Triangles Test

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Triangles Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

To prove that $$\Delta DEF$$ and $$\Delta ABC$$ are congruent by SAS, what additional information is needed?

A
$$EF \cong BC$$

B
$$ \angle DFE \cong \angle ABC$$

C
$$DE \cong AB$$

D
$$\angle DFE \cong \angle ACB$$

Solution

Given $$FD=CA$$ and $$\angle D=\angle A$$
Now for triangles to be congruent by $$SAS$$ we need a third side that include the given angles with the given equal sides.
So $$DE=AB$$ $$\implies DE\cong AB$$

Hence, option $$C$$ is correct.
Question 2
1 / -0

In the figure, $$\angle BCD = \angle ADC$$ and $$\angle ACB = \angle BDA$$.

A
$$\angle B = \angle B$$

B
$$\angle A = \angle D$$

C
$$\angle A = \angle B$$

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

Solution

In $$\bigtriangleup ACD$$ and $$\bigtriangleup BDC$$
$$\angle BCD=\angle ADC$$ [Given].

Given,$$\angle ACB=\angle BDA$$.
Adding these two above equation , we get,
$$\angle BCD+\angle ACB=\angle ADC+\angle BDA$$
$$\implies$$ $$\angle ACD=\angle BDC$$.

Also, $$CD=CD$$ [Common]
$$\bigtriangleup ACD \cong \bigtriangleup BDC$$ .....(By $$ASA$$ congruency criterion).

Therefore, by $$CPCT$$ rule, the corresponding parts are equal.
Then, $$AD=BC$$ and $$\angle A=\angle B$$.

Hence, option $$C$$ is correct.
Question 3
1 / -0

In a quadrilateral $$ACBD, AC = AD$$ and $$AB$$ bisect $$\angle A$$, then:

A
$$BC \cong AD$$

B
$$BD \cong AC$$

C
$$BC \cong BD$$

D
$$AB \cong AC$$

Solution

In $$\bigtriangleup ABC$$ and $$ \bigtriangleup ABD$$ ,

$$AB=AB$$ (Common)
$$\angle CAB=\angle DAB$$ ($$AB$$ bisect $$\angle A$$)
$$AC=AD$$ (Given)

By using $$SAS$$ rule of congruence,
$$\triangle ABC\cong \triangle ABD$$
$$\Rightarrow BC\cong BD$$ [by $$CPCT$$]

Therefore, option $$C$$ is correct.
Question 4
1 / -0

Consider isosceles triangle $$ABC$$, in which $$\angle ABC=\angle ACB$$ ,then which of the two sides are similar?

A
$$AB=AC$$

B
$$AB=BC$$

C
$$AC=BC$$

D
All of the above

Solution

Here, $$\triangle ABC$$ is an isosceles triangle.
Given, $$\angle ABC=\angle ACB$$.
We know, by isosceles triangle property, angles opposite to equal sides are equal.
Thus, if $$\angle ABC=\angle ACB$$, then $$AB=AC$$ (as $$\angle ACB$$ is opposite to side $$AB$$ and $$\angle ABC$$ is opposite to side $$AC$$).

Hence, option $$A$$ is correct.
Question 5
1 / -0

Consider isosceles triangle $$ABC$$, in which $$\angle ABC=\angle ACB$$ , $$AB=2BC$$ and $$AB=8cm$$ . What is the perimeter of the $$\triangle ABC$$?

A
$$24cm$$

B
$$32cm$$

C
$$20cm$$

D
$$18cm$$

Solution

$$\triangle ABC$$ is an isosceles triangle.
Gi en that $$\angle ABC=\angle ACB$$, $$AB=2BC$$ and $$AB=8$$ cm$$
So, $$AB=AC=8$$ cm
$$\Rightarrow BC=\dfrac {AB}{2}=4$$ cm
Perimeter of $$\triangle ABC= 8+8+4=20$$ cm.

Here, $$\triangle ABC$$ is an isosceles triangle.
Given, $$\angle ABC=\angle ACB$$, $$AB=2BC$$ and $$AB=8 cm$$.
We know, by isosceles triangle property, angles opposite to equal sides are equal.
Then, So, $$AB=AC=8 cm$$.
Since $$AB=2BC$$,
$$\Rightarrow BC=\dfrac {AB}{2}=4 cm$$.
Thus, perimeter of $$\triangle ABC= 8+8+4=20 cm$$.

Hence, option $$C$$ is correct.
Question 6
1 / -0

Which congruence criteria can be used to state that $$\triangle$$XOY $$\cong$$ $$\triangle$$POQ?

A
ASA

B
SAS

C
SSS

D
RHS

Solution

In $$\triangle XOY$$ and $$\triangle POQ$$

$$\Rightarrow$$ $$\angle YXO=\angle QPO=65^o$$ [Given]

$$\Rightarrow$$ $$OX=OP$$ [Given]

$$\Rightarrow$$ $$\angle XOY=\angle POQ$$ [Vertically opposite angles]

$$\therefore$$ $$\triangle XOY\cong\triangle POQ$$ [By ASA criteria]
Question 7
1 / -0

In Fig if $$\angle C >\angle A,\angle D> \angle E$$ then

A
$$AE>CD$$

B
$$AE < CD$$

C
$$AB < BC$$

D
$$BE < BD$$

Solution

Given,
$$\angle C > \angle A\implies AB>BC$$ ------ Side opposite to largest angle is longest
Similarly, $$\angle D > \angle E\implies EB>BD$$

From above,
$$(AB+EB)>(BC+BD)$$
$$\implies AE>CD$$ Option A
Question 8
1 / -0

In the given figure, which of the following is correct?

A
$$\Delta PQR \cong \Delta RSP$$

B
$$\Delta PQR \cong \Delta SPR$$

C
$$\Delta PQR \cong \Delta RPS$$

D
$$\Delta PQR \cong \Delta PSR$$

Solution

In $$\triangle PQR$$ and $$\triangle RSP$$
$$\angle QPR = \angle SRP = 45^o$$
$$PQ = RS = 5.5 cm$$
$$PR = RP$$ (common)
$$\triangle PQR \cong \triangle RSP$$ [By SAS congruency]
Hence option (A) is correct
Question 9
1 / -0

In given figure,
CF and AE are equal perpendiculars on BD, BF = FE = ED.
$$\angle$$BAE = ______.

A
$$\angle$$BCD

B
$$\angle$$CBA

C
$$\angle$$ADC

D
$$\angle$$DCF

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]
$$\therefore$$ $$\angle BAE=\angle DCF$$ [By CPCT]
Question 10
1 / -0

In given figure $$\triangle PQR \cong \triangle XYZ$$ by _______ congruency rule.

A
$$SSS$$

B
$$AAA$$

C
$$SAS$$

D
$$ASA$$

Solution

In $$\triangle PQR$$ and $$\triangle XYZ$$,

$$\angle$$ $$QPR$$ $$=$$ $$\angle$$ $$YXZ$$ (given)
$$PQ = XY $$ (given)
$$\angle$$ $$PQR$$ $$=$$ $$\angle$$ $$XYZ$$ (given)

Therefore, $$\triangle PQR \cong \triangle XYZ$$ by $$ASA$$ rule of congruency.