Maharashtra Board Class 10 Maths 2 Sample Paper Set 6 with Answers

Maharashtra Board SSC Class 10 Maths 2 Sample Paper Set 6 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 2 Model Paper Set 6 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. A line makes an angle of 30° with the positive direction of X-axis. So the slope of the line is _________. (C)
Answer:
(C) \(\frac{1}{\sqrt3}\)

ii. If the volume of a cube is 1728 cm³, then the length of its edge is equal to (A)
(A) 12 cm
(B) 14 cm
(C) 16 cm
(D) 18 cm
Answer:
(A) 12 cm

iii. In a cyclic , ABCD, twice the measure of ∠A is thrice the measure of ∠C. Find the measure of ∠C? (B)
(A) 36°
(B) 72°
(C) 90°
(D) 108°
Answer:
(B) 72°

iv. 9 sec² A- 9 tan² A = (D)
(A) 0
(B) 3
(C) -9
(D) 9
Answer:
(D) 9
Hints:
i. Slope = tan 30° = \(\frac{1}{\sqrt3}\)

ii. Volume of a cube = (side)³
∴ 1728 = (side)³
∴ side = 12 cm

iii. ∠A + ∠C = 180° …[Theorem of cyclic quadrilateral]
∴ 2∠A + 2∠C = 2 × 180° …[Multiplying both sides by 2]
∴ 3∠C + 2∠C = 360° …[∵ 2∠A = 3∠C]
∴ 5∠C = 360°
∴ ∠C = 72°

iv. 9 sec²A – 9 tan²A = 9(sec²A – tan²A)
= 9(1) = 9

Question 1.
(B) Solve the following sub-questions: [4]
i. In the adjoining figure, BC ⊥ AB, AD ⊥ AB,
BC = 4. AD = 8, then find \(\frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{ADB})}\)

ii. If tan θ = 5. w here θ is an acute angle, find sec θ using the identity.
iii. Find the diagonal of a square whose side is 16 cm.
iv. Find the distance betweeen the points O(0, 0) and P (-3, 4).
Answer:
i. ∆ABC and ∆ADB have same base AB.

ii. 1 + tan²θ = sec²θ
∴ 1 + (5)² = sec²θ …[∵ tanθ = 5]
∴ 1 + 25 = sec²θ
∴ sec² θ = 26
∴ sec θ = √26 …[Taking square root of both sides]

iii. Diagonal of a square = √2 × (side) = √2 × (16)
∴ Diagonal of a square = 16√2 cm

iv.

Question 2.
(A) Complete the following activities and rewrite it (Any two): [8]
i. In the given figure, M is the midpoint of seg AB and seg CM is a median of ∆ABC.

Point C’ is the common vertex of ∆AMC and ∆BMC.
∴ The height of the two triangles is equal.

ii. Read the following flow chart to draw the tangents to the circle at points A and B.

iii. In the adjoining figure, seg AB is a diameter of a circle with centre O. The bisector of ∠ACB intersects the circle at point D.
Prove that, seg AD ≅ seg BD.

Complete the following proof by filling the blanks.
Construction: Draw seg OD.
Proof: ∠ACB = 90° …[Angle inscribed in a semicircle]

∴ seg AD ≅ seg BD
(B) Solve the following sub-questions (Any four):
i. If cot θ = \(\frac{40}{9}\), find the value of cosec θ.
ii. Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.
iii. Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.
iv. In the adjoining figure, chords AC and DE intersect at B. If ∠ABE = 108°, m(arc AE) = 95°, find m (arc DC).

v. Find the slope of the line passing through the points L(-2, -3) and M(-6, -8).
Answer:
(A) i. Draw CD ⊥ AB, A-D-B.
seg CM is the median of ∆ABC. …[Given]

ii.

iii. ∠ACB = 90°

∴ seg AD ≅ seg BD

(B) i. cot θ = \(\frac{40}{9}\)
We know that,
1 + cot² θ = cosec² θ
∴ 1 + (\(\frac{40}{9}\))² = cosec² θ

ii. 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.

iii. Given: Radius (r) = 3.5 cm,
length of arc (I) = 2.2 cm
To find: Area of the sector.
Solution:

∴ The area of the sector is 3.85 cm²

iv. Chords AC and DE intersect internally at point B.
∴ ∠ABE = \(\frac{1}{2}\) [m(arc AE) + m(arc DC)]
∴ 108° = \(\frac{1}{2}\) [95° + m(arc DC)]
∴ 108° × 2 = 95° + m(arc DC)
∴ 95° + m(arc DC) = 216°
∴ m(arc DC) = 216° – 95°
∴ m(arc DC) = 121°

v. Here, X₁ = – 2, x₂ = -6, y₁ = – 3, y₂ = -8

Question 3.
(A) Complete the following activity and rewrite it (Any one): [3]
In the figure, seg AB ǁ Y-axis and seg CB ǁ X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below.
In ∆ABC, ∠B = 90°

ii. In trapezium PQRS, side PQ ǁ side SR, AR = 5 AP, AS prove that SR = 5 PQ. Complete the following activity.
side PQ ǁ side SR …[Given]
and seg SQ is their transversal.

(B) Solve the following sub-questions (Any two): [6]
i. Prove that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides.
ii. ∆AMT ~ ∆AHE. In ∆AMT, AM = 6.3 cm. ∠TAM = 50°, AT = 5.6 cm, \(\frac{AM}{AH}=\frac{7}{5}\).
Construct ∆AHE.
iii. A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.
iv. In the adjoining 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
c. arc AB
Answer:
(A) i. In ∆ABC, ∠B = 90°

ii. side PQ ǁ side SR …[given]
and seg SQ is their transversal.

But, AS = 5 AQ …[Given]

(B) i. Given: In ∆ABC, ∠ABC = 90°
To prove: AC² = AB² + BC²
Construction: Draw seg BD ⊥ AC, A-D-C.
Proof: In ∆ABC, ∠ABC = 90° …[Given]

seg BD ⊥ hypotenuse AC [Construction]
∴ ∆ABC ~ ∆ADB …[Similarity of right angled triangles]
∴ \(\frac{AB}{AD}=\frac{AC}{AB}\) …[Corresponding sides of similar triangles]
∴ AB² = AD × AC …(i)
Also, ∆ABC ~ ∆BDC …[Similarity of right angled triangles]
∴ \(\frac{BC}{DC}=\frac{AC}{BC}\) …[Corresponding sides of similar triangles]
∴ BC² = DC × AC …(ii)
AB² + BC² = AD × AC + DC × AC …[Adding (i) and (ii)]
= AC (AD + DC)
= AC × AC …[A-D-C]]
∴ AB² + BC² = AC²
i.e. AC² = AB² + BC²

ii.

Steps of construction:
a. Draw ∆AMT of given measure. Draw ray AB making an acute angle with side AM.
b. Taking convenient distance on the compass, mark 7 points A₁, A₂, A₃, A₄, A₅, A₆ and A₇, such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
c. Join A₇M. Draw lines parallel to A₇M through A₁, A₂, A₃, A₄, A₅ and A₆. The line through A₅ intersects seg AM at H.
d. Draw a line parallel to side TM through H. Name the point of intersection of this line and seg AT as E.
∆AHE is the required triangle similar to ∆AMT.

iii. Let AB represent the height of the tree.
Suppose the tree broke at point C and its top touches the ground at D.
BD = 20 m
AC is the broken part of the tree which takes position CD such that ∠CDB = 60°
∴ AC = CD …(i)

∴ The height of the tree was 40 + 20√3 m.

iv. a. seg OA = seg OB = radius …(i)[Radii of the same circle]
seg AB = radius …(ii)[Given]
∴ seg OA = seg OB = seg AB …[From (i) and (ii)]
∴ ∆OAB is an equilateral triangle.
∴ m∠AOB = 60° …[Angle of an equilateral triangle]

b. m∠ACB = \(\frac{1}{2}\) m∠AOB …[Measure of an angle subtended by an arc at … a point on the circle is half of the measure of the angle subtended by the arc at the centre]
= \(\frac{1}{2}\) × 60°
∴ m∠ACB = 30°

c. m(arc AB) m∠AOB …[Definition of measure of minor arc]
∴ m(are AB) = 60°

Question 4.
Solve the following sub-questions (Any two): [8]
i. In □ABCD, side BC ǁ side AD. Diagonal AC and diagonal BD intersect in point Q. If AQ = \(\frac{1}{3}\) AC, then show that DQ = \(\frac{1}{2}\) BQ.

ii. The radius of the internal and external surface of the hollow spherical shell are 3 cm and 5 cm respectively. After melting, it is recasted into a solid cylinder of height 2\(\frac{2}{3}\) cm. What is the radius of cylinder?
iii. A circle with centre P is inscribed in the ∆ABC. Side AB, side BC and side AC touch the circle at points L, M and N respectively . Radius of the circle is r.
Prove that:
A(∆ABC) = \(\frac{1}{2}\)(AB + BC + AC) × r.

Answer:
i. side AD ǁ side BC and seg BD is their transversal.
∠ADB = ∠CBD …(i)[Alternate angles]
In ∆AQD and A∆CQB,
∠ADQ ≅ ∠CBQ …[From (i), B-Q-D]
∠AQD ≅ ∠CQB …[Vertically opposite angles]
∴ ∆AQD ~ ∆CQB …[AA test of similarity]

ii. Volume of sphere = \(\frac{4}{3}\) πr³
∴ External volume of spherical shell = \(\frac{4}{3}\) π(5)³ …(i)
∴ Internal volume of spherical shell = \(\frac{4}{3}\) π(3)³ …(ii)
∴ Volume of metal = (External volume of spherical shell) – (Internal volume of spherical shell)

Volume of the cylinder = πr²h
Volume of cylinder = Volume of metal
∴ Radius of the cylinder is 7 cm.

iii. Given: Side AB, side BC and side AC touch the circle at points L, M and N respectively. Radius of the circle is r.
To prove: A(∆ABC) = \(\frac{1}{2}\) (AB + BC + AC) × r
Construction: Join seg PM, seg PN, seg PL, seg AP, seg BP and seg CP.
Proof: seg BC is a tangent to circle at M.

Question 5.
Solve the following sub-questions (Any one): [3]
i. 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?
ii. In a trapezium ABCD,
seg AB ǁ seg DC, seg BD ⊥ seg AD, seg AC ⊥ seg BC. If AD = 15, BC = 15, AB = 25 and CD = 7, then find A(□ABCD).

Answer:
i. Here, radius (r) = 6 m
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.

ii. Construction: Draw seg CF ⊥ seg AB. DC

∴ A(□ABCD) = 192 sq. units

SSC Maharashtra Board Maths 2 Question Paper with Solutions