M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip

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M.C.Q (1 Marks)

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36 questions · 30 auto-graded MCQ + 6 self-marked written.

MCQ 11 Mark

Some questions and their alternative answers are given. Select the correct alternative.
In D ABC, AB = $6\sqrt 3$ cm, AC = 12 cm, BC = 6 cm. Find measure of $\angle A$.

✓
$30^\circ$
B
$60^\circ$
C
$90^\circ$
D
$45^\circ$

Answer

Correct option: A.

$30^\circ$

As,

$(6 \sqrt{ }3)^2+6^2=12^2$
$\Rightarrow A B^2+B C^2=A C^2$
i.e. sides of the triangle $A B C$ satisfy the Pythagoras theorem
$(\text { Hypotenuse })^2=(\text { base })^2+(\text { Perpendicular })^2$
$\therefore \mathrm{ABC}$ is a right-angled triangle with hypotenuse as AC
Now,
$B C=\frac{1}{2} A C$
By converse of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle theorem i.e.
In a right-angled triangle, if one side is half of the hypotenuse then the angle opposite to that side is $30^{\circ}$.
$\angle A=30^{\circ}$

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Question 21 Mark

Some questions and their alternative answers are given. Select the correct alternative.
Height and base of a right angled triangle are $24\ cm$ and $18\ cm$ find the length of its hypotenuse

Answer

By Pythagoras theorem
$(\text { Hypotenuse })^2=(\text { base })^2+(\text { Perpendicular })^2$
Given,
Base $=18 cm$
Perpendicular $=$ Height $=24 cm$
$\Rightarrow$ Hypotenuse $^2=24^2+18^2$
$\Rightarrow$ Hypotenuse $^2=576+324$
$\Rightarrow$ Hypotenuse $^2=900$
$\Rightarrow$ Hypotenuse $=30\ cm$

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Question 31 Mark

Some questions and their alternative answers are given. Select the correct alternative.
Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.
A. 9 cm
B. 4 cm
C. 6 cm
D. $2\sqrt 6$ cm

Answer

Let $A B C$ be a right-angled triangle, at $B$, and $B P$ be the altitude on hypotenuse that divides it in two parts such that,
$\mathrm{AP}=4 \mathrm{~cm}$
$\mathrm{PC}=9 \mathrm{~cm}$
As, $A B C, A B P$ and $C B P$ are right-angled triangles, therefore they all satisfy Pythagoras theorem i.e.
$(\text { Hypotenuse })^2=(\text { base })^2+(\text { Perpendicular })^2$
$\therefore \ln \triangle A B C$
$A B^2+B C^2=A C^2$
$\Rightarrow A B^2+B C^2=(A P+C P)^2$
$\Rightarrow A B^2+B C^2=(4+9)^2=13^2$
$\Rightarrow A B^2+B C^2=169[1]$
$\therefore \ln \triangle A B P$
$A P^2+B P^2=A B^2$
$A P^2+4^2=A B^2[2]$
$\therefore \ln \triangle C B P$
$C P^2+B P^2=B C^2$
$\Rightarrow 9^2+B P^2=B C^2[3]$
Adding [2] and [3], we get
$A P^2+4^2+9^2+B P^2=A B^2+B C^2$
$\Rightarrow 2 A P^2+16+81=169[\text { From 1] }$
$\Rightarrow 2 A P^2=72$
$\Rightarrow A P^2=36$
$\Rightarrow A P=6 \mathrm{~cm}$
Hence, length of Altitude is 6 cm .

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Question 41 Mark

Some questions and their alternative answers are given. Select the correct alternative.
Find perimeter of a square if its diagonal is 10 2 cm.
A. 10 cm
B. $40\sqrt 2$ cm
C. 20 cm
D. 40 cm

Answer

We know that,
Diagonal of a square = √2 a
Where 'a' is the side of the triangle.
⇒ √2 a = 10√2
⇒ a = 10 cm
Also, we know
Perimeter of square = 4a
Where 'a' is the side of the triangle
∴ Perimeter of given square = 4(10) = 40 cm

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MCQ 51 Mark

Some questions and their alternative answers are given. Select the correct alternative.
If $a, b, c$ are sides of a triangle and $a^2 + b^2 = c^2$, name the type of triangle.

A
Obtuse angled triangle
B
Acute angled triangle
C
Right angled triangle
D
Equilateral triangle

Answer

As, the sides of right-angled triangles satisfies the Pythagoras theorem, i.e.
$(Hypotenuse)^2 = (base)^2 + (Perpendicular)^2$

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MCQ 61 Mark

Some questions and their alternative answers are given. Select the correct alternative.
Out of the dates given below which date constitutes a Pythagorean triplet?

A
$15/08/17$
B
$16/08/16$
C
$ 3/5/17$
D
$ 4/9/15$

Answer

A Pythagorean triplet consists of three positive integers $(l, b, h)$ such that
$l^2 + b^2 = h^2$
and 15/08/17 is a Pythagorean triplet as,
$15^2 + 8^2 = 17^2$
i.e. $225 + 64 = 289$

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MCQ 71 Mark

Some questions and their alternative answers are given. Select the correct alternative. In a right $-$ angled triangle, if sum of the squares of the sides making right angle is $169$ then what is the length of the hypotenuse?

A
$15$
✓
$13$
C
$5$
D
$12$

Answer

Correct option: B.

$13$

Given,
Sum of the squares of the sides making right angle $=169$
$\Rightarrow(\text { base })^2+(\text { perpendicular })^2=169$
But we know, By Pythagoras's theorem
$(\text { Hypotenuse })^2=(\text { base })^2+(\text { Perpendicular })^2$
$\Rightarrow(\text { Hypotenuse })^2=169$
$\Rightarrow$ Hypotenuse $=13$ units.

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MCQ 81 Mark

Some questions and their alternative answers are given. Select the correct alternative.
Out of the following which is the Pythagorean triplet?

A
$(1, 5, 10)$
✓
$(3, 4, 5)$
C
$(2, 2, 2)$
D
$(5, 5, 2)$

Answer

Correct option: B.

$(3, 4, 5)$

A Pythagorean triplet consists of three positive integers $(l, b, h)$ such that
$ l^2 + b^2 = h^2 $
and $(3, 4, 5)$ is a Pythagorean triplet as,
$5^2 = 3^2 + 4^2$

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MCQ 91 Mark

In which of the following quadrilateral sum of squares of all sides is equal to the sum of squares of diagonals?

A
Parallelogram
B
Rhombus
C
Square
✓
(A), (B) and (C)

Answer

Correct option: D.

(A), (B) and (C)

D

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MCQ 101 Mark

In $\Delta$PQR, m$\angle$PQR = 90°, seg QS $\perp$ hyp PR then.

✓
$QS ^2= PS \times RS$
B
$PS ^2= OS \times PR$
C
$PR ^2= QS \times PS$
D
$PR ^2= QS ^2 \times PS ^2$

Answer

Correct option: A.

$QS ^2= PS \times RS$

A

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MCQ 111 Mark

In the adjoining figure, $A B^2+A C^2=122, B C=10$, then find AQ _______

A
3
✓
6
C
12
D
36

Answer

Correct option: B.

6

B

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MCQ 121 Mark

Appollonius theorem is a theorem relating the length _____ of a triangle.

A
Altitude
B
Angle bisector
C
Perpendicular bisector
✓
Median and sides

Answer

Correct option: D.

Median and sides

D

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MCQ 131 Mark

In $\Delta$QSR, m$\angle$Q = 45°, m$\angle$S = 90° and SR = 4, find QS.

A
3
✓
4
C
5
D
6

Answer

Correct option: B.

4

B

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MCQ 141 Mark

Which of the following is a Pythagorean triplet?

✓
60, 61, 11
B
40, 41, 42
C
11, 12, 15
D
9, 15, 17

Answer

Correct option: A.

60, 61, 11

A

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MCQ 151 Mark

In $\Delta$PQR, $\angle$Q = 90°, PQ = QR = $5 \sqrt{2}$ PR = 10 then ∠P _______

A
$30^{\circ}$
✓
$45^{\circ}$
C
$60^{\circ}$
D
Data not sufficient

Answer

Correct option: B.

$45^{\circ}$

B

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MCQ 161 Mark

In $\Delta$ABC, $\angle B=90^{\circ}, \angle C=30^{\circ}$, AB = 6cm then AC = _____

A
$3 \sqrt{3} cm$
B
$2 \sqrt{3} cm$
C
$12 \sqrt{3} cm$
✓
12cm

Answer

Correct option: D.

12cm

D

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MCQ 171 Mark

In $\Delta$PQR, $\angle$PQR = $90^{\circ}$, seg QM $\perp$ hyp PR, PM = 16 and RM = 9 then QM = ______

✓
12
B
25
C
7
D
$16 \times 9$

Answer

Correct option: A.

12

A

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MCQ 181 Mark

In $\triangle RST , \angle S =90^{\circ}, RT =12 m$, $ST =8 m$ then $RS =$ ______

A
$10 \sqrt{8} m$
B
$5 \sqrt{4} m$
✓
$4 \sqrt{5} m$
D
5m

Answer

Correct option: C.

$4 \sqrt{5} m$

C

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MCQ 191 Mark

In $D A B C, A B=6 \sqrt{3} cm , A C=12 cm , B C=6 cm$. Find measure of $\angle A$.

✓
$30^{\circ}$
B
$60^{\circ}$
C
$90^{\circ}$
D
$45^{\circ}$

Answer

Correct option: A.

$30^{\circ}$

(A) $30^{\circ}$As,
$\begin{array}{l}
(6 \sqrt{3})^2+6^2=12^2 \\
\Rightarrow A B^2+B C^2=A C^2
\end{array}$
i.e. sides of the triangle $ABC$ satisfy the Pythagoras theorem
$(\text { Hypotenuse })^2=(\text { base })^2+(\text { Perpendicular })^2$
$\therefore ABC$ is a right-angled triangle with hypotenuse as $AC$
Now,
$BC =\frac{1}{2} AC$
By converse of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle theorem i.e.
In a right-angled triangle, if one side is half of the hypotenuse then the angle opposite to that side is $30^{\circ}$.
$\angle A=30^{\circ}$

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MCQ 201 Mark

Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse

A
24 cm
✓
30 cm
C
15 cm
D
18 cm

Answer

Correct option: B.

(B) 30 cm
By Pythagoras theorem
|(Hypotenuse)² = (base)² + (Perpendicular)²Given,
Base = 18 cm
Perpendicular = Height = 24 cm
⇒ Hypotenuse² = 24² + 18²⇒ Hypotenuse² = 576 + 324
⇒ Hypotenuse² = 900
⇒ Hypotenuse = 30 cm

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MCQ 211 Mark

Find perimeter of a square if its diagonal is $10 \sqrt{2}$ cm.

A
10 cm
B
$40 \sqrt{2}$ cm
C
20 cm
✓
40 cm

Answer

Correct option: D.

(D) 40 cm
We know that,
Diagonal of a square = √2 a
Where 'a' is the side of the triangle.
⇒ √2 a = 10√2
⇒ a = 10 cm
Also, we know
Perimeter of square = 4a
Where 'a' is the side of the triangle
∴ Perimeter of given square = 4(10) = 40 cm

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MCQ 221 Mark

If a, b, c are sides of a triangle and a² + b² = c², name the type of triangle.

A
Obtuse angled triangle
B
Acute angled triangle
✓
Right angled triangle
D
Equilateral triangle

Answer

Correct option: C.

Right angled triangle

(C) Right angled triangle
As, the sides of right-angled triangles satisfies the Pythagoras theorem, i.e.
(Hypotenuse)² = (base)² + (Perpendicular)²

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MCQ 231 Mark

Out of the dates given below which date constitutes a Pythagorean triplet?

✓
15/08/17
B
16/08/16
C
3/5/17
D
4/9/15

Answer

Correct option: A.

(A)15/08/17
A Pythagorean triplet consists of three positive integers (l, b, h) such that
l² + b² = h²
and 15/08/17 is a Pythagorean triplet as,
15² + 8² = 17²
i.e. 225 + 64 = 289

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MCQ 241 Mark

In a right-angled triangle, if sum of the squares of the sides making right angle is 169 then what is the length of the hypotenuse?

A
15
✓
13
C
5
D
12

Answer

Correct option: B.

(B) 13
Given,
Sum of the squares of the sides making right angle = 169
⇒ (base)² + (perpendicular)² = 169
But we know, By Pythagoras's theorem
(Hypotenuse)² = (base)² + (Perpendicular)²
⇒ (Hypotenuse)² = 169
⇒ Hypotenuse = 13 units.

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MCQ 251 Mark

Out of the following which is the Pythagorean triplet?

A
(1, 5, 10)
✓
(3, 4, 5)
C
(2, 2, 2)
D
(5, 5, 2)

Answer

Correct option: B.

(B) (3, 4, 5)
A Pythagorean triplet consists of three positive integers (l, b, h) such that
l² + b² = h²
and (3, 4, 5) is a Pythagorean triplet as,
5² = 3² + 4²

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MCQ 261 Mark

Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.

A
9 cm
B
4 cm
C
6 cm
D
$2 \sqrt{6}$ cm

Answer

Let ABC be a right-angled triangle, at B, and BP be the altitude on hypotenuse that divides it in two parts such that,
AP = 4 cm
PC = 9 cm
As, ABC, ABP and CBP are right-angled triangles, therefore they all satisfy Pythagoras theorem i.e.
(Hypotenuse)² = (base)² + (Perpendicular)²∴ In ΔABC
AB² + BC² = AC²⇒ AB² + BC² = (AP + CP)²⇒ AB² + BC² = (4 + 9)² = 13²⇒ AB² + BC² = 169 [1]
∴ In ΔABP
AP² + BP² = AB²AP² + 4² = AB² [2]
∴ In ΔCBP
CP² + BP² = BC²⇒ 9² + BP² = BC² [3]
Adding [2] and [3], we get
AP² + 4² + 9² + BP² = AB² + BC²⇒ 2AP² + 16 + 81 = 169 [From 1]
⇒ 2AP² = 72
⇒ AP² = 36
⇒ AP = 6 cm
Hence, length of Altitude is 6 cm.

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MCQ 271 Mark

Which of the following is a Pythagorean triplet?

✓
60, 61, 11
B
40, 41, 42
C
11, 12, 15
D
9, 15, 17

Answer

Correct option: A.

60, 61, 11

A

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MCQ 281 Mark

In which of the following quadrilateral sum of squares of all sides is equal to the sum of squares of diagonals?

A
Parallelogram
B
Rhombus
C
Square
✓
(A), (B) and (C)

Answer

Correct option: D.

(A), (B) and (C)

D

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MCQ 291 Mark

In the adjoining figure, $A B^2+A C^2=122, B C=10$, then find AQ _______

A
3
✓
6
C
12
D
36

Answer

Correct option: B.

6

B

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MCQ 301 Mark

In $\Delta$QSR, m$\angle$Q = 45°, m$\angle$S = 90° and SR = 4, find QS.

A
3
✓
4
C
5
D
6

Answer

Correct option: B.

4

B

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MCQ 311 Mark

In $\Delta$PQR, m$\angle$PQR = 90°, seg QS $\perp$ hyp PR then.

✓
$QS ^2= PS \times RS$
B
$PS ^2= OS \times PR$
C
$PR ^2= QS \times PS$
D
$PR ^2= QS ^2 \times PS ^2$

Answer

Correct option: A.

$QS ^2= PS \times RS$

A

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MCQ 321 Mark

In $\Delta$PQR, $\angle$Q = 90°, PQ = QR = $5 \sqrt{2}$ PR = 10 then ∠P _______

A
$30^{\circ}$
✓
$45^{\circ}$
C
$60^{\circ}$
D
Data not sufficient

Answer

Correct option: B.

$45^{\circ}$

B

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MCQ 331 Mark

In $\Delta$PQR, $\angle$PQR = $90^{\circ}$, seg QM $\perp$ hyp PR, PM = 16 and RM = 9 then QM = ______

✓
12
B
25
C
7
D
$16 \times 9$

Answer

Correct option: A.

12

A

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MCQ 341 Mark

In $\Delta$ABC, $\angle B=90^{\circ}, \angle C=30^{\circ}$, AB = 6cm then AC = _____

A
$3 \sqrt{3} cm$
B
$2 \sqrt{3} cm$
C
$12 \sqrt{3} cm$
✓
12cm

Answer

Correct option: D.

12cm

D

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MCQ 351 Mark

In $\triangle RST , \angle S =90^{\circ}, RT =12 m$, $ST =8 m$ then $RS =$ ______

A
$10 \sqrt{8} m$
B
$5 \sqrt{4} m$
✓
$4 \sqrt{5} m$
D
5m

Answer

Correct option: C.

$4 \sqrt{5} m$

C

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MCQ 361 Mark

Appollonius theorem is a theorem relating the length _____ of a triangle.

A
Altitude
B
Angle bisector
C
Perpendicular bisector
✓
Median and sides

Answer

Correct option: D.

Median and sides

D

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