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Exam: Maharashtra State Board Class 10
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Q326
#326
Geometry → Pythagoras Theorem → Pythagoras Theorem and its Converse
·
Easy
Add
Find the length of the hypotenuse of a right-angled triangle if its remaining sides are 9 cm and 12 cm.
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[QQ2(B)(2) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q327
#327
Geometry → Circle → Tangent and Secant to a Circle
·
Moderate
·
Has image
Add
In the figure,
m
(
arc
N
S
)
=
125
∘
m(\text{arc } NS) = 125^\circ
m
(
arc
N
S
)
=
12
5
∘
and
m
(
arc
E
F
)
=
37
∘
m(\text{arc } EF) = 37^\circ
m
(
arc
E
F
)
=
3
7
∘
. Find the measure of
∠
N
M
S
\angle NMS
∠
N
M
S
.
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[QQ2(B)(3) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q328
#328
Geometry → Co-ordinate Geometry → Slope of a Line
·
Easy
Add
Find the slope of the line passing through the points
A
(
2
,
3
)
A(2, 3)
A
(
2
,
3
)
and
B
(
4
,
7
)
B(4, 7)
B
(
4
,
7
)
.
Show model answer
[QQ2(B)(4) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q329
#329
Geometry → Mensuration → Surface Area and Volume of Solids
·
Easy
Add
Find the surface area of a sphere of radius 7 cm.
(
π
=
22
7
)
\left(\pi = \dfrac{22}{7}\right)
(
π
=
7
22
)
Show model answer
[QQ2(B)(5) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q330
#330
Geometry → Similarity → Basic Proportionality Theorem
·
Moderate
·
Has image
Add
In
△
A
B
C
\triangle ABC
△
A
B
C
, ray BD bisects
∠
A
B
C
\angle ABC
∠
A
B
C
with
A
−
D
−
C
A - D - C
A
−
D
−
C
, and seg
D
E
∥
DE \parallel
D
E
∥
side BC with
A
−
E
−
B
A - E - B
A
−
E
−
B
. Complete the following activity to show that
A
B
B
C
=
A
E
E
B
\dfrac{AB}{BC} = \dfrac{AE}{EB}
B
C
A
B
=
E
B
A
E
.
Show model answer
[QQ3(A)(1) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q331
#331
Geometry → Circle → Theorems on Chords and Tangents
·
Moderate
·
Has image
Add
Chords AB and CD of a circle with centre P intersect at point E. Draw seg AC and seg BD, then complete the following activity to prove that
A
E
×
E
B
=
C
E
×
E
D
AE \times EB = CE \times ED
A
E
×
E
B
=
C
E
×
E
D
.
Show model answer
[QQ3(A)(2) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q332
#332
Geometry → Co-ordinate Geometry → Slope of a Line
·
Moderate
Add
Determine whether the points
A
(
1
,
−
3
)
A(1, -3)
A
(
1
,
−
3
)
,
B
(
2
,
−
5
)
B(2, -5)
B
(
2
,
−
5
)
,
C
(
−
4
,
7
)
C(-4, 7)
C
(
−
4
,
7
)
are collinear.
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[QQ3(B)(1) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q333
#333
Geometry → Geometric Constructions → Construction of a Similar Triangle
·
Moderate
Add
△
A
B
C
∼
△
L
M
N
\triangle ABC \sim \triangle LMN
△
A
B
C
∼
△
L
M
N
. In
△
A
B
C
\triangle ABC
△
A
B
C
,
A
B
=
5.5
AB = 5.5
A
B
=
5.5
cm,
B
C
=
6
BC = 6
B
C
=
6
cm,
C
A
=
4.5
CA = 4.5
C
A
=
4.5
cm. Construct
△
A
B
C
\triangle ABC
△
A
B
C
and
△
L
M
N
\triangle LMN
△
L
M
N
such that
B
C
M
N
=
5
4
\dfrac{BC}{MN} = \dfrac{5}{4}
M
N
B
C
=
4
5
.
Show model answer
[QQ3(B)(2) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q334
#334
Geometry → Pythagoras Theorem → Applications of Pythagoras Theorem
·
Moderate
Add
Seg PM is a median of
△
P
Q
R
\triangle PQR
△
P
QR
,
P
M
=
9
PM = 9
P
M
=
9
and
P
Q
2
+
P
R
2
=
290
PQ^2 + PR^2 = 290
P
Q
2
+
P
R
2
=
290
. Then find QR.
Show model answer
[QQ3(B)(3) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q335
#335
Geometry → Similarity → Basic Proportionality Theorem
·
Moderate
Add
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 remaining sides in the same proportion.'
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[QQ3(B)(4) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q336
#336
Geometry → Trigonometry → Trigonometric Ratios and Identities
·
Hard
Add
If
1
sin
2
θ
−
1
cos
2
θ
−
1
tan
2
θ
−
1
cot
2
θ
−
1
sec
2
θ
−
1
cosec
2
θ
=
−
3
\dfrac{1}{\sin^2\theta} - \dfrac{1}{\cos^2\theta} - \dfrac{1}{\tan^2\theta} - \dfrac{1}{\cot^2\theta} - \dfrac{1}{\sec^2\theta} - \dfrac{1}{\operatorname{cosec}^2\theta} = -3
sin
2
θ
1
−
cos
2
θ
1
−
tan
2
θ
1
−
cot
2
θ
1
−
sec
2
θ
1
−
cosec
2
θ
1
=
−
3
, then find the value of
θ
\theta
θ
.
Show model answer
[QQ4(1) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q337
#337
Geometry → Mensuration → Surface Area and Volume of Solids
·
Moderate
Add
A cylinder of radius 12 cm contains water up to a height of 20 cm. A spherical iron ball is dropped into the cylinder and the water level rises by 6.75 cm. What is the radius of the iron ball?
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[QQ4(2) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q338
#338
Geometry → Geometric Constructions → Construction of a Tangent to a Circle
·
Moderate
Add
Draw a circle with centre O and radius 3 cm. Draw tangent segments PA and PB through a point P outside the circle such that
∠
A
P
B
=
70
∘
\angle APB = 70^\circ
∠
A
P
B
=
7
0
∘
.
Show model answer
[QQ4(3) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q339
#339
Geometry → Similarity → Tests of Similarity of Triangles
·
Moderate
Add
□
A
B
C
D
\square ABCD
□
A
B
C
D
is a trapezium with
A
B
∥
C
D
AB \parallel CD
A
B
∥
C
D
. The diagonals of the trapezium intersect at point P. (a) Draw the figure using the given information. (b) Write any one pair of alternate angles and one pair of opposite (vertically opposite) angles. (c) Write the names of the similar triangles with the test of similarity.
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[QQ5(1) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q340
#340
Geometry → Circle → Tangent and Secant to a Circle
·
Moderate
Add
AB is a chord of a circle with centre O. AOC is a diameter of the circle and AT is a tangent at A. (a) Draw the figure using the given information. (b) Find the measures of
∠
C
A
T
\angle CAT
∠
C
A
T
and
∠
A
B
C
\angle ABC
∠
A
B
C
with reasons. (c) Are
∠
C
A
T
\angle CAT
∠
C
A
T
and
∠
A
B
C
\angle ABC
∠
A
B
C
congruent? Justify your answer.
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[QQ5(2) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2024 board paper · 2024]
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Q341
#341
Geometry → Pythagoras Theorem → Pythagoras Theorem and its Converse
·
Easy
Add
If
a
,
b
,
c
a, b, c
a
,
b
,
c
are sides of a triangle and
a
2
+
b
2
=
c
2
a^2 + b^2 = c^2
a
2
+
b
2
=
c
2
, name the type of triangle:
A
Obtuse angled triangle
B
Acute angled triangle
C
Right angled triangle
D
Equilateral triangle
Tap an option to check your answer.
Show solution
[QQ1(A)(1) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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Q342
#342
Geometry → Circle → Theorems on Chords and Tangents
·
Easy
Add
Chords AB and CD of a circle intersect inside the circle at point E. If
A
E
=
4
AE = 4
A
E
=
4
,
E
B
=
10
EB = 10
E
B
=
10
,
C
E
=
8
CE = 8
C
E
=
8
, then find ED:
A
7
B
5
C
8
D
9
Tap an option to check your answer.
Show solution
[QQ1(A)(2) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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Q343
#343
Geometry → Co-ordinate Geometry → Coordinates and the Cartesian Plane
·
Easy
Add
Co-ordinates of the origin are ________.
A
(
0
,
0
)
(0, 0)
(
0
,
0
)
B
(
0
,
1
)
(0, 1)
(
0
,
1
)
C
(
1
,
0
)
(1, 0)
(
1
,
0
)
D
(
1
,
1
)
(1, 1)
(
1
,
1
)
Tap an option to check your answer.
Show solution
[QQ1(A)(3) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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Q344
#344
Geometry → Mensuration → Surface Area and Volume of Solids
·
Easy
Add
If the radius of the base of a cone is 7 cm and its height is 24 cm, then find its slant height:
A
23 cm
B
26 cm
C
31 cm
D
25 cm
Tap an option to check your answer.
Show solution
[QQ1(A)(4) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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Q345
#345
Geometry → Similarity → Theorem of Areas of Similar Triangles
·
Easy
Add
If
△
A
B
C
∼
△
P
Q
R
\triangle ABC \sim \triangle PQR
△
A
B
C
∼
△
P
QR
and
A
(
△
A
B
C
)
A
(
△
P
Q
R
)
=
16
25
\dfrac{A(\triangle ABC)}{A(\triangle PQR)} = \dfrac{16}{25}
A
(
△
P
QR
)
A
(
△
A
B
C
)
=
25
16
, then find
A
B
:
P
Q
AB : PQ
A
B
:
P
Q
.
Show model answer
[QQ1(B)(1) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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Q346
#346
Geometry → Pythagoras Theorem → 30-60-90 Triangle Theorem
·
Easy
Add
In
△
R
S
T
\triangle RST
△
R
S
T
,
∠
S
=
90
∘
\angle S = 90^\circ
∠
S
=
9
0
∘
,
∠
T
=
30
∘
\angle T = 30^\circ
∠
T
=
3
0
∘
,
R
T
=
12
RT = 12
R
T
=
12
cm, then find RS.
Show model answer
[QQ1(B)(2) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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Q347
#347
Geometry → Circle → Theorems on Chords and Tangents
·
Easy
Add
If the radius of a circle is 5 cm, then find the length of the longest chord of the circle.
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[QQ1(B)(3) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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Q348
#348
Geometry → Co-ordinate Geometry → Distance Formula
·
Easy
Add
Find the distance between the points
O
(
0
,
0
)
O(0, 0)
O
(
0
,
0
)
and
P
(
3
,
4
)
P(3, 4)
P
(
3
,
4
)
.
Show model answer
[QQ1(B)(4) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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Q349
#349
Geometry → Circle → Inscribed Angle and Intercepted Arc
·
Easy
·
Has image
Add
In the given figure, points L, M, N lie on a circle and
∠
L
=
35
∘
\angle L = 35^\circ
∠
L
=
3
5
∘
. Complete the activity to find (i)
m
(
arc
M
N
)
m(\text{arc } MN)
m
(
arc
M
N
)
and (ii)
m
(
arc
M
L
N
)
m(\text{arc } MLN)
m
(
arc
M
L
N
)
.
Show model answer
[QQ2(A)(1) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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Q350
#350
Geometry → Trigonometry → Trigonometric Ratios and Identities
·
Easy
Add
Show that
cot
θ
+
tan
θ
=
csc
θ
×
sec
θ
\cot\theta + \tan\theta = \csc\theta \times \sec\theta
cot
θ
+
tan
θ
=
csc
θ
×
sec
θ
.
Show model answer
[QQ2(A)(2) · Maharashtra State Board Class 10 (SSC) — Geometry (Mathematics Part II), March 2023 board paper · 2023]
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