Congruence of Triangles Test

Result

Congruence of Triangles Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

In the figure, AB and CD are two walls of equal height. A ladder BO is shifted to reach the other wall at D. Then, the triangles are congruent by the postulate

A
SSS

B
SAS

C
ASA

D
RHS

Solution
Question 2
1 / -0

In the figure, $$\displaystyle AB=CD$$ and $$\displaystyle \angle A={ 90 }^{ o }=\angle D$$. Then

A
$$\displaystyle \Delta ABC\cong \Delta DBC$$ by SAS postulate

B
$$\displaystyle \Delta ABC\cong \Delta DCB$$ by RHS postulate

C
$$\displaystyle \Delta ABC\cong \Delta DBC$$ by AAS postulate

D
$$\displaystyle \Delta ABC\cong \Delta DCB$$ by SSS postulate

Solution

In $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DBC$$
$$\displaystyle \angle A={ 90 }^{ o }=\angle D$$ given
$$\displaystyle BC=BC$$ common side
$$\displaystyle AB=CD$$ given
$$\displaystyle \therefore \Delta ABC\cong \Delta DCB$$ (By RHS congruence postulate)
Question 3
1 / -0

State the property by which$${\triangle }$$ADB $${\cong }$$ $${\triangle }$$ADC in the following figure.

A
SAS property

B
SSS property

C
RHS property

D
ASA property

Solution

$$AB=AC$$ (Given)
$$\angle BAD=\angle CAD$$ (given)
$$AD=AD$$ (common)
$$\therefore \triangle ADB\cong \triangle ADC$$
By $$SAS$$ property.
Question 4
1 / -0

In the figure, $$AD$$ and $$BC$$ are perpendicular to $$AB$$, and $$AD=BC$$. Then , by $$SAS$$ congruence postulate, $$\displaystyle \Delta ABC\cong $$

A
$$\displaystyle \Delta ABD$$

B
$$\displaystyle \Delta BDA$$

C
$$\displaystyle \Delta BAD$$

D
$$\displaystyle \Delta ADB$$

Solution

$$AD\perp AB$$ and $$BC\perp AB$$ [ Given ]

$$\therefore$$ $$\angle BAD=\angle ABC=90^o$$ ---- ( 1 )

In $$\triangle ABC$$ and $$\triangle BAD$$

$$BC=AD$$ [ Given ]
$$\angle ABC=\angle BAD=90^0$$ [ From ( 1 ) ]
$$AB=BA$$ [ Common side ]

$$\therefore$$ $$\triangle ABC\cong\triangle BAD$$
Question 5
1 / -0

If three sides of a triangle are equal to three sides of another triangle, then the two triangles are congruent. This is _____ condition for congruence.

A
$$ASA$$

B
$$AAS$$

C
$$SSS$$

D
$$SAS$$

Solution

Side-Side-Side $$(S.S.S)$$ condition states that if three sides of one triangle are equal to three sides of another triangle then the two triangles are congruent.
Question 6
1 / -0

If the hypotenuse and one of the other two sides of a right angles triangle is equal to the hypotenuse and one of the other two sides of the other right-angled triangle respectively, then the two right-angled triangles are ___.

A
congruent

B
unequal

C
equilateral

D
None of the these

Solution

Given,
Both the hypotenuse of the triangles are equal
One of other two sides are equal
Both are right angled triangles.
So by Side Angle Side we can say that both the angles are congruent.
Hence option A is the correct answer.
Question 7
1 / -0

AC is a diagonal of a rectangle ABCD. Which triangle is congruent to $$\triangle$$ ADC?

A
$$\triangle BCA$$

B
$$\triangle ACB$$

C
$$\triangle CBA$$

D
$$\triangle BCD$$

Solution

In $$\triangle ADC$$ and $$\triangle CBA$$,

$$AD=CB$$ (Opposite sides of rectangle)

$$DC=BA$$ (Opposite sides of the rectangle)

$$AC=CA$$ (Common side)

$$\triangle ADC\cong \triangle CBA$$ by $$SSS$$ criteria.

So option $$C$$ is correct.
Question 8
1 / -0

When two triangles have corresponding sides equal in length, then the two triangles are congruent.

A
SAS congruency Theorem

B
AA congruency Theorem

C
AAA congruency Theorem

D
SSS congruency Theorem

Solution

All $$3$$ sides of a triangle are equal with all $$3$$ sides of another triangle then these two triangles are said to be congruent SSS congruency Theorem.
Therefore, D is the correct answer.
Question 9
1 / -0

Which of the following can be used to prove that $$\Delta ABC \cong \Delta SRT$$?

A
ASA

B
SAS

C
RHS

D
AAA

Solution

Two triangles triangle are congruent, if the hypotenuse and one side of the one triangle are respectively equal to the hypotenuse and one side of the other.
Therefore, $$\Delta ABC \cong \Delta SRT$$ by RHS congruence condition.
Question 10
1 / -0

By $$SAS$$ congruence rule, $$\triangle PQR$$ is congruent to

A
$$\triangle BCA$$

B
$$\triangle CAB$$

C
$$\triangle ABC$$

D
$$\triangle ACB$$

Solution

In $$\triangle PQR$$ and $$\triangle CAB,$$

$$PQ=CA$$ [given]
$$PR=CB$$ [given]
and, $$\angle QPR=\angle ACB$$ [given]

Thus, by $$SAS$$ congruence rule,
$$\triangle PQR \cong \triangle CAB$$

Hence, $$Op-B$$ is correct.