Triangles Test

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Triangles Test - 42

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Question 1
1 / -0

In the above figure, $$ \text AD $$ = $$ \text DC $$= $$ \text CB $$ and the measure of $$\angle DAC$$ = $$40^o$$ . What is the measure of $$\angle DCB$$?

A
$$ 80^o$$

B
$$ 40^o$$

C
$$ 20^o$$

D
$$ 300^o$$

Solution

$$\because AD=DC$$

$$\implies$$ $$\angle DAC=\angle ACD $$ (By isosceles triangle property, angles opposite equal sides are equal).

In $$\Delta ACD:$$

$$ { 40 }^{ \circ }+{ 40 }^{ \circ }+\angle CDA={ 180 }^{ \circ }$$ (By angle sum property of a triangle)
$$ \implies { 80 }^{ \circ }+\angle CDA={ 180 }^{ \circ }$$
$$\implies \angle CDA={ 180 }^{ \circ }-80^o$$
$$ \implies \angle CDA={ 100 }^{ \circ }$$.

$$ \therefore \angle CDB={ 180 }^{ \circ }-{ 100 }^{ \circ }={ 80 }^{ \circ }$$ $$(\because$$ linear pair, i.e. sum of angles on a straight line $$={ 180 }^{ \circ })$$.

In $$\Delta DCB$$ :
$$ \angle BDC=\angle DBC$$ (By isosceles triangle property, angles opposite equal sides are equal).
$$ \implies \angle DBC={ 80 }^{ \circ }$$.

$$ In\quad \Delta DBC$$
$$\implies$$ $${ 80 }^{ \circ }+{ 80 }^{ \circ }+\angle DCB={ 180 }^{ \circ }$$ (By angle sum property of a triangle)
$$\implies$$ $$ \angle DCB={ 180 }^{ \circ }-{ 160 }^{ \circ }$$$$={ 20 }^{ \circ }$$.

Therefore, option $$C$$ is correct.
Question 2
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In, $$\triangle DEF$$ , if $$\angle $$ $$\text D$$ $$\equiv$$ $$\angle F$$, then:

A
$$\text DE$$ $$\equiv$$ $$ \text EF $$ $$\equiv$$ $$ \text DF $$

B
$$\text DE$$ $$\equiv$$ $$ \text DF $$

C
$$\text DE$$ $$\equiv$$ $$ \text EF $$

D
All the above

Solution

By isosceles triangle property,
the angles opposite to equal sides in a triangle are equal.
Then, by converse of isosceles triangle property,
the sides opposite to equal angles in a triangle are equal.
That is, from figure, $$DE\equiv EF$$.

Therefore, option $$C$$ is correct.
Question 3
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Which of the following will form the sides of a triangle?

A
$$23\,cm,\,17\,cm,\,8\,cm$$

B
$$12\,cm,\,10\,cm,\,25\,cm$$

C
$$6\,cm,\,7\,cm,\,16\,cm$$

D
$$8\,cm,\,7\,cm,\,16\,cm$$

Solution

Triangle inequality theorem states that the sum of two sides of a triangle is always greater than the third side.

$$(a)\ 17+8>23$$ is true
Hence, the sides with the measure $$23\ cm,17\ cm,8\ cm$$ will form the sides of a triangle.

$$(b)\ 10+12>25$$ is not true
Hence, the sides with the measure $$10\ cm,12\ cm,25\ cm$$ will not form the sides of a triangle.

$$(c)\ 6+7>16$$ is not true
Hence, the sides with the measure $$6\ cm,7\ cm,16\ cm$$ will not form the sides of a triangle.

$$(d)\ 8+7>16$$ is not true
Hence, the sides with the measure $$8\ cm,7\ cm,16\ cm$$ will not form the sides of a triangle.
Question 4
1 / -0

In $$\Delta PQR$$ and $$\Delta SQR$$, $$\overline{PQ}$$ =$$\overline{SR}$$ and $$\angle PQR = \angle QRS$$ then:

A
$$\Delta PQR \sim \Delta SRQ$$

B
$$\Delta PQR \equiv \Delta SRQ$$

C
$$\Delta PQR = \Delta SRQ$$

D
None of these

Solution

In $$\Delta PQR$$ and $$\Delta SQR$$,
$$\overline{PQ} =\overline{SR}$$ (Given)
And $$\angle PQR = \angle QRS$$
$$QR=QR$$ [common side]
$$\Delta PQR \sim \Delta SRQ$$ (SAS property)
Question 5
1 / -0

Take any point O in the interior of a triangle XYZ. Which option is/are correct?

A
$$OX+OY>XY?$$

B
$$OY+OZ>YZ?$$

C
$$OZ+OX>XZ?$$

D
All of the above.

Solution
Question 6
1 / -0

In a triangle $$ABC, \angle ABC=50^{o}, \angle BAC=30^{o}$$, then the shortest sides is

A
$$AB$$

B
$$BC$$

C
$$CA$$

D
$$None$$

Solution

The side opposite to the largest angle is the longest side of the triangle and the side opposite to the smallest angle is the shortest side of the triangle.
By angle sum property,$$ \angle{A}+\angle{B}+\angle{C}={180}^{\circ}$$
$$\Rightarrow {30}^{\circ}+{50}^{\circ}+\angle{C}={180}^{\circ}$$
$$\Rightarrow \angle{C}={180}^{\circ}-{80}^{\circ}={100}^{\circ}$$
The side opposite to the smallest angle $$\left({30}^{\circ}\right)$$ is the side $$BC$$
Question 7
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If $$\Delta ABC$$ and $$\Delta XYZ$$ are equilateral triangles and $$AB = XY,$$ the condition under which $$\Delta ABC \cong\Delta XYZ$$ is

A
$$ASA$$

B
$$RHS$$

C
$$SSS$$

D
$$AAS$$

Solution

Given that $$\Delta ABC$$ and $$\Delta XYZ$$ are equilateral triangles.

i.e., $$AB=BC=CA$$
and $$XY=YZ=ZX$$ …………….$$(1)$$
and
here $$AB=XY$$ ………………$$(2)$$

By equation $$(1)$$ & $$(2)$$

$$BC=YZ$$ and $$CA=ZX$$ …………….$$(3)$$

We have to show that

$$\Delta ABC \cong\Delta XYZ$$

In $$\Delta ABC$$ and $$\Delta XYZ$$
$$AB=XY$$ (Given)
$$BC=YZ$$ (by equation $$(3)$$)
$$CA=ZX$$ (by equation $$(3)$$)
So $$\Delta ABC \cong\Delta XYZ$$ (By $$SSS$$ rules).
Question 8
1 / -0

In $$\Delta ABC$$,

A
$$AB + BC > AC$$

B
$$AB + BC < AC$$

C
$$AB + AC > BC$$

D
$$AC + BC > AB$$

Solution

In a triangle ABC,
$$\Rightarrow AB + BC > AC$$
$$\Rightarrow BC+ AC > AB$$
$$\Rightarrow AC + AB > BC$$
Since the third side of the triangle is less than the sum of any two sides of its.
Question 9
1 / -0

In $$\Delta PQR$$,

A
$$PQ - QR > PR$$

B
$$PQ + QR < PR$$

C
$$PQ - QR < PR$$

D
$$PQ + PR > QR$$

Solution

In a triangle PQR,
$$\Rightarrow \,PQ + QR > PR$$
$$\Rightarrow \,QR + PR > PQ$$
$$\Rightarrow \,PR + PQ > QR$$
Since the third side of the triangle is less then the sum of any two sides of its.
Now,
$$\Rightarrow \,PQ - QR < PR$$
$$\Rightarrow \,QR - PR < PQ$$
$$\Rightarrow \,PR - PQ < QR$$
Since the length of third side of the triangle is always greater than the difference between two sides.
Question 10
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If $$\Delta PQR$$ in congruent to $$\Delta STU$$, then what is the length of $$TU$$?

A
$$5 \,cm$$

B
$$6 \,cm$$

C
$$7 \,cm$$

D
cannot be determined

Solution

Given,
$$ \Delta PQR \cong \Delta STU$$.
Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $$PQ=ST$$, $$QR=TU$$, $$PR=SU$$, $$\angle P=\angle S$$, $$\angle Q=\angle T$$ and $$\angle R=\angle U$$.

Thus, $$QR=TU$$.
$$\therefore TU = 6 \ cm$$.

Therefore, option $$B$$ is correct.