Maharashtra Board Class 10 Maths 2 Sample Paper Set 5 with Answers

Maharashtra Board SSC Class 10 Maths 2 Sample Paper Set 5 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 2 Model Paper Set 5 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.
For each of the following sub-questions four alternative answers are given. Choose the correct alternative and write its alphabet: [4]
i. In ∆ABC and ∆PQR, in a one to one correspondence \(\frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ}\), then (B)

(A) ∆PQR ~ ∆ABC
(B) ∆PQR ~ ∆CAB
(C) ∆CBA ~ ∆PQR
(D) ∆BCA ~ ∆PQR
Answer:
(B) ∆PQR ~ ∆CAB

ii. A frustum of a cone has height 8 cm and radii of upper and lower circular surfaces are 22 cm and 16 cm respectively. Its slant height is (A)
(A) 10 cm
(B) 8√5cm
(C) 4√6 cm
(D) 6√3 cm
Answer:
(A) 10 cm

iii. Two circles intersect each other such that each circle passes through the centre of the other. If the distance between their centres is 12, what is the radius of each circle? (B)
(A) 6 cm
(B) 12 cm
(C) 24 cm
(D) can’t say
Answer:
(B) 12 cm

iv. \(\frac{1}{1+\tan ^2 \theta}\) = (C)
(A) sec²θ
(B) cot²θ
(C) cos²θ
(D) cosec²θ
Answer:
(C) cos²θ
Hints:

Question 1.
(B) Solve the following sub-questions:
i. In the given figure, ∠ABC = ∠DCB = 90°,
AB = 6, DC = 8, then \(\frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{DCB})}\) = ?

ii. In the following figure, in ∆ABC, ∠B = 90°, ∠C = 60°, ∠A = 30°, AC = 16 cm. Find BC.

iii. Find the distance between the points O (0, 0) and P (-3, 4).
iv. What is the value of 6 tan²θ – 6 sec²θ?
Answer:
i. ∆ABC and ∆DCB have same base BC.

ii. ∆ABC is a 30°- 60°- 90° triangle.
BC = \(\frac{1}{2}\)AC
∴ BC = \(\frac{1}{2}\)  16
∴ BC = 8 cm

iii.

iv. 6 tan – sec = 6(tan – sec)
= 6(-1)
= -6

Question 2.
Complete the following activities and rewrite it (Any two): [4]
i. In the adjoining figure, chord EF ǁ chord GH. Prove that, chord EG ≅ chord EH. Fill in the blanks and write the proof.

∴ chord EG ≅ chord EH …[The chords corresponding to congruent arcs of a circle are congruent]
ii. In the given figure, side of square ABCD is 7 cm. With centre D and radius DA, sector D-AXC is drawn. Fill in the following boxes properly and fmd out the area of the shaded region.
Area of square = side²

iii. Complete the following flow chart to draw the tangents to the circle at points A and B.

(B) Solve the following sub-questions (Any four): [8]
i. For the given figure, identify if ray PM is the bisector of ∠QPR.

ii. Find RP and PS using the information given in ∆PSR.

iii. In the given figure, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.

iv. Draw a circle of radius 2.7 cm. Draw a tangent to the circle at any point on it.
v. Prove that sec²θ + cosec²θ = sec²θ – cosec²θ.
Answer:
(A) i. ∠EFG = ∠FGH …(i) [Alternate angles]

∴ chord EG ≅ chord EH …[The chords corresponding to congruent arcs of a circle are congruent]

ii. Area of a square = side²
= 7²
= 49 cm²

∴ A (shaded region) = Area of a square – Area of sector
= 49 cm2 – 38.5 cm²
= 10.5 cm²

iii.

(B) i. In ∆PQR,

∴ Ray PM is the bisector of ∠QPR. …[Converse of angle bisector theorem]

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

∴ RP = 12 units, PS = 6√3 units.
SMART TIP
To verify our answer, we use Pythagoras Theorem.
If /(SR)² + /(SP)² = /(PR)², then our answer is correct.
/(SR)² + /(SP)² = (6)² + (6√3)² = 36 + 108 = 144
/(PR)² = (12)² = 144
∴ /(SR)² + /(SP)² = /(PR)²
Hence, our answer is correct.

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
∴ PS = \(\frac{144}{8}\)
∴ PS = 18 units
Now, PS – PR + RS …[P-R-S]
∴ 18 – 8 + RS
∴ RS – 18 – 8
∴ RS = 10 units

iv.

Steps of construction:
a. With centre O, draw a circle of radius 2.7 cm.
b. Take any point M on the circle and draw ray OM.
c. Draw line l ⊥ ray OM at point M.
Line l is the required tangent to the circle at point M.

v.

Question 3.
(A) Complete the following activity and rewrite it (Any one): [3]
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC, then prove that ED ǁ BC by filling the boxes.
In ∆ABC, ray BD bisects ∠ABC. …[Given]


ii. Given A (4, -3), B (8, 5). Find the co-ordinates of the point that divides segment AB in the ratio 3 : 1 by completing the following activity.
Suppose point C divides seg AB in the ratio 3:1.

(B) Solve the following sub-questions (Any two): [6]
i. In the figure circle with centre D touches the sides ∠ACB at A and B. If ∠ACB = 52°, find measure ∠ADB.

ii. 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.
iii. Draw a circle of radius 3.3 cm. Draw a chord PQ of length 6.6 cm. Draw tangents to the circle at points P and Q. Write your observation about the tangents.
iv. From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metres, then find how far the ship is from the lighthouse. (√3 = 1.73)
Answer:
i. In ∆ABC, ray BD bisects ∠ABC. …[Given]

ii. Suppose point C divides seg AB in the ratio 3 : 1.
Here, (x₁, y₁) = (4,-3)
(X₂, y₂) = (8,5)
∴ By section formula,

(B) i. The sum of all angles of a quadrilateral is 360°.
∴ ∠ACB + ∠CAD + ∠CBD + ∠ADB = 360°
∴ 52° + 90° + 90° + ∠ADB = 360° …[Tangent theorem]
∴ ∠ADB + 232° = 360°
∴ ∠ADB = 360° – 232° = 128°

ii. Given: In ∆ABC, 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]

iii. Analysis:
seg OP ⊥ line l
seg OQ ⊥ line m …[Tangent is perpendicular to radius]
The perpendicular to seg OP and seg OQ at points P and Q respectively will give the required tangents at P and Q.

Radius = 3.3 cm
∴ Diameter = 2 × 3.3 = 6.6 cm
∴ Chord PQ is the diameter of the circle.
∴ The tangents through points P and Q (endpoints of diameter) are parallel to each other.

iv. Let AB represent the height of lighthouse and point C represent the position of the ship.
AB = 90 m

Angle of depression = ∠PAC = 600
Now, ray AP ǁ seg BC
∠ACB = ∠PAC …[Alternate angles]
∠ACB = 60°
In right angled ∆ABC,

∴ The ship is 51.90 m away from the lighthouse.

Question 4.
Solve the following sub-questions (Any two): [8]
i. In a right angled ∆ABC, the perpendicular drawn from the point C on the hypotenuse meets the hypotenuse at D, and the bisector of ∠C meets the hypotenuse at E.

Prove that \(\frac{AD}{DB}=\frac{AE^2}{EB^2}\).
ii. A test tube has lower part hemispherical and upper part cylindrical with same radius. If \(\frac{5159}{6}\) cm³ of water is added, the test tube will be just completely filled. But if \(\frac{2002}{3}\) cm³ water is added, 5 cm of height will remain empty. Find the radius and height of the cylindrical part.
iii. AB and AC are the two chords of a circle whose radius is r. If p and q are the distance of chord AB and CD, from the centre respectively and if AB = 2AC, then prove that 4q² = p² + 3r².
Answer:
i. In ∆ACB, ray CE bisects ∠C. …[Given]

ii.

r = 3.5 cm
Total volume of the test tube = Volume of cylindrical part + Volume of hemispherical part

∴ Radius of test tube is 3.5 cm and height of cylindrical part is 20 cm.

iii. Given: AB and AC are two chords of circle with centre O and radius r.

AB = 2AC
ON = p and OM = q
To Prove: 4q² = p² + 3r²
Proof: Seg OM ⊥ chord AC and
Seg ON ⊥ chord AB
∴ AM = MC = \(\frac{1}{2}\) AC …(i)
and AN = NB = \(\frac{1}{2}\) AB …[Perpendicular drawn from the centre to the chord bisects the chord]
∴ AN = NB = \(\frac{1}{2}\) (2AC) …[∵ AB = 2AC]
∴ AN = NB = AC …(ii)
In ∆ONA, ∠ONA = 90°
∴ OA² = AN² + NO² …[Pythagoras theorem]
∴ r² = AC² + p² …[From (ii)]
∴ AC² = r² – p² …(iii)
Similarly, In ∆OMA, ∠OMA = 90°
AO² = AM² + MO² …[Pythagoras theorem]
∴ r² = [\(\frac{1}{2}\)AC]² + q² …[From (i)]
∴ \(\frac{1}{4}\) AC² = r² – q²
∴ AC² = 4(r² – q²) …(iv)
∴ r² – p² = 4(r² _ q²) …[From (iii) and (iv)]
∴ r² – p² = 4r² – 4q²
∴ 4q² = 3r² + p²

Question 5.
Solve the following sub-questions (Any one): [3]
i. A tent is in the shape of triangular prism resting on a rectangular base. If AB = AC, AD = 0.8 m, BC = 3 m and, a length of the tent = 6 m, ∠ABC = 42°, then find,

a. ∠ACB
b. AB
c. Volume of the tent
ii. Show that the points (2, 0), (-2, 0) and (0, 2) are the vertices of a triangle. Also state with reason the type of the triangle.
Answer:
i. In ∆ABC,
AB = BC …[Given]
∴ ∠ACB = ∠ABC …[Isosceles triangle theorem]
∴ ∠ACB = 42°
In ∆ABD and ∆ACD,
seg AB ≅ seg AC …[Given]
seg AD ≅ seg AD …[Common side]
∠ADB ≅ ∠ADC …[Each is of measure 90°]
∴ ∆ABD ≅ ∆ACD …[Hypotenuse side test]
∴ seg BD ≅ seg DC …[c.s.c.t]
∴ BD = \(\frac{1}{2}\) BC = \(\frac{1}{2}\) × 3 = 1.5 m
In ∆ADB, ∠ADB = 90°
∴ AB² = AD² + BD² …[Pythagoras theorem]
= 0.8² + 1.5²
= 0.64 + 2.25
= 2.89
∴ AB = √289 …[Taking square root of both sides]
∴ AB = 1.7 m
A(∆ABC) = \(\frac{1}{2}\) × BC × AD
\(\frac{1}{2}\) × 3 × 0.8 = 1.2 m²
Volume of tent = A(∆ABC) × CG
= 1.2 × 6
= 7.2 m³
∴ The volume of the tent is 7.2 m³.

ii. Let A(2, 0), B(-2, 0) and C(0, 2) be the vertices of the triangle.
By distance formula,


On adding (ii) and (iii),
d (B, C) + d (A, C) = 2√2 + 2√2 = 4√2
2√2 + 2√2 > 4
∴ Sum of any two sides of a triangle is greater than the third side.
∴ A, B and C are the vertices of a triangle.
Also, BC = AC = 2√2 and AB² = AC² + BC²
∴ ∆ABC is an isosceles right-angled triangle.
SMART TIP
To verify our answer, we plot the given points on the graph.
In the graph we can see that given points form an isosceles right angled triangle.

SSC Maharashtra Board Maths 2 Question Paper with Solutions