Triangles(9th Grade > Mathematics ) Questions and answers for exam Preparation

9th Grade > Mathematics

TRIANGLES MCQs

Total Questions : 56 | Page 4 of 6 pages

Question 31.

In the figure, if AB = CD and $∠$ ABO = $35^{\circ}$ . What is the value of $∠$ DCO in degrees?

___

Discuss Question

Answer: Option A. ->
:

In $△$ AOB and $△$ DOC,
OB = OC (Radii of the same circle)

OA = OD (Radii of the same circle)

AB = CD (Given)

$△$ AOB $≅$ $△$ DOC (By SSS congruence condition)

So, $∠$ ABO = $∠$ DCO = $35^{\circ}$ (Corresponding parts of congruent triangles)

Question 32.

In the figure below, if AC $>$ AB and D is point on AC such that AB = AD, then the relation between BC and CD is BC $>$ CD.

True
False
CA
All are same in length

Discuss Question

Answer: Option A. -> True
:
A

In $△$ ABC,

AB+BC $>$ AC

$\Rightarrow$ AB + BC $>$ AD + CD

$\Rightarrow$ AB + BC $>$ AB + CD (Since AB = AD)

$\Rightarrow$ BC $>$ CD

Question 33.

$In the given figure, Δ A B C and Δ O B C are both isosceles.$
$It can be concluded that:$

$Δ A O B ≅ Δ A O C$
AB = AO
$∠ A B O = ∠ A C O$
$Δ A O B is not congruent to Δ A O C$

Discuss Question

Answer: Option A. ->

Δ A O B ≅ Δ A O C

:
A and C

$In Δ A O B and Δ A O C, A B = A C (Sides of isosceles Δ A B C) . O B = O C (Sides of isosceles Δ O B C) . A O i s c o m m o n ∴ Δ A O B ≅ Δ A O C (by SSS congruence)$
And $∠ A B O = ∠ A C O$ ( By CPCT)

Question 34.

In the given figure, AM $⊥$ BC and AN is the bisector of $∠$ A. Then $∠$ MAN is

$32 {\frac{1}{2}}^{\circ}$
$16 {\frac{1}{2}}^{\circ}$
$16^{\circ}$
$32^{\circ}$

Discuss Question

Answer: Option C. ->

16^{\circ}

:
C

In given figure,
$∠ B A C + ∠ A B C + ∠ A C B = 180^{\circ} (Angles of same triangle) \Rightarrow ∠ B A C + 65^{\circ} + 33^{\circ} = 180^{\circ} \Rightarrow ∠ B A C = 82^{\circ}$
$∠ C A N = \frac{∠ B A C}{2} = 41^{\circ} (Given that AN is an angle bisector) . Now, ∠ C A N + ∠ A N C + ∠ A C N = 180^{\circ} (Angles of same triangle) \Rightarrow 41^{\circ} + ∠ A N C + 33^{\circ} = 180^{\circ} \Rightarrow ∠ A N C = 106^{\circ}$
$∠ A N C + ∠ A N M = 180^{\circ} (Linear pair) \Rightarrow 106^{\circ} + ∠ A N M = 180^{\circ} \Rightarrow ∠ A N M = 74^{\circ}$
$∠ A N M + ∠ A M N + ∠ M A N = 180^{\circ} \Rightarrow 74^{\circ} + 90^{\circ} + ∠ M A N = 180^{\circ} \Rightarrow ∠ M A N = 16^{\circ} .$

Question 35.

$In parallelogram A B C D, A P and C Q are perpendicular to diagonal B D . Statement 1: Δ A P B ≅ Δ C Q D Statement 2: SSS congruence is applied$

Pick the correct option.

Both statements are correct
Both statements are false
Statement 1 is correct but 2 is false
Statement 1 is false but 2 is correct

Discuss Question

Answer: Option C. -> Statement 1 is correct but 2 is false
:
C

In Δ A P B and Δ C Q D, A B = C D (opposite sides of a parallelogram) . ∠ A B P = ∠ C D Q (alternate angles) . ∠ A P B = ∠ C Q D (both are right angles) . ∴ Δ A P B ≅ Δ C Q D

(using AAS criterion)
Hence, statement 1 is correct but statement 2 is false

Question 36.

O is any point on the bisector of the acute angle $∠$ XYZ. From O, a line is extended to join XY such that OP is parallel to ZY. Then, $△$ YPO is:

Scalene
Isosceles but not right angled
Equilateral
Right & isosceles

Discuss Question

Answer: Option B. -> Isosceles but not right angled
:
B

$∠$ POY = $∠$ OYZ (alternate angles)

$∠$ Y is bisected, so $∠$ POY = $∠$ PYO
Hence, PY = PO
So, $△$ YPO is isosceles
Also, it is given that $∠$ XYZ is acute, so any angle which is half of it (bisected by OY) is less than $45^{\circ}$ .
or $∠$ PYO + $∠$ OYZ < $90^{\circ}$
Hence, the third angle of the $△$ YPO i.e. $∠$ YPO will be obtuse to satisfy angle-sum property of a triangle.

Hence, $△$ YPO is isosceles but not right angled.

Question 37.

In $△ N V Y$ , if $N V = V Y$ and $V Y \neq Y N$ , then which of the following is true?

$∠ N V Y$ = $∠ V Y N$
$∠ N V Y$ = $∠ Y N V$
$∠ N V Y$ + $2 ∠ V Y N$ = $180^{\circ}$
$∠ V Y N$ + $2 ∠ N V Y$ = $180^{\circ}$

Discuss Question

Answer: Option C. ->

∠ N V Y

2 ∠ V Y N

180^{\circ}

:
C

Given, in $△ N V Y$ , NV = VY.

Then, $∠ V Y N = ∠ Y N V$ .
[angles opposite to equal sides of a triangle are equal]
Using angle sum property of a triangle, we get:
$∠ N V Y + ∠ V Y N + ∠ Y N V = 180^{\circ}$
So, $∠ N V Y + 2 ∠ V Y N = 180^{\circ}$ .
It is also given that $V Y \neq Y N$ . Hence the options $∠ N V Y = ∠ V Y N$ and $∠ N V Y = ∠ Y N V$ are not correct.

Question 38.

In the given figure, O is equidistant from the sides AC and AB. Then the value of x - 3 is
___.

Discuss Question

Answer: Option C. ->

∠ N V Y

2 ∠ V Y N

180^{\circ}

$In Δ A O C and Δ A O B, O C = O B (Given) ∠ O B C = ∠ O C A (Both angles are right angles) And A O is common side . Hence, Δ A O C ≅ Δ A O B (RHS congruence) ∴ ∠ C A O = ∠ B A O (CPCT) \Rightarrow 2 x + 24 = 30 \Rightarrow x = 3 \Rightarrow x - 3 = 0$

Question 39.

Which of the following is sufficient for two triangles denoted by $Δ 1$ and $Δ 2$ to be congruent ?

Any two sides of $Δ 1$ should be equal to any two sides of $Δ 2$
All angles of $Δ 1$ should be equal to all angles of $Δ 2$
Any two sides of $Δ 1$ and one angle should be equal to any two sides and one angle of $Δ 2$
Any two sides of $Δ 1$ and the included angle should be equal to any two sides and the included angle of $Δ 2$

Discuss Question

Answer: Option D. -> Any two sides of

Δ 1

and the included angle should be equal to any two sides and the included angle of

Δ 2

:
D

Two triangles can't be congruent if any two sides and one angle of one are equal to any two sides and one angle of the other.
They will be congruent when the angle is included between the equal pair of sides. This is the SAS condition of congruency of triangles.
The SAS congruence rule states:
Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.