SSC Maharashtra Board Maths 2 Question Paper 2019 with Solutions

Maharashtra Board SSC Class 10 Maths 2 Question Paper 2019 with Solutions Answers Pdf Download.

SSC Maths 2 Question Paper 2019 with Solutions Pdf Download Maharashtra Board

Time : 2 Hours
Max. Marks : 40

General Instructions :

1. All questions are compulsory.
2. Use of calculator is not allowed.
3. Figures to the right of qusetions indicate full marks.
4. Draw proper figures for answers wherever necessary.
5. The marks for construction should be clear and distinct. Do not erase them.
6. While writing any proof, drawing relevant figure is necessary. Also the proof should be consistent with the figure.

Question 1.
(A) Solve the following questions (Any four): [5]

(i) If ∆ABC ~ ∆PQR and ∠A = 60°, then ∠P = ?
Solution:
Given, ∆ABC ~ ∆PQR
If ∠A = 60°
∠A = ∠P
∴ ∠P = 60°

(ii) In right-angled ∆ABC, if ∠B = 90°, AB = 6, BC = 8, then find AC.
Solution:
Given, In ∆ABC, ∠B = 90°, AB = 6, BC = 8.

Using Pythagoras theorem,
∵ AB² + BC² = AC²
6² + 8² = AC²
36 + 64 = AC²
100 = AC²
10 = AC
AC = 10 units

(iii) Write the length of largest chord of a circle with radius 3.2 cm.
Solution:
Given, radius = 3.2 cm.
Largest chord of a circle is the diameter of that circle.
∴ Length of largest chord = diameter
= 2 × radius
= 2 × 3.2
= 6.4 cm

(iv) From the given number line, find d(A, B) :

Solution:
Given, y > x
d(A, B) = y – x
= 3 – (-3)
= 3 + 3
= 6 units

(v) Find the value of sin 30° + cos 60°.
Solution:
Sin 30° + cos 60° = \(\frac{1}{2}+\frac{1}{2}\) + [∵ sin 30° = \(\frac{1}{2}\), cos 60° = \(\frac{1}{2}\)]
= \(\frac{2}{2}\)
= 1

(vi) Find the area of a circle of radius 7 cm.
Solution:
Given, r = 7 cm.
Area of circle = πr²
= \(\frac{22}{7}\) × (7)²
= \(\frac{22}{7}\) × 7 × 7
= 154 sq. cm.

(B) Solve the following questions (Any two): [4]

(i) Draw seg AB of length 5.7 cm and bisect it.
Solution:

(i) Steps of construction:

(1) Draw a line segment AB of length 5.7 cm.
(2) Keep the compass at one end (A) of the line segment AB (It can be either of the ends).

(3) Adjust the compass length to over more than half the line segment.
(4) With this length, draw arcs on each side of the line segment (at the top and at the bottom).
(5) Without changing the width of the compass, keep the compass on point B and draw arcs on each side of the segment AB.
(6) Now, we get two arc intersections (P and Q).
(7) Draw line PQ.
(8) Line PQ is the required perpendicular bisector of line segment AB.

(ii) In right-angled triangle PQR, if ∠P = 60°, ∠R = 30° and PR = 12, then find the values of PQ and QR.
Solution:
Given, In ∆PQR, ∠P = 60°, ∠Q = 90°, ∠R = 30°.

(iii) In a right circular cone, if perpendicular height is 12 cm and radius is 5 cm, then find its slant height.
Solution:
Given, h = 12 cm, r = 5 cm.

l² = r² + h² (Pythagoras theorem)
l² = 5² + 12²
= 25 + 144
l² = 169
l = 13
∴ Slant height. l = 13 cm

Question 2.
(A) Choose the correct alternative: [4]
(i) ∆ ABC and ∆ DEF are equilateral triangles. If ar(∆ ABC) : ar(∆ DEF) = 1 : 2 and AB = 4, then what is the length of DE?
(a) 2\(\sqrt{2}\)
(b) 4
(c) 8
(d) 4\(\sqrt{2}\)
Solution:

(i) Given, ∆ABC and ∆DEF are equilateral triangles.
∴ ∆ABC ~ ∆DEF

(ii) Out of the following which is a Pythagorean triplet?
(a) (5, 12, 14)
(b) (3, 4, 2)
(c) (8, 15, 17)
(d) (5, 5, 2)
Solution:
17² = 289
and 15² + 8² = 225 + 64
= 289
Hence, 17² = 15² + 8²
By the converse of Pythagoras theorem, option (c).

(iii) ∠ACB is inscribed in arc ACB of a circle with centre O. If ∠ACB = 65°, find m (arc ACB) :
(a) 130°
(b) 295° .
(c) 230°
(d) 65°
Solution:

\(\frac{1}{2}\)m(arcAXB) = ∠ACB
m (arc AXB) = 2 × 65°
= 130°
m(arc ACB) = 360° – m(ärc AXB)
= 360° – 130°
= 230°, option (c)

(iv) 1 + tan² θ = ?
(a) sin² θ
(b) sec² θ
(c) cosec² θ
(d) cot² θ
Solution:
1 + tan²θ = sec²θ, option (b)

(B) Solve the following questions (Any two): [4]

(i) Construct tangent to a circle with centre A and radius 3.4 cm at any point P on it.
Solution:
Steps of construction :

(1) Draw a circle with centre A and radius AP, l(AP) = 3.4 cm.
(2) Draw a ray AX.
(3) Take a point Q on ray AX such that l(PA) = l(PQ)
(4) Draw a perpendicular bisector of AQ (Name it as l).
(5) Line l is a required tangent to the circle at point P on it.

(ii) Find slope of a line passing through the points A(3, 1) and B(5, 3).
Solution:
Given, A (3, 1) and B (5, 3).
∴ x₁ = 3, y₁ = 1, x₂ = 5, y₂ = 3.
∴ slope of line = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{3-1}{5-3}\)
= \(\frac{2}{2}\)
= 1

(iii) Find the surface area of sphere of radius 3.5 cm.
Solution:
Given, r = 3.5 cm
∵ Surface area of a sphere = 4πr²
= 4 × \(\frac{22}{7}\) × 3.5 × 3.5
= 4 × 11.0 × 3.5
= 44 × 3.5
= 154 cm²

Question 3.
(A) Complete the following activities (Any two):

(i) In ∆ ABC, ray BD bisects ∠ABC.
If A – D – C, A – E – B and seg ED || side BC, then prove that:
\(\frac{A B}{B C}\) = \(\frac{A E}{E B}\)

Solution:

(ii) Prove that, angles inscribed in the same arc are congruent.
Given: ∠PQR and ∠PSR are inscribed in the same arc.
Arc PXR is intercepted by the angles.

To prove:
∠PQR = ∠PSR
Proof:

Solution:
Prove that, angles inscribed in the same arc are congruent.
Given: ∠PQR and ∠PSR are inscribed in the same

Arc PXR is intercepted by the angles.
To prove:
∠POR ≅ ∠PSR
Proof:

(iii) How many solid cylinders of radius 6 cm and height 12 cm can be made by melting a solid sphere of radius 18 cm?
Activity: Radius of the sphere, r = 18 cm
For cylinder, radius R = 6 cm, height H = 12 cm

Solution:
Activity : Radius of the sphere r = 18 cm
For cylinder, radius R = 6 cm, height H = 12 cm

(B) Solve the following questions (Any two):

(i) In right-angled ∆ ABC, BD ⊥ AC.
If AD = 4, DC = 9, then find BD.
Solution:
Given, In right-angled ∆ ABC, BD ⊥ AC, AD = 4 and DC =9.

In ∆ ABC,
m∠B = 90°
Seg BD ⊥ hypo AC
∴ BD² = AD × DC (Geometric mean theorem)
= 4 × 9
BD² = 36
BD = 6

(ii) Verify whether the following points are collinear or not:
A(1, -3), B(2, -5), C(-4, 7).
Solution:
(ii) Given, A(1, -3),B (2, -5), C (-4, 7).
Consider, A(1, -3) and B (2, -5).
x₁ = 1, y₁ = -3, x₂ = 2, y₂ = -5.
Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{-5-(-3)}{2-1}\)
= \(\frac{-5+3}{1}\)
= -2
Now consider, B(2, -5) and C (-4, 7).
x₂ = 2, y₂ = -5, x₃ = -4, y₃ = 7
Slope of BC = \(\frac{7-(-5)}{-4-2}\)
= \(\frac{12}{-6}\)
= -2
Slope of AB = Slope of BC
Also, both lines have a common point B.
∴ A, B, C are collinear points.

(iii) If sec θ = \(\frac{25}{7}\), then find he value of tan θ.
Solution:
Given,
sec θ = \(\frac{25}{7}\)
1 + tan² θ = sec² θ
1 + tan² θ = \(\left(\frac{25}{7}\right)^2\)
1 + tan² θ = \(\frac{625}{49}\)
tan² θ = \(\frac{625}{49}\) – 1
tan² θ = \(\frac{625-49}{49}\)
tan² θ = \(\frac{576}{49}\)
tan² θ = \(\frac{24}{7}\)

Question 4.
Solve the following questions (Any three):

(i) In ∆ PQR, seg PM is a median, PM = 9 and PQ² + PR² = 290. Find the length of QR.
Solution:
Given, In ∆ PQR, seg PM is a median.
Also, PM = 9, PQ² + PR² = 290.
PQ² + PR² = 2PM² + 2QM² (Apollonius theorem)

290 = 2(9)² + 2QM²
290 = 2 × 81 + 2QM²
290 = 162 + 2QM²
290 – 162 = 2QM²
128 = 2QM²
64 = QM²
QM = 8 units
∵ PM is a median of the ∆ PQR.
∴ M is the mid point of QR.
QR = 2QM
= 2 × 8
QR = 16 unit

(ii) In the given figure, O is centre of circle, ∠QPR = 70° and m(arc PYR) = 160°, then find the value of each of the following:
(a) m(arc QXR)
(b) ∠QOR
(c) ∠PQR

Solution:

Given : In fig., O is the centre of circle.
∠QPR = 70°
m(arc PYR) = 160°
∠QPR is inscribed in the arc QXR.
∴ m∠QPR = \(\frac{1}{2}\)m(arc QXR) (Inscribed angle theorem)
70° = \(\frac{1}{2}\)m(arc QXR)
140° = m(arc QXR)
m(arc QXR) = 140°

(b) Measure of the central angle is equal to the corresponding minor arc.
m∠QOR = m(arc QXR)
∴ m∠QOR = 140°

(c) ∠PQR is inscribed in the arc PYR
∴ m∠PQR = \(\frac{1}{2}\)m(arc PYR) (Inscribed angle theorem)
= \(\frac{1}{2}\) × 160°
m∠PQR = 80°

(iii) Draw a circle with radius 4.2 cm. Construct tangents to the circle from a point at a distance of 7 cm from the centre.
Solution:
Steps of construction:

(1) Draw a circle with centre O and radius 4.2 cm.
(2) Draw ray OX.
(3) Take a point P on ray OX such that l(OP) = 7 cm.

(4) Draw a perpendicular bisector of OP which intersect ray OP at point M.
(5) Taking M as a centre and OM as radius, draw a circle, which cuts the circle with centre O at point A and B.
(6) Draw ray PA and ray PB.
Line PA and line PB are the required tangents to the circle from point P at a distance of 7 cm from the centre.

(iv) When an observer at a distance of 12 m from a tree looks at the top of the tree, the angle of elevation is 60°. What is the height of the tree? (\(\sqrt{3}\) = 1.73)
Solution:
Let, seg AB is tree, and an observer is at point C.
BC = 12 cm
∠ACB is an angle of elevation
m∠ACB = 60°
In ∆ ABC
tan C = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
tan 60° = \(\frac{\mathrm{AB}}{12}\)
\(\sqrt{3}\) = \(\frac{\mathrm{AB}}{12}\)
12\(\sqrt{3}\) = AB
AB = 20.78
∴ height of the tree = 20.78 m

Question 5.
Solve the following questions (Any one):

(i) A circle with centre P is inscribed in the ∆ ABC. Side AB, side BC and side AC touch the circle at points L, M and N respectively. Radius of the circle is r.
Prove that:

ar(∆ABC) = \(\frac{1}{2}\)(AB + BC + AC) × r.
Solution:
Given, In the figure, the circle with centre P is inscribed in the ∆ABC. Side AB, side BC, side AC touch the circle at points L, M and N respectively. Radius of the circle is r.
To prove that:
A (∆ ABC) = \(\frac{1}{2}\) (AB + BC + AC) × r

Construction : We draw the radii PM, PL, and PN.
Also, we draw seg PA, seg PB, seg PC.
Proof: In a circle with centre P, seg PL, seg PM and seg PN are the radii and seg. AB, seg BC, seg AC are the tangents at point L, M, N respectively.
∴ seg PL ⊥ side AB
seg PM ⊥ side BC
seg PN ⊥ side AC (By the theorem, tangent is perpendicular to the radius)
We know,
Area of triangle = \(\frac{1}{2}\) × base × height
A(∆ APB) = \(\frac{1}{2}\) × AB × PL
A(∆ APB) = \(\frac{1}{2}\) × AB × r …… (i)
Similarly, A(∆ BPC) = \(\frac{1}{2}\) × BC × r …… (ii)
A(∆ APC) = \(\frac{1}{2}\) × AC × r …… (iii)
Adding equations (i), (ii) and (iii), we get
ar(∆ APB) + ar(∆ BPC) + ar(∆ APC) = \(\frac{1}{2}\) × AB × r + \(\frac{1}{2}\) × BC × r + \(\frac{1}{2}\) × AC × r ar(∆ ABC) = \(\frac{1}{2}\) × r (AB + BC + AC)
ar(∆ ABC) = \(\frac{1}{2}\)(AB + BC + AC) × r
Hence Proved.

(ii) In ∆ABC, ∠ACB = 90°. seg CD ⊥ side AB and seg CE is angle bisector of ∠ACB.
Prove that:
\(\frac{\mathrm{AD}}{\mathrm{BD}}\) = \(\frac{\mathrm{AE}^2}{\mathrm{BE}^2}\)

Solution:
Given, In ∆ACB,
m∠ACB = 90°

Question 6.
Solve the following questions (Any one):

(i) Show that the points (2, 0), (- 2, 0) and (0, 2) are the vertices of a triangle. Also state with the reason the type of the triangle.

Solution:

5.64 > 4 unit
∴ d(B, C) + d(A, C) > d(A, B) …….(ii)
d(A, B) + d(A, C) = 4 + 2.82
= 6.82 unit
6.82 > 2.82 unit
∴ d(A, B) + d(A, C) > d(B, C) ……. (ii)
From equations (i), (ii) and (iii),
The three points (2, 0) (-2, 0) (0, 2) are the vertices of triangle as sum of length of any two sides is greater than the third side of the triangle. Hence Proved.
This triangle is an isosceles triangle as the two sides are equal.
d(BC) = (AC) = 2.82 units

(ii) In the above figure, ▭XLMT is a rectangle. LM = 21 cm, XL = 10.5 cm. Diameter of the smaller semicircle is half the diameter of the larger semicircle. Find the area of non-shaded region.
Solution:
Given: ▭XLMT is a rectangle.
∴ l (XL) = l(TM) = 10.5 cm
l(XT) = l(L M) = 21 cm
Let, the diameter of the smaller semicircle be x.
∴ Diameter of the larger semicircle will be 2x
From figure,
Sum of diameters of the semicircies = l(LM) = 21 cm
∴ x + 2x = 21
3x = 21
x = 7 cm
∴ Diameter of smaller semicircle = 7 cm
Radius of smaller semicircle = \(\frac{7}{2}\) cm
Diameter of larger semicircle = 2x
= 2 × 7 = 14 cm
Radius of larger semicircle = \(\frac{14}{2}\) = 7 cm
We know,
Area of semicircle = \(\frac{1}{2}\)πr²

Area of shaded region = 96.25 sq. cm
Area of rectangle = length × breadth
ar(▭XLMT) = 21 × 10.5 = 220.5 sq. cm
Area of non-shaded region = ar(▭XLMT) – area of shaded region
= 220.5 – 96.25 = 124.25 sq. cm

SSC Maharashtra Board Maths 2 Question Paper with Solutions