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

9th Grade > Mathematics

TRIANGLES MCQs

Total Questions : 56 | Page 3 of 6 pages

Question 21. Given that ACBD is a kite. By which congruency property are the triangles ACB and ADB congruent?

SAS property
SSS property
RHS property
ASA property

Discuss Question

Answer: Option B. -> SSS property
:
B
In the

△ A B C

and

△ A B D

,
AC = AD (Given)
CB = DB (Given)
AB is the common side.

⟹ △ A B C ≅ △ A B D

[SSS congruency]
Thus, by SSS congruencyrule, the two triangles (ABC and ABD) are congruent.

Question 22.

In the given figure, AC = CE and AB

∥

ED. The value of

x

is ___ units.

Discuss Question

In Δ A B C and Δ E D C, A C = C E (given) ∠ B A C = ∠ D E C (since AB||DE and AE is a transversal, so they are alternate angles) ∠ A C B = ∠ E C D (vertically opposite angles) ∴ Δ A B C ≅ Δ E D C (A.S.A. congruence criteria) ∴ A B = D E (sides of congruent triangles) ∴ x + 10 = 2 x - 5 \Rightarrow x = 15 u n i t s

Question 23.

If Δ A B C ≅ Δ P Q R and it is given that ∠ A = 60^{\circ}, ∠ B = 70^{\circ} . Then the value of ∠ R is

___

^{\circ}

Discuss Question

If Δ A B C ≅ Δ P Q R, ∠ A = ∠ P = 60^{\circ} ∠ B = ∠ Q = 70^{\circ} Now, ∠ P + ∠ Q + ∠ R = 180^{\circ} (angles of a triangle) \Rightarrow 60^{\circ} + 70^{\circ} + ∠ R = 180^{\circ} \Rightarrow ∠ R = 50^{\circ}

Question 24. In

Δ A B D

, AB = AD and AC is perpendicular to BD. State the congruence rule by which

Δ A C B ≅ Δ A C D

In ΔABD, AB = AD and AC is Perpendicular To BD. State The...

SAS congruence rule
SSS congruence rule
RHS congruence rule
ASA congruence rule

Discuss Question

Answer: Option C. -> RHS congruence rule
:
C

In the given triangle, A B = A D (Given) A C is the common side ∠ A C B = ∠ A C D = 90^{\circ}

∴ Δ A C B ≅ Δ A C D

[ RHS criteria]
Hence, by RHS congruence rule, the two triangles are congruent.

Question 25. In the figure, AB = AC and AP

⊥

BC. Then

AB = AP
AB < AP
AB > AP
AB < BP

Discuss Question

Answer: Option C. -> AB > AP
:
C

A P ⊥ B C

⟹ △ A P B

is aright-angled triangle.

⟹

AB is the hypotenuse and hence the longest side, which makes is greater thanAP.

∴ A B > A P

Question 26. In the given figure it is given that AB = CF, EF = BD and

∠

AFE =

∠

DBC. Then by which criterion

△ A F E ≅ △ C B D

In The Given Figure It Is Given That AB = CF, EF = BD And �...

SAS
SSS
ASA
None of these

Discuss Question

Answer: Option A. -> SAS
:
A
In the given figure,
AB = CF
AB + BF = CF + BF
[Adding BF on both the sides]
AF = CB
In

△ A F E

and

△ C B D

AF = CB [Proved above]
EF = BD [Given]

∠ A F E = ∠ D B C

[Given]

∴ △ A F E ≅ △ C B D

[SAS criterion]

Question 27.

If $△ N V Y ≅ △ A J U$ and both the triangles are scalene, then which of the following is true?

NV = JU
VY = AJ
$∠ V N Y = ∠ A U J$
$∠ V Y N = ∠ J U A$

Discuss Question

Answer: Option D. ->

∠ V Y N = ∠ J U A

:
D

Corresponding sides and angles of congruent triangles must be equal.

So, if $△ N V Y ≅ △ A J U$ , then

NV = AJ
VY = JU
NY = AU
$∠ N V Y = ∠ A J U$
$∠ V Y N = ∠ J U A$
$∠ V N Y = ∠ J A U$

Question 28.

In a $△$ ABC, $∠$ ABC = $30^{\circ}$ and $∠$ ACB = $45^{\circ}$ . Which is the longest side in this triangle?

AB
BC
CA
All are same in length

Discuss Question

Answer: Option B. -> BC
:
B

Sum of all angles in a triangle is $180^{\circ}$ .
$∠$ BAC + $∠$ CAB + $∠$ CBA = $180^{\circ}$
$∠$ BAC + $30^{\circ}$ + $45^{\circ}$ = $180^{\circ}$
$∠$ BAC = $180^{\circ} - (30^{\circ} + 45^{\circ}) = 105^{\circ}$ .
The side opposite to the largest angle will be the longest.
Side opposite to $∠$ BAC = BC.
Hence, BC is the longest side.

Question 29.

In the figure below, $△$ ABC $≅$ $△$ BED, $∠$ DEB = $115^{\circ}$ and $∠$ CAB = $25^{\circ}$ . It can be concluded that BC||ED.

True
False
CA
All are same in length

Discuss Question

Answer: Option A. -> True
:
A

Since $Δ A B C ≅ Δ B E D, ∠ D B E = ∠ C A B = 25^{\circ} (CPCT) And, ∠ D E B = ∠ C B A (CPCT) Now, ∠ B D E + ∠ D B E + ∠ B E D = 180^{\circ} \Rightarrow ∠ B D E + 25^{\circ} + 115^{\circ} = 180^{\circ} \Rightarrow ∠ B D E = 40^{\circ} Also, ∠ C B A + ∠ C B D + ∠ D B E = 180^{\circ} (Linear set of angles) \Rightarrow ∠ C B A + 25^{\circ} + 115^{\circ} = 180^{\circ} \Rightarrow ∠ C B A = 40^{\circ} \Rightarrow ∠ B D E = ∠ C B D and they are alternate angles, so B C | | E D .$