm, ∠B = 45° and ∠A = 100°, then construct a triangle PBQ whose sides are \(\frac{4}{7}\) times the corresponding sides of AABC.
iii. In the given figure, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q.
Prove that, ∠PRQ + ∠PSQ =180°.

Answer:
i. In ∆GED and ∆ABD,
∠GED ≅ ∠ABD ..[Corresponding angles, GE ǁ AB on transversal BC]
∠GDE ≅ ∠ADB …[Common angle]
∴ ∆GED ~ ∆ABD …[By AA test of similarity]

∴ A(∆GEF) : A(∆ABC) = 1 : 9

ii. Analysis :
In ∆ABC,
BC = 6 cm, ∠B – 45°, ∠A = 100°
∴ ∠C = 35° …[Remaining angle of ∆ABC]

Steps of construction:
a. Draw ∆ABC of the given measure. Draw ray BX making an acute angle with side BC.
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 B₇C. Draw a line parallel to B₇C through B₄. The line through B₄ 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. Given: Two circles intersect each other at points S and R. line PQ is a common tangent.
To prove: ∠PRQ + ∠PSQ = 180°
Construction: Draw seg SR.

Proof:
Line PQ is the tangent at point P and seg PR is a secant.
∠RPQ = ∠PSR …(i) [”Tangent secant theorem]
∠PQR = ∠QSR …(ii)
In ∆PQR,
∠PQR + ∠PRQ + ∠RPQ = 180° …[Sum of the measures of angles of a triangle is 180°]
∠QSR + ∠PRQ + ∠PSR = 180° …[From (i) and (ii)]
∠PRQ + ∠QSR + ∠PSR = 180°
∠PRQ + ∠PSQ = 180° …[Angle addition property]

Question 5.
Solve the following sub-questions (Any one): [3]
i. If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a² + b² = m² + n².
ii. In the given figure, □XLMT is a rectangle. LM = 21 cm, XL = 10.5 cm. Diameter of the smaller semicircle is half the diameter of the larger semicircle. Find the area of non-shaded region.

Answer:
i. m = a cos θ + b sin θ and n = a sin θ – b cos θ …[Given]
Consider,
m² + n² = (a cos θ + b sin θ)² + (a sin θ – b cos θ)²
= a²cos²θ + 2ab-sin θ cos θ + b²sin²2 θ + a²sin² θ – 2ab-sin θ cos θ + b²cos² θ …[∵ (a + b)² = a² + 2ab + b², (a -b)² = a² -2ab + b²]
= a cos² θ + a2sin² θ + b²sin² θ + b²cos² θ
= a²(cos² θ + sin² θ) + b²(sin² θ + cos² θ)
= a²(l) + b²(l)
= a² + b²
∴ a² + b² = m² + n²

ii. Let diameter of the smaller semicircle = 2r,
diameter of the larger semicircle = 4r
LM = 21cm
∴ LM = 2r + 4r
∴ 21 = 6r
∴ r = \(\frac{21}{6}\)
∴ r = 3.5 cm
∴ Radius of smaller semicircle = r = 3.5 cm
and radius of larger semicircle (R) = 2 × 3.5 = 7cm

∴ Area of non-shaded region
= Area of rectangle – (Area of smaller semicircle + Area of larger semicircle)
= 220.50 – (19.25 + 77)
= 220.50 – 96.25
= 124.25 cm²

SSC Maharashtra Board Maths 2 Question Paper with Solutions