Triangles Test

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Triangles Test - 27

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Weekly Quiz Competition

Question 1
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Consider isosceles triangle $$ABC$$, in which $$AB=AC$$ and$$\angle ABC =50^o$$. Find the $$\angle BAC$$.

A
$$50^o$$

B
$$65^o$$

C
$$80^o$$

D
$$130^o$$

Solution

Here, $$\triangle ABC$$ is an isosceles triangle.
Given, $$AB=AC$$ and $$\angle ABC=50^o$$.
We know, by isosceles triangle property, angles opposite to equal sides are equal.
Thus, $$\angle ABC=\angle ACB=50^o$$.
Therefore, by angle sum property of triangle,
$$\angle BAC+\angle ABC + \angle ACB=180^o$$
$$\implies$$ $$\angle BAC+50^o + 50^o=180^o$$
$$\implies$$ $$\angle BAC=180^o-(50^o+50^o)=80^o$$.

Hence, $$\angle BAC=80^o$$.
Question 2
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Which of the following is congruent to the above figure?

A

B

C

D
None of these

Solution

Two figures are said to be congruent if they have exactly same shape and size or one can exactly overlap the other.
Here, if we rotate the figure in option $$B$$, and place it over each other, we get the same figure given, i.e. they will overlap.

Therefore, option $$B$$ is correct.
Question 3
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In the given figure, $$PA$$ $$\perp$$ $$AB$$, $$QB$$ $$\perp$$ $$AB$$ and $$\Delta$$ $$OAP$$ $$\cong \Delta$$ $$OBQ$$, then:

A
$$PA$$ $$=$$ $$OB$$

B
$$AP$$ $$=$$ $$QB$$

C
$$OP$$ $$=$$ $$BQ$$

D
$$OA$$ $$=$$ $$OQ$$

Solution

It is given that $$\Delta OAP $$ $$\cong $$ $$\Delta OBQ$$ both triangles are congruent.
That is by CPCT rule, corresponding parts of congruent triangles are equal.
Then, $$OA = OQ , AP = QB , OP = OQ$$, $$\angle O=\angle O$$, $$\angle A=\angle B$$ and $$\angle P=\angle Q$$.

So, option $$B$$ is correct.
Question 4
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If for $$\triangle$$ABC and $$\triangle$$DEF, the correspondence CAB $$\leftrightarrow$$ EDF gives a congruence, then which of the following is NOT true?

A
AC = DE

B
AB = EF

C
$$\angle$$A = $$\angle$$D

D
$$\angle$$C = $$\angle$$E

Solution

In $$\triangle ABC$$ and $$\triangle DEF$$
$$\Rightarrow$$ The correspondence $$CAB\leftrightarrow EDF$$ gives a congruence.
$$\Rightarrow$$ So, $$CA=ED,\,AB=DF,\,CB=EF$$
$$\Rightarrow$$ $$\angle CAB=\angle EDF,\,\angle ACB=\angle DEF,\,\angle ABC=\angle DFE$$
$$\therefore$$ $$AB=EF$$ is not true.
Question 5
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Which congruence criterion can be used to conclude $$\triangle$$ $$XYZ$$ $$\cong$$ $$\triangle$$ $$QPR$$?

A
SAS

B
SSS

C
RHS

D
None of these

Solution

In $$\triangle$$ $$XYZ$$ and $$\triangle$$ $$PQR$$,
$$\angle$$ $$YXZ$$ $$=$$ $$\angle$$ $$QPR$$ (given)
$$XY = PQ$$ (given)
$$\angle$$ $$XYZ =$$ $$\angle$$ $$PQR$$ (given).
Therefore, $$\triangle$$ $$XYZ$$ $$\cong$$ $$\triangle$$ $$PQR$$ (By $$ASA$$ congruency criterion).
Hence, option $$D$$ is correct.
Question 6
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If $$a, b$$ and $$c$$ are the sides of a triangle, then __________.

A
$$a-b > c$$

B
$$c > a +b$$

C
$$c =a+b$$

D
$$b< c+a$$

Solution

This problem is totally based on the theorem which says any triangle with sides let say $$a , b ,c $$, then
$$a+b > c$$ and
$$c+b > a$$ and
$$c+a > b$$
Therefore, option D is correct.
Question 7
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Which of the following pair of triangles are congruent by RHS criterion?

A
(i) and (ii)

B
(iii) and (iv)

C
(i) and (iii)

D
(ii) and (iv)

Solution

In $$\triangle ABC$$ and $$\triangle QRP$$
$$\Rightarrow$$ $$AB=QR=4\,cm$$
$$\Rightarrow$$ $$BC=RP=5\,cm$$ [Hypotenuse]
$$\Rightarrow$$ $$\angle A=\angle Q=90^o$$
$$\therefore$$ $$\triangle ABC\cong\triangle QRP$$ [By RHS criteria]
Question 8
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Converse of isosceles triangle theorem states that ...........

A
The two angles that lie opposite to the two congruent sides of the isosceles triangle would also be congruent

B
If two angles of a triangle are congruent then the sides opposite to these angles are congruent.

C
If two angles of a triangle are congruent then the sides opposite to these angles are notcongruent.

D
None of these

Solution

Converse of isosceles triangle theorm states that if two angles of a triangle are congruent then the sides opposite to these angles are congruent.
If $$\angle A =\angle C$$ then $$AB=AC$$.

Therefore, option $$B$$ is correct.
Question 9
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Direction (14 - 15) : Study the figure and information given below carefully and answer the following questions.
CF and AE are equal perpendiculars on BD, BF = FE = ED.
$$\triangle$$ABE is congruent to

A
$$\triangle$$AED

B
$$\triangle$$BFC

C
$$\triangle$$CDF

D
$$\triangle$$BCD

Solution

In $$\triangle ABE$$ and $$\triangle CDF$$
$$\Rightarrow$$ $$AE = CF$$ [given]
$$\Rightarrow$$ $$\angle AEB=\angle CFD$$ [given]
$$\Rightarrow$$ $$BF + FE = DE + EF$$
$$\Rightarrow$$ $$BE = DF$$
$$\therefore$$ $$ \triangle ABE \cong \triangle CDF$$ [By SAS criteria]
Question 10
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By which congruency criterion, $$\triangle$$PQR $$\cong$$ $$\triangle$$PQS?

A
RHS

B
ASA

C
SSS

D
SAS

Solution

In $$\triangle PQR$$ and $$\triangle PQS$$
$$\Rightarrow$$ $$PR = PS = a \,cm$$ [given]
$$\Rightarrow$$ $$RQ = SQ = b\,cm $$ [given]
$$\Rightarrow$$ $$PQ = PQ$$ [common side]
$$\Rightarrow$$ $$\triangle PQR \cong \triangle PQS$$ (By SSS)