Maharashtra Board Class 10 Maths 2 Sample Paper Set 7 with Answers

Maharashtra Board SSC Class 10 Maths 2 Sample Paper Set 7 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 2 Model Paper Set 7 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. In ∆ABC, AB = 6√3 cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A. (A)
(A) 30°
(B) 60°
(C) 90°
(D) 45°
Answer:
(A) 30°

ii. The slope of the line parallel to Y-axis (D)
(A) is 0
(B) is 1
(C) is -1
(D) cannot be determined
Answer:
(D) cannot be determined

iii. The curved surface area of a cylinder is 440 cm² and its radius is 5 cm. Find its height. (A)
(A) \(\frac{44}{\pi}\) cm
(B) 22π cm
(C) 44π cm
(D) \(\frac{22}{\pi}\) cm
Answer:
(A) \(\frac{44}{\pi}\) cm

iv. sin 45° + cos 45° = (A)

Answer:
(A) √2
Hints:
i. We know that, 6 = \(\frac{1}{2}\)(12)

BC = \(\frac{1}{2}\) AC
∠A = 30° …[Converse of 30° – 60° – 90° theorem]

ii. line l ǁ Y axis

iii. Curved surface area of cylinder = 2πrh
∴ 440 = 2 × π × 5 × h
∴ h = \(\frac{440}{10 \pi}=\frac{44}{\pi}\) cm

iv.

Question 1.
(B) Solve the following sub-questions: [4]
i. In the following figure, RP : PK = 3 : 2, find the value of A(∆TRP) : A(∆TPK).

ii. Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.
iii. Write the slope of X-axis.
iv. If 5 sin θ – 12 cos θ = 0, then find the value of tan θ.
Answer:
(B) i. RP : PK = 3 : 2 …[Given]

ii. Since, the distance between the centres of the circles touching externally is equal to the sum of their radii.
∴ Distance between their centres = 5.5 + 4.2 = 9.7 cm

iii. Inclination of X-axis is zero.
∴ Slope = tan θ = tan 0 = 0

iv. 5 sin θ – 12 cos θ = 0
∴ 5 sin θ = 12 cos θ

Question 2.
(A) Complete the following activities and rewrite it (Any two): [4]
i. Complete the following flow chart to draw a tangent to a circle at a point on the circle.

ii. If the line passing through points P (-12, -3) and Q (4, k) has slope\(\frac{1}{2}\), complete the activity to find the value of k.


iii. In the given figure, ray PQ touches the circle at point Q. PQ = 12, PR = 8, complete the following activity to find PS and RS.
Ray PQ is a tangent to the circle at point Q and seg PS is the secant. …[Given]

(B) Solve the following sub-questions (Any four): [8]
i. In ∆LMN, ray MT bisects ∠LMN. If LM = 6. MN = 10, TN = 8, then find LT.

ii. If tan θ = \(\frac{3}{4}\), find the value of sec θ.
iii. Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm². Find the area of its corresponding major sector, (π = 3.14 )
iv. In, the adjoining figure, points G, D, E, F are concyclic points of a circle with centre C. ∠ECF = 70°, m(arc DGF) = 200°. Find m(arc DE) and m(arc DEF).

v. Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.
Answer:
(A) i

iii. Ray PQ is a tangent to the circle at
point Q and seg PS is the secant. …[Given]
∴ PR × PS = PQ² …[Tangent secant segments theorem]
∴ 8 × PS = 12²
∴ 8 × PS = 144

(B) i.

ii.

iii. Given: Radius (r) = 10 cm,
area of minor sector = 100 cm²
To find: Area of major sector.
Area of circle = πr²
= 3.14 × (10)²
= 3.14 × 100
= 314 cm²
Now, area of major sector _ area of circle – area of minor sector
= 314 – 100
= 214 cm²
∴ The area of the corresponding major sector is 214 cm².

iv. m(arc EF) = mZECF …[Definition of measure of arc]
∴ m(arc EF) = 70°
m(arc DE) + m(arc DGF) + m(arc EF) = 360° …[Measure of a circle is 360°]
∴ m(arc DE) = 360° – m(arc DGF) – m(arc EF)
= 360° – 200° – 70°
∴ m(arc DE) = 90°
m(arc DEF) = m(arc DE) + m(arc EF) …[Property of sum of measures of arcs]
= 90° + 70°
∴ m(arc DEF) = 160°

v.

Steps of construction:
a. Draw a circle of any radius and take any point A on it.
b. Draw chord AB of any length and an inscribed ∠BCA of any measure.
c. By taking C as centre and any convenient distance on compass draw an arc intersecting the arms of ∠BCA in points P and Q.
d. With A as centre and the same distance in the compass, draw an arc intersecting the chord AB at point S.
e. Taking radius equal to PQ and S as centre draw an arc intersecting the previously drawn arc. Name the point of intersection as R.
f. Draw line AR. Line AR is the required tangent to the circle.

Question 3.
(A) Complete the following activity and rewrite it (Any one):
i. In ∆ABC, ∠C is an acute angle, seg AD ⊥ seg BC. A
Prove that: AB² = BC² + AC² – 2 BC × DC.

Complete the following activity.
Let AB = c, AC = b, AD = p,
BC = a, DC = x
BD = BC – DC …[B-D-C]

In ∆ADC, ∠D = 90°

c² = a² + b² – 2ax
AB- = BC² + AC² – 2 BC × DC
ii. Find the type of the quadrilateral if points A (-4, -2), B (-3, -7), C (3, -2) and D (2, 3) are joined serially by completing the following activity.

(B) Solve the following sub-questions (Any two):
i. Prove that if a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.
ii. 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.
iii.

iv. In the given figure, if WX = 25, YT = 8, YZ = 26, find WT.

Answer:
(A) i. Let AB = c, AC = b, AD = p,
BC = a, DC = x
BD = BC – DC …[B-D-C]

∴ c² = a² + b² – 2ax
∴ AB² = BC² + AC² – 2 BC × DC

ii.

lope of AB = slope of [CD]
∴ line AB ǁ line CD
slope of [BC slope of AD
∴ line BC ǁ line AD
Both the pairs of opposite sides of DABCD are parallel.
∴ □ABCD is a parallelogram.
∴ The quadrilateral formed by joining the points A, B, C and D is a [parallelogram].

(B) i. Given: In AABC, line l ǁ side BC.
Line l intersects sides AB and AC at points P and Q respectively.
To prove: \(\frac{AP}{PB}=\frac{AQ}{QC}\)
Construction: Draw seg BQ and seg CP.
Proof:

∆APQ and ∆BPQ have a common vertex Q and their bases AP and BP lie on the same line AB, hence they have equal heights.

Also, ∆APQ and ∆CPQ have a common vertex P and their bases AQ and QC lie on the same line AC, hence they have equal heights.

seg PQ is the common base of ∆BPQ and ∆CPQ.
∆BPQ and ∆CPQ lie between parallel lines PQ and BC.
∴ ∆BPQ and ∆CPQ have equal height.
∴ A(∆BPQ) = A(∆CPQ) …(iii)[Areas of two triangles having equal base and height are equal]

ii.

Steps of construction:
a. With centre P, draw a circle of radius 4.1 cm.
b. Take point Q such that PQ = 7.3 cm.
c. Draw the perpendicular bisector of seg PQ. It intersects PQ in point M.
d. With M as centre and radius equal to PM, draw an arc intersecting the circle in points R and S.
e. Draw rays QR and QS.
Rays QR and QS are the required tangents to the circle.

iii.

iv. Let the value of WT be x.
WT + TX = WX …[W-T-X]
∴ x + TX = 25
∴ TX = 25 – x
Also, YT + TZ = YZ …[Y-T-Z]
∴ 8 + TZ = 26
∴ TZ = 26 – 8
= 18 units
But, WT x TX = YT x TZ …[Theorem of internal division of chords]
∴ x × (25 – x) = 8 × 18
∴ 25x – x² = 144
∴ x² – 25x + 144 = 0
∴ (x – 16)(x – 9) = 0
∴ x = 16 or x = 9
∴ WT = 16 units or WT = 9 units

Question 4.
Solve the following sub-questions (Any two):
i. In the adjoining figure, ∆ABC is an equilateral triangle. Bisector of ∠B intersects circumcircle of ∆ABC in point P. Prove that, CQ = CA.

ii. 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.

iii. A cylindrical jar of radius 10 cm is filled with water up to a height of 15 cm. 14 spherical balls of radius 3 cm each are immersed in the jar. Find the new level to which water is filled in the jar.
Answer:
i. Given: ∆ABC is an equilateral triangle. Bisector of ∠B intersects circumcircle of ∆ABC in point P.
To prove: CQ = CA.
Proof:
∆ABC is an equilateral triangle.
∴ ∠ABC = ∠ACB = ∠BAC = 60° …(i)[Angles of an equilateral triangle]
∠CBP = \(\frac{1}{2}\)∠ABC …[seg BP bisects ZB]
∴ ∠CBP = \(\frac{1}{2}\) × 60° …[From (i)]
∴ ∠CBP = 30°
∠CBP = ∠CAP = 30° …[Angles inscribed in the same arc]
∴ ∠CAQ = 30°
In ∆ABQ, …(ii)[A-P-Q]
∠BAQ = ∠BAC + ∠CAQ …[Angle addition property]
∴ ∠BAQ = 60° + 30° …[From (i) and (ii)]
∴ ∠BAQ = 90°
Also, ∠ABQ = 60° …[From (i) and B-C-Q]
∴ ∠BQA = 30° …[Remaining angle of AABQ]
∴ ∠CQA = 30° …(iii)[B-C-Q]
In ∆CQA,
∠CAQ = ∠CQA …[From (ii) and (iii)]
∴ CQ = CA …[Converse of isosceles triangle theorem]

ii. Given: In DABCD, side BC ǁ side AD. Diagonal AC and diagonal BD intersect in point Q.
AQ = \(\frac{1}{3}\)AC
To Prove: DQ = \(\frac{1}{2}\)BQ Proof:
side AD ǁ side BC and seg BD is their transversal.
∠ADB ≅ ∠CBD …(i)[Alternate angles]
In ∆AQD and ∆CQB,
∠ADQ ≅ ∠CBQ …[From (i), B-Q-D]
∠AQD ≅ ∠CQB …[Vertically opposite angles]
∆AQD ~ ∆CQB …[By AA test of similarity]

iii. Radius of spherical ball (r) = 3 cm

Volume of 14 spheres = 14 × 36π = 504π
For cylindrical jar,
radius (R) = 10 cm, height (H) = 15 cm
Volume of water in the jar = πR²H = π × (10)² × 15 = 1500π
Total volume of water + Volume of 14 spheres
= 1500π + 504π = 2004π
Let the new height of water be h
Volume of water in the cylinder when spherical balls are immersed = 2004π

∴ New level up to which water is filled in the jar is 20.04 cm.

Question 5.
Solve the following sub-questions (Any one):
i. If a and b are natural numbers and a > b. then show that (a² + b²), (a² – b²), (2ab) is a Pythagorean triplet. Find two Pythagorean triplets using any convenient values of a and b.
ii. As shown in the adjacent figure, a sphere is placed in a cylinder.
It touches the top, bottom and the curved surface of the cylinder.
If radius of the base of the cylinder is ‘r’,

a. what is the ratio of the radii of the sphere and the cylinder?
b. what is the ratio of the curved surface area of the cylinder and the surface area of the sphere?
c. what is the ratio of the volumes of the cylinder and the sphere?
Answer:
i. (a² + b²)² = a⁴ + 2a²b² + b⁴ …(i)
(a² – b²)² = a⁴ – 2a²b² + b⁴ …(ii)
(2ab)² = 4a²b² …(iii)
Now, (a⁴ + 2a²b² + b⁴) = (a⁴ – 2a²b² + b⁴) + 4a²b²
∴ (a² + b²)² = (a² – b²)² + (2ab)² …[From (i), (ii) and (iii)]
∴ [(a² + b²), (a² – b²), (2ab)] is a Pythagorean triplet.
Assigning different values to a and b:
Let a = 4, b = 1
a² + b² = 4² + 1² = 16 + 1 = 17
a² – b² = 4² – 1² = 16 – 1 = 15
2ab = 2 × 4 × 1 = 8
∴ (17, 15, 8) is a Pythagorean triplet.
Let a = 2, b = 1
a² + b² = 2² + 1² = 4 + 1 = 5
a² – b² = 2² – 1² = 4 – 1 = 3
2ab = 2 × 2 × 1 = 4
∴ (5, 3, 4) is a Pythagorean triplet.

ii. Radius of base of cylinder = radius of sphere
∴ Radius of sphere = r
Also, height of cylinder = diameter of sphere
∴ h = d
∴ h = 2r …(i)

SSC Maharashtra Board Maths 2 Question Paper with Solutions