Maharashtra Board Class 10 Maths 2 Sample Paper Set 2 with Solutions

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 BD² = 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
⇒ BD² = 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 cm²
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 cm²
Area of shaded region = Area of semi-circle PABQ – Area of rectangle ABCD
= 15714 – 98 = 59.14 cm²
Hence, the area of the shaded region is 59.14 cm².

(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 (r₁) = 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
[R² + r² + 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 cm³
= 9152 × \(\frac{1}{1000}\)litre ……[∵ 1000 cm³ = 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 QR² = 4 (PS² – PQ²).
Solution:
Given: In triangle PQR, ∠PQR = 90° and S is the mid-point of QR.
To prove : QR² = 4(PS² – PQ²)
□ In right-angled ∆PQS, by Pythagoras theorem,
PQ² + QS² = PS²
⇒ QS² = PS² – PQ²
Since S is the mid-point of side QR,
∴ QR = \(\frac{\mathrm{QR}}{2}\)
Substituting the value of QS in equation (i),
\(\left(\frac{\mathrm{QR}}{2}\right)^2\) = PS² – PQ²
\(\frac{\mathrm{QR}^2}{4}\) = PS² – PQ²
QR² = 4(PS² – PQ²)
Hence Proved.

SSC Maharashtra Board Maths 2 Question Paper with Solutions