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

## Maharashtra Board Class 10 Maths 2 Model Paper Set 2 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) If ∆ABC ~ ∆DEF such that ∠A = 92° and ∠B = 40°, then ∠F = ?

(A) 52°

(B) 92°

(C) 40°

(D) 48°

Solution:

Given : ∠A = 92°, ∠B = 40°.

Since, ∆ ABC ~ ∆DEF

So, ∠A, ≅ ∠D, ∠B ≅ ∠E and ∠C ≅ ∠F

⇒ ∠D = 92°, ∠E = 40°

In ∆DEF,

∠D + ∠E + ∠F = 180°

⇒ 92° + 40° + ∠F = 180°

⇒ ∠F = 180° – 132° = 48°

Thus, ∠F = 48°

Hence, the correct option is (D).

(ii) Find the value of sin 0° + cos 0° + tan 0° + sec 0°.

(A) 2

(B) 1

(C) 3

(D) 0

Solution:

sin0° + cos0° + tan0° + sec0° = 0 + 1 + 0 + 1 = 2

Thus, the value of sin 0° + cos 0° + tan 0° + sec 0° is 2.

Hence, the correct option is (A).

(iii) Find the centroid of the ABC whose vertices are A (-2, 0), B (7, -3) and C (6, 2).

(A) (\(\frac{11}{3}, \frac{1}{3}\))

(B) (\(\left(\frac{11}{3}, \frac{-1}{3}\right)\))

(C) \(\left(\frac{-11}{3}, \frac{-1}{3}\right)\)

(D) \(\left(\frac{-11}{3}, \frac{1}{3}\right)\)

Solution:

Given points are A(-2, 0), B(7, -3) and C (6, 2)

Coordinates of centroid = \(\left(\frac{-2+7+6}{3}, \frac{0-3+2}{3}\right)\) = \(\left(\frac{11}{3}, \frac{-1}{3}\right)\)

Thus, the coordinates of centroid is \(\left(\frac{11}{3}, \frac{-1}{3}\right)\)

Hence, the correct option is (B).

(iv) If two tangents TL and TM are drawn to a circle with centre C such that ∠LTM = 70°, then find ∠MCT.

(A) 30°

(B) 60°

(C) 45°

(D) 55°

Solution:

In the given figure, TC bisects ∠LTM.

∠CTM = \(\frac{1}{2}\)∠LTM

= \(\frac{1}{2}\) × 70° = 35°

Also, ∠CTM = 90° ……… [Tangent theorem]

In ∆TCM,

∠MCT + ∠CMT + ∠CTM = 180°

∠MCT + 90° + 35° = 180°

∠MCT = 180° – 90° – 35° = 55°

Thus, ∠MCT = 55°

Hence, the correct option is (D).

(B) Solve the following questions.

(i) If the coordinate of point A on the number line is -1 and that of point B is 6, then find d (A, B).

Solution:

The coordinate of A on the number line is -1 and that of B is 6.

∵ 6 > -1

∴ d(A, B) = 6 – (-1) = 6 + 1 = 7.

(ii) Two opposite angles of a parallelogram are (2x + 60)° and (4x)°. Find the value of x.

Solution:

Opposite angles of a parallelogram are (2x + 60)° and (4x)°.

We know, opposite angles of a parallelogram are equal.

∴ 2x + 60° = 4x

⇒ 4x – 2x = 60°

⇒ 2x = 60°

⇒ x = 30°

Hence, the value of x is 30°.

(iii) In which quadrant, the abscissa and ordinate of a point have same sign ?

Solution:

The abscissa and ordinate of a point has same sign in quadrant I and III.

(iv) Find the value of sin 45° + cos 45° + tan 45°.

Solution:

sin 45° + cos 45° + tan 45° = \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) + 1

= \(\frac{1+1}{\sqrt{2}}\) + 1

= \(\frac{2}{\sqrt{2}}\) + 1

= \(\sqrt{2}\)

Hence, the value of sin 45° + cos 45° + tan 45° is \(\sqrt{2}\) + 1.

Question 2.

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

(i) If the points P (1, 2), Q (0, 0) and R (x, y) are collinear, then find the relation between x and y.

Given points are P (1, 2), Q (0, 0) and R (x, y).

The given points are collinear, so the area of triangle formed by them is __.

Solution:

Given points are P(1, 2), Q (0, 0) and R (x, y).

The given points are collinear, so the area of triangle formed by them is __zero__.

(ii) AB, BC and AC are three sides of a right-angled triangle having lengths 6 cm, 8 cm and 10 cm, respectively. To verify the Pythagoras theorem for this triangle, fill in the boxes :

∆ ABC is a right-angled triangle and ∠ABC = 90°.

So, by the Pythagoras theorem,

Solution:

∆ ABC is right-angled triangle and ∠ABC = 90°

So, by the Pythagoras theorem,

(iii) If the length of diagonal of a cube is 5\(\sqrt{3}\) cm, find the total surface area.

Solution:

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

(i) In the adjoining figure, ∆ ADB ~ ∆ BDC. Prove that BD^{2} = AD × DC.

Solution:

Given: ∆ ADB ~ ∆ BDC

∴ By the corresponding sides criterion of similarity,

\(\frac{\mathrm{AD}}{\mathrm{BD}}\) = \(\frac{\mathrm{BD}}{\mathrm{DC}}\)

⇒ AD × DC = BD × BD

⇒ BD^{2} = AD × DC

Hence Proved.

(ii) A tangent JK is drawn to a circle with centre C such that CK = 6 cm and ∠CKJ = 60°. Find the length of the tangent JK.

Solution:

JK is a tangent at the point J and CJ is a radius.

So, CK = 6 cm and ∠CKJ = 60° . .. [Given]

Now, ∠CJK = 90° .. . [Tangent theorem]

In ∆ CJK,

cos 60° \(=\frac{\text { Base }}{\text { Hypotenuse }}\)

⇒ cos 60° = \(\frac{\mathrm{JK}}{\mathrm{KC}}\) = \(\frac{\mathrm{JK}}{6}\)

⇒ \(\frac{1}{2}\) = \(\frac{\mathrm{JK}}{6}\)

⇒ JK = 3

Hence, the length of JK is 3 cm.

(iii) A pole of height 30 m is observed from a point. The angle of depression of the point is 30°. Find the distance of the point from the base of the pole.

Solution:

Let the pole be XY and the point be Z from where the pole is observed.

MXZ and ∠XZY are alternate angles.

Now, in ∆ XYZ,

tan 30° = \(\frac{X Y}{Y Z}\)

⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{30}{Y Z}\)

⇒ YZ = 30\(\sqrt{3}\)

Hence, the distance of point Z from the base of the pole is 30\(\sqrt{3}\) m.

(iv) A point C divides the line segment whose points are A (4, -6) and B (5, 9) in the ratio 2 : 1. Find the coordinates of C.

Solution:

Given points are A (4, -6) and B (5, 9) and the ratio is 2 : 1.

Let the ccordinates of C be (x, y).

(v) Draw a circle of radius 4 cm. Draw a point 8 cm away from its centre and construct a pair of tangeñts.

Solution:

Steps of construction:

(i) Draw a circle with centre O and radius 4 cm.

(ii) Take a point L in the exterior of the circle such that OL = 8 cm.

(iii) Join the centre O and the point L and bisect seg OL to get its mid-point P.

(iv) Draw a circle, taking P as centre and radius OP that intersects the previous circle at points M and N.

(v) Join LM and LN.

Thus, the required tangents are LM and LN.

Question 3.

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

(i) Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Proof:

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Solution:

Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

(ii) Determine whether the following points are collinear. A (-1, -1), B (0, 1), C (1, 3)

Slope of line AB = Slope of line BC and B is the common point.

∴ Points A, B and C are colliear.

Solution:

Given : Points A(-1, -1) B(0, 1) and C(1, 3)

Slope of line AB = Slope of line BC and B is the common point.

∴ Points A, B and C are collinear.

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

(i) In the given figure, a rectangle ABCD is inscribed Inside a semi circle of radius 10 cm. Using the dimensions given in the figure, determine the area of the shaded region.

Solution:

From the figure,

ABCD is the rectangle inscribed in the semi-circle with centre O.

Area of rectangle ABCD = l × b

= AB × BC

= 14 × 7 = 98 cm^{2}

Radius of semi-circle PABQ = 10 cm

Area of semi-circle PABQ = \(\frac{1}{2} \pi r^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times(10)^2\)

= \(\frac{2200}{14}[/late] = 157.14 cm^{2}

Area of shaded region = Area of semi-circle PABQ – Area of rectangle ABCD

= 15714 – 98 = 59.14 cm^{2}

Hence, the area of the shaded region is 59.14 cm^{2}.

(ii) Construct any ∆ ABC. Construct ∆ A’BC’ such that AB : A’B = 5:3 and ∆ ABC ~ ∆ A’BC’.

Solution:

Steps of construction:

(i) Draw ∆ABC of any measure.

(ii) Extend rays AB and CB.

(iii) Divide segment AB in 5 equal parts.

(iv) Take measure of one equal part (of segment AB) on the compass and mark 3 arcs on extended ray AB starting from point B.

(v) Name the intersection of the third arc and exten-ded ray AB as A’.

(vi) From A’, draw a line parallel to AC.

(vii) Name the point of intersection of this line and extended ray CB as C’.

Thus, ∆A’BC’ is the required triangle similar to ∆A’BC.

(iii) The top of a banquet hall has an angle of elevation of 45° from the foot of a transmission tower and the angle of elevation of the topmost point of the tower from the foot of the banquet hail is 60°. If the tower is 60 m high, find the height of the banquet hall in decimals.

Solution:

In the figure, PQ is the banquet hail and SR is the transmission tower.

So, SR = 60 m, ∆SQR = 60° and ∠PRQ = 45°.

(iv) In the given figure, ∆PQR is a right-angled triangle with ∠PQR = 90°. QS is perpendicular to PR. Prove that pq = rx.

Solution:

Given: ∠PQR = 90°and QS ⊥ PR.

So, ∠QSR = ∠QSP = 90°

Now, in ∆PQR and ∆QSR,

∠QRP ≅ ∠SRQ . . . [Common angle]

∠PQR ≅ ∠QSR . .. [Each angle is equal to 90°]

So, by AA criterion of similarity,

∆PQR – ∆QSR

∴ [latex]\frac{\mathrm{PR}}{\mathrm{QR}}\) = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)

⇒ \(\frac{r}{q}\) = \(\frac{p}{x}\)

⇒ x × r = p × q

⇒ pq = rx

Question 4.

Solve the following sub-questions. (Any two)

(i) Use area theorem of similar triangles to prove congruency of two similar triangles with equal areas.

Solution:

Let two similar triangles be ∆ABC and ∆PQR.

So, ∆ABC ~ ∆PQR

By area theorem of similar triangles,

Since sides of one triangle are equal to corresponding sides of another triangle

So, by SSS rule of congruence,

∆ABC ≅ ∆PQR

(ii) If two consecutive angles of cyclic quadrilateral are congruent, then prove that one pair of opposite sides is congruent and other is parallel.

Solution:

Given : ABCD is a cyclic quadrilateral and ∠ABC ≅ ∠BCD.

To prove : Side DC Side AB, AD II BC

Construction: Draw seg AM and seg DN both perpendicular to side BC.

Proof:

∠ABC ≅ ∠BCD …..(i) [Given]

∠ABC + ∠ADC = 180° ……(ii)

[Opposite angles of a cyclic quadrilateral are supplementary]

From equations (i) and (ii),

∠BCD + ∠ADC = 180°

∴ Side AD || Side BC …… [Interior angles test]

In ∆ DNC and ∆ AMB,

seg DN ≅ segAM . . . [Perpendicular distance between two parallel lines]

∠DNC ≅ ∠AMB …… [Each is 90°]

∠DCN ≅ ∠ABM ……. [Given]

So, by SAA test of congruence

∆ DNC ≅ ∆ AMB

∴ Side DC ≅ Side AB

Hence, side AD || side BC and side DC ≅ side AB.

(iii) The radius of a metallic sphere is 8 cm. It was melted to make a wire of diameter 6 mm. Find the length of the wire.

Solution:

The radius of a metallic sphere (r) = 8 cm

The diameter of a wire = 6 mm

∴ Its radius (r_{1}) = 3 mm = \(\frac{3}{10}\) cm [∵ 1 cm = 10 mm]

Let the length be h.

sphere is melted to make a wire

∴ volume of wire = volume of sphere

∴ \(\pi r_1^2 h\) = \(\frac{4}{3} \pi r^3\)

∴ \(\frac{3}{10} \times \frac{3}{10}\) × h = \(\frac{4}{3}\) × 8 × 8 × 8

h = \(\frac{4 \times 8 \times 8 \times 8 \times 10 \times 10}{3 \times 3 \times 3}\)

h = \(\frac{204800}{27}\) = 7585.1851 cm

h = 75.85 cm

The length of the wire is 75.85 m [∵ 1 m = 100 cm]

Question 5.

Solve the following questions. (Any one)

(i) A milk container of height 16 cm is made of metal sheet in the form of frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of 22 per litre which the container can hold.

Solution:

Height of container, h = 16 cm

Radius of lower end, r = 8 cm

Radius of upper end, R = 20 cm

Volume of milk in the container = \(\frac{1}{3}\)πh

[R^{2} + r^{2} + Rr]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 14[(20)^{2} + (8)^{2} + 20 × 8]

= \(\frac{1}{3}\) × 22 × 2[400 + 64 + 160]

= \(\frac{1}{3}\) × 22 × 624

= 9152 cm^{3}

= 9152 × \(\frac{1}{1000}\)litre ……[∵ 1000 cm^{3} = 1 litre]

= 9.152 litre

Now, cost of 1 litre of milk = ₹ 22

So, cost of 9.152 litre of milk = ₹ 22 × 9.152

= ₹ 201.344

Hence, the cost of milk is R 201.34.

(ii) In the given figure, triangle PQR is right-angled at Q. S is the mid point of side QR. Prove that QR^{2} = 4 (PS^{2} – PQ^{2}).

Solution:

Given: In triangle PQR, ∠PQR = 90° and S is the mid-point of QR.

To prove : QR^{2} = 4(PS^{2} – PQ^{2})

□ In right-angled ∆PQS, by Pythagoras theorem,

PQ^{2} + QS^{2} = PS^{2}

⇒ QS^{2} = PS^{2} – PQ^{2}

Since S is the mid-point of side QR,

∴ QR = \(\frac{\mathrm{QR}}{2}\)

Substituting the value of QS in equation (i),

\(\left(\frac{\mathrm{QR}}{2}\right)^2\) = PS^{2} – PQ^{2}

\(\frac{\mathrm{QR}^2}{4}\) = PS^{2} – PQ^{2}

QR^{2} = 4(PS^{2} – PQ^{2})

Hence Proved.