Maharashtra Board SSC Class 10 Maths 2 Sample Paper Set 1 with Solutions Answers Pdf Download.

## Maharashtra Board Class 10 Maths 2 Model Paper Set 1 with Solutions

Question 1.

(A) Four alternative answers are given for every sub-questions. Select the correct alternative and write the alphabet of that answers :

(i) In right-angled triangle ABC, Hypotenuse AC = 10 and side AB = 5, then what is the measure of ∠A ?

(A) 30°

(B) 60°

(C) 90°

(D) 45°

Solution:

Given : Hypotenuse AC = 10 and side AB = 5

cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)

⇒ cos A = \(\frac{5}{10}\) = \(\frac{1}{2}\)

⇒ cos A = cos 60°

⇒ A = 60°

Hence, the correct option is (B).

(ii) If tan θ = \(\frac{12}{5}\), then 5 sin θ – 12 cos θ = ?

(A) = \(\frac{119}{13}\)

(B) 0

(C) 1

(D) \(\frac{1}{13}\)

Solution:

Here, tan θ = \(\frac{12}{5}\)

Now, by Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

AC^{2} = 12^{2} + 5^{2}

⇒ AC^{2} = 144 + 25

⇒ AC^{2} = 169

⇒ AC = 13

Hence, sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{13}\)

and cos θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) = \(\frac{5}{13}\)

So, 5 sin θ – 12 cos θ = 5 × \(\frac{12}{13}\) – 12 × \(\frac{5}{13}\)

= \(\frac{60}{13}\) – \(\frac{60}{13}\) = 0

Hence, the correct option is (B).

(iii) If x = \(\frac{\theta}{360}\) × 2πr then what is x in the formula ?

(A) Length of the arc of measure θ

(B) Area of sector of measure θ

(C) Area of the segment of measure θ

(D) The circumference of the circle

Solution:

Here, x is the formula of length of the arc.

Hence, the correct option is (A).

(iv) In the figure, line l is parallel to X-axis. Which of the following statement is true ?

(A) The slope is zero.

(B) The slope cannot be determined.

(C) The slope is positive.

(D) The slope is negative.

Solution:

If a line parallel to X-axis, then the slope is zero.

Hence, the correct option is (A).

(B) Solve the following questions. (Marks 4)

(i) Which of the following conditions is not sufficient to determine the congruence of two triangles ?

SSS, SAS, SSA, ASA.

Solution:

SSA is not sufficient condition.

(ii) The diagonals bisect each other at right angles.” In which of the following quadrilaterals is the given property observed ?

Rectangle, Rhombus, Parallelogram, Trapezium.

Solution:

Rhombus

(iii) A line is parallel to Y-axis and is at a distance of 5 units from the Y- axis. Write the equation of that line.

Solution:

The equation of line is x = 5.

(iv) In ∆ABC, ∠ABC = 90° and ∠ACB = θ.

Then write the ratios of sin 0 and tan 0 from the figure.

Solution:

sin θ \(=\frac{\mathrm{AB}}{\mathrm{AC}}\) and tan θ \(=\frac{\mathrm{AB}}{\mathrm{BC}}\)

Question 2.

(A) Complete and write the following activities. (Any two)

(i) From the information in the figure, complete the following activity to find the length of the hypotenuse AC.

Solution:

(ii) Prove that, The areas of two triangles with same height are in the proportion of their

corresponding bases. To prove this theorem start as follows:

(i) Draw two triangles, give names of all points, show heights.

Solution:

The triangles are :

(ii) Write ‘Given’ and ‘To prove’ from the figure drawn.

Solution:

Given : AP = DQ

To Prove: \(\frac{\mathrm{A}(\Delta \mathrm{ABC})}{\mathrm{A}(\Delta \mathrm{DEF})}\) = \(\frac{\mathrm{BC}}{\mathrm{EF}}\)

(iii) Given: In the figure, point A is in the exterior of the circle with centre P. AB is the tangent segment and secant through A intersects the circle in C and D.

To prove : AB^{2} = AC × AD

Construction: Draw segment BC and BD.

Write the proof by completing the activity.

Solution:

Proof : In ∆ ABC and ∆ ADB,

(B) Solve the following sub-questions. (Any four)

(i) If length of the circular arc is 10 cm and the radius 35 cm, find the area of the sector of the circle.

Solution:

Length of the circular arc = 10 cm.

Radius = 3.5 cm.

(ii) From the information given in the figure, determine whether MP is the bisector of ∠KMN.

Solution:

We know that the bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

Now, \(\frac{\mathrm{KP}}{\mathrm{PN}}\) = \(\frac{2}{3}\)

and \(\frac{\mathrm{MK}}{\mathrm{MN}}\) = \(\frac{5}{6}\)

∵ \(\frac{2}{3}\) ≠ \(\frac{5}{6}\)

Hence, MP is not the bisector of ∠KMN.

(iii) In the figure, the centre of the circle is O and ∠STP = 40°.

(i) m (arc SP) = ? By which theorem?

(ii) m ∠SOP = ? Give reason.

Solution:

Given: ∠STP = 40°

(i) m (arc SP) = 2 ∠STP

⇒ m (arc SP) = 2 × 40° = 80°

Hence, m (arc SP) is 80° by inscribed angle theorem.

(ii) m∠SOP = 2 × m ∠STP

= 2 × 40° = 80°

Hence, m∠SOP is 80° because the measure of an angle subtended by an arc at the centre of the circle is double the measure of the angle subtended by the arc at any point on the circle.

(iv) Draw a circle of suitable radius. Take point T on it. Draw a tangent through point T.

Solution:

Steps of construction:

(i) Draw a circle with centre O.

(ii) Take any point T on the circle and join OT and extend it to X.

(iii) Take points A and B on OT and TX respectively i.e. TA = AB.

(iv) Taking A as centre and radius greater than AT, draw two arcs on both side of OX.

(v) Taking B as centre and radius greater than BT, draw two arcs intersecting the arcs drawn in step (iv).

(vi) Join the arcs to obtain the line LM, passing through point T.

Line LM is the required tangent to the circle at point T.

(v) Find the slope of the line passing through given points G (3, 7) and

K (-2, -3).

Solution:

Given points are G (3, 7) and K (- 2, – 3).

Let x_{1} = 3, x_{2} = -2, y_{1} = 7, y_{2} = -3

∴ Slope of line GK = \(\frac{y_2-y_1}{x_2-x_1}\)

⇒ Slope of line GK = \(\frac{-3-7}{-2-3}\)

= \(\frac{-10}{-5}\) = 2

Hence, the slope of the line GK is 2.

Question 3.

(A) Complete and write the following activities: (Any one)

(i) Prove that secθ + tanθ = \(\frac{\cos \theta}{1-\sin \theta}\)

Proof:

LH.S. = secθ + tanθ

Solution:

(ii) Find distance between points P(-5, -7) and Q (0, 3).

By distance formula,

Solution:

By distance formula,

(B) Solve the following sub-questions. (Any two) (Marks 6)

(i) If one looks from a tower 10 m high at the top of a flag staff, the

depression angle of 30° is made. Also, looking at the bottom of the

staff from the tower, the angle of the depression made is of 60°.

Find the height of the flag staff.

Solution:

In the figure, PQ is the tower of height 10 m and AB is the flag staff.

Hence, the height of flag staff is 667 m.

(ii) In the figure, PQRS is a square with side 10 cm. The sectors drawn

with P and R as centres form the shaded figure. Find the area of the

shaded figure. (Use π = 3.14)

Solution:

Sides of square = 10 cm

Radius of quadrant SEQR = 10 cm

∴ Area = \(\frac{1}{4}\) × \(\frac{22}{7}\) × 10 × 10 = 78.57 cm^{2}

Area of triangle SQR = \(\frac{1}{2}\) × QR × SR

= \(\frac{1}{2}\) × 10 × 10

= 50 cm^{2}

Area of segment SEQ = Area of quadrant SEQR – Area of ∆SQR

= 78.57 – 50 = 28.57 cm^{2}

Total area of shaded portion = 2 × 28.57 = 57.14 cm

Hence, the area of shaded portion is 57.14 cm^{2}.

(iii) Draw ∆ RSP ~ ∆ TQP. In ∆ TQP, TP = 5 cm, ∠P = 50°, PQ = 4.5 cm and \(\frac{\mathrm{RS}}{\mathrm{TQ}}\) = \(\frac{2}{3}\).

Solution:

Steps of construction:

(i) Draw segment PQ = 45 cm.

(ii) Draw ray PX such that ∠XPQ = 500.

(iii) Take point T on ray PX such that PT = 5 cm.

(iv) Join TQ.

∆ TQP is obtained.

(v) Divide segment PQ into 3 equal parts.

(vi) Fix point S on PQ such that length of PS is of length of segment PQ.

(vii) Draw a line parallel to side TQ through S.

(viii) Name the point of intersection of the line and ray PT as R.

Thus, we get required ∆ RSP similar to ∆ TQP.

(iv) In the figure with z ABC, P, Q R are the mid-points of AB, AC and BC respectively. Then prove that the four triangles formed are congruent to each other.

Solution:

Given: ABC is a triangle and P, Q and R are respectively the mid-points of sides AB, AC and BC.

To prove : ∆ ABC is divided into 4 congruent triangles.

Proof:

P and Q are the mid-points of AB and AC of ∆ABC.

So, PQ || BC …..[Line segment joining the mid-points of two sides of a triangle is parallel to the third side]

Similarly, PR || AC and QR || AB

In PQRB, PB || QR and PQ || BR ……[Parts of parallel lines are parallel]

i.e., both pairs of opposite sides are parallel.

Thus, PQRB is a parallelogram.

So, PR is a diagonal of parallelogram PQRB.

So, ∆PBR ≅ ∆PQR …….(i)

…… [A diagonal of a parallelogram divides it into two congruent triangles.]

Similarly, APRQ is a parallelogram

So, ∆APQ ≅ ∆PQR …….(ii)

and PQRC is a parallelogram

So, ∆PQR ≅ ∆QRC …….(iii)

From equations (i), (ii) and (iii),

∆PBR ≅ ∆PQR ≅ ∆APQ ≅ ∆ QRC

Hence, all 4 triangles are congruent to each other.

Hence Proved.

Question 4.

Solve the following sub-questions. (Any two)

(i) Given: O is the centre of the circle, AB is a diameter, OA = AP, O – A – P, PC is a tangent through C. A tangent through point A intersects PC in E and BC in D.

To prove : ∆ CED is a equilateral triangle.

Solution:

Given: OA = AP, PC is a tangent at C, AD is tangent at A.

To prove : ∆ CED is an equilateral angle.

Construction: Join OC and OE.

Proof: AD is a tangent to the circle at A.

∴ ∠ACP = ∠ABC …… (i) [Alternate segment theorem]

Since, the 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

∴ ∠ABC = \(\frac{1}{2}\)∠AOC

In ∆ AOE and ∆ COE

OA ≅ OC …….[Radii of circle]

OE ≅ OE …[Common side]

AE ≅ CE …… [Tangents to the circle from point E]

∴ By SSS criterion of congruence, .

∆ AOE ≅ ∆ COE

∴ ∠AOE = ∠COE ……. [C.A.C.T.]

or ∠AOE = ∠COE = \(\frac{1}{2}\) ∠AOC …… (iii)

From equations (ii) and (iii),

∠ABC = ∠AOE ……(iv)

Now, in ∆ OEP

AE is the bisector of ∠OEP

\(\frac{\mathrm{OA}}{\mathrm{AP}}\) = \(\frac{\mathrm{OE}}{\mathrm{EP}}\)

⇒ OE = EP …… [∵ OA = AP, Given]

∴ ∠AOE = ∠APE …… (v) [Angles opposite to equal sides]

or ∠ABC = ∠APE ……(vi) [Using equation (iv)]

From the figure,

∠DCE = 90° – ∠ACP ……(vii)

∠CDE = 90° – ∠ABC …… (viii)

∠AEP = ∠CED = 90° – ∠APE ……(ix)

[∠AEP = ∠CED; Vertically opposite angles]

From equations (vii), (viii) and (ix)

∠DCE = ∠CDE = ∠CED

or DE = CE = CD ….. [Sides opposite to equal angles]

Hence, ∆ CED is an equilateral triangle.

Hence Proved.

(ii) ∆ PQR, is a right angled triangle with ∠Q = 90°, QR = b, and A (∆ PQR) = a.

If QN ⊥ PR, then prove that

QN = \(\frac{2 a b}{\sqrt{b^4+4 a^2}}\)

Solution:

We have

(iii) A cylinder and a cone have equal bases. The height of the cylinder is 2 cm and the area of its base is 64 cm^{2}. The cone is placed upon the cylinder volume of the solid figure so formed is 400 cm^{3}. Find the total height of the figure.

Solution:

A cylinder and a cone have equal bases.

∴ They have equal radii.

Let the radius be r.

Area of its base = 64 cm^{2}

∴ πr^{2} = 64 cm^{2}

and height of cylinder h1 = 2 cm

Volume of solid figure = 400 cm^{3}

∴ Volume of solid figure = Volume of cylinderical Part + Volume of conical part

Question 5.

Solve the following sub-questions. (Any one)

(i) If m and n are real numbers and m > n, if m^{2} + n^{2}, m^{2} – n^{2} and 2 mn are the sides of the triangle, then prove that the triangle is right angled. (Use the converse of the Phythogoras theorem). Find out two Pythagorian triplets using convenient values of m and n.

Solution:

Given : m > n, sides of the triangle are m^{2} + n^{2}, m^{2} – n^{2} and 2 mn.

To prove : The triangle is right-angled triangle

Proof: (m^{2} + n^{2})^{2} = (m^{2} – n^{2})^{2} + (2 mn)^{2}

(m^{2})^{2} + (n^{2})^{2}+ 2 m^{2}n^{2} = (m^{2})^{2} + (n^{2})^{2} – 2 m^{2}n^{2} + 4 m^{2}n^{2}

m^{4} + n^{4} + 2 m^{2}n^{2} = m^{4} + n^{4} + 2 m^{2}n^{2}

Since, both sides are equal,

So, by the converse of Pythagoras theorem, given triangle is right-angled triangle.

Hence, two Pythagorean triplets are (6, 8, 10) and (8, 15, 17).

(ii) A tent is of a shape of a triangular ‘prism’ resting on a rectangular side PQ = PR, PT = 1.5 m, QR = 1.8 m, length of the tent = 3 m.

Find:

(i) ∠PQR

(ii) Volume of the tent

Solution:

Given : QR = 18 m, PT = 1.5 m, PQ = PR.

Since, PQ = PR

So, ∆ PQR is an isoceles triangle.

So, PT bisects QR.

∴ QT = \(\frac{1 \cdot 8}{2}\)

(i) tan Q \(=\frac{\text { Perpendicular }}{\text { Base }}\)

tan Q = \(\frac{1.5}{0 \cdot 9}\)

tan Q = 16666666667

Q = 59036°

Hence, Z PQR = 59036°.

(ii) Volume of tent,

V = \(\frac{1}{2}\)LWH

Here, L = 3 m, W = 18 m and H = 15 m

V = \(\frac{1}{2}\) × 3 × 1.8 × 1.5 = 4.05 m^{3}

Hence, the volume of tent is 4.05 m^{3}.