Triangles Test

Result

Triangles Test - 31

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In a triangle $$ PQR,\:QR= PR $$ and $$ \angle P= 36^{0} .$$ Which is the largest side of the triangle?

A
$$QR$$

B
$$PR$$

C
$$PQ$$

D
None of these

Solution

Since $$ QR = PR $$, the angles opposite to them will be equal as well, $$ => \angle P = \angle Q = {36}^{0}$$
Since sum of angles in a triangle $$ = {180}^{o} $$
$$ \angle R= {180}^{o} - {36}^{o} - {36}^{o} = {108}^{o} $$
Out of the three angles, $$ \angle R $$ is the largest.
This means, the side opposite to it, which is $$PQ$$ will be the greatest side of the triangle.
Question 2
1 / -0

In a $$\triangle$$ $$PQR, QR = PR$$ and $$\angle\,P\,=\,36^{\circ}.$$ The largest side of the triangle is:

A
$$PR$$

B
$$QR$$

C
$$PQ$$

D
Incomplete data

Solution

Since, $$ QR = PR $$, their opposite angles are also equal.
Thus $$ \angle P = \angle Q = {36}^{0} $$
Since sum of angles in a triangle PQR $$ = {180}^{o} $$
$$ \Rightarrow \angle R = 180 - 2(36) = {108}^{o} $$
Thus, $$ \angle R $$ is the largest angle of the triangle, the side opposite to it, which is $$PQ$$ will be the greatest side of the triangle.
Question 3
1 / -0

Directions For Questions

The following figure shows a triangle $$ABC$$ with exterior angles as $$x, y$$ and $$z.$$

...view full instructions

In the same figure, if $$y > x > z;$$ arrange sides $$AB, BC$$ and $$AC $$ in descending order according their lengths.

A
$$BC > AB > AC$$

B
$$AB > BC > AC$$

C
$$BC > AB < AC$$

D
None of these

Solution

$$\text {Since, }
y>x>z;\\ \\ \angle A\quad =\quad 180\quad -\quad x\quad \\ \angle B\quad =\quad 180\quad -\quad y\quad (\quad y\quad is\quad the\quad largest,\quad so\quad \angle B\quad \quad is\quad the\quad smallest)\\ \angle C\quad =\quad 180\quad -\quad z\quad (\quad z\quad is\quad the\quad smallest,\quad So\quad \angle C\quad is\quad the\quad largest)\\ \\ so\quad \angle C\quad >\quad \angle A\quad >\quad \angle B\\ So\quad AB\quad >\quad BC\quad >\quad AC\quad (\quad Angle\quad side\quad relationship)\\ \\ \quad
$$
Question 4
1 / -0

By which congruency are the following pair of triangles congruent:
In $$\Delta\,ABC$$ and $$\Delta \,DEF$$, $$\angle\,B = \angle\,E = 90\,^{\circ}, AC = DF$$ and $$BC = EF.$$

A
RHS

B
SAS

C
AAA

D
SSS

Solution

In $$\triangle ABC$$ and $$DEF$$,

$$\angle B = \angle E = 90^{\circ}$$ .......... (Given)

$$AC = DF$$ ........ (Given)

$$BC = EF$$ ......... (Given)

Thus, $$\triangle ABC \cong \triangle DEF$$ (RHS postulate)
Question 5
1 / -0

In $$\triangle ABC,$$ $$\angle\,B\,=\,30^{\circ}$$ and $$\angle\,C\,=\,70^{\circ}$$. The greatest side of the triangle is:

A
$$AC$$

B
$$AB$$

C
$$BC$$

D
Incomplete data

Solution

Since the sum of angles in a triangle is equal to $$ {180}^{o}. $$

$$ \angle A = {180}^{o} - {30}^{o} - {70}^{o} $$
$$= {80}^{o} $$

Out of the three angles, $$ \angle A $$ is the largest.
This means, the side opposite to it, which is $$BC$$ will be the greatest side of the triangle.
Question 6
1 / -0

If $$AB > AC > BC,$$ arrange the angles $$x, y$$ and $$z$$ in decending order of their values.

A
$$y< z < x$$

B
$$x>y>z$$

C
$$z>y>x$$

D
Incomplete question

Solution

Given $$AB > AC > BC,$$
This means that angles opposite these sides are in the same order.
So$$,$$ $$\angle C > \angle B > \angle A$$
$$\angle z = 180 - \angle C\\ \angle y = 180 - \angle B \\ \angle x = 180 - \angle A$$
From the above statements it is clear that $$x > y > z.$$
Hence$$,$$ $$B$$ is the answer.
Question 7
1 / -0

$$l$$ and $$m$$ are two parallel lines intersected by another pair of parallel lines $$p$$ and $$q$$. Then:

A
$$\triangle ABC\cong \triangle CDA$$.

B
$$\triangle ABC\cong \triangle ADC$$.

C
$$\triangle ABC\cong \triangle DCA$$.

D
$$\triangle ABC\cong \triangle CAD$$.

Solution

In $$\triangle ABC$$ and $$\triangle CDA$$
$$\angle CAB=\angle ACD$$ (Pair of alternate angle)
$$\angle BCA=\angle DAC$$ (Pair of alternate angle)
$$AC=AC$$ (Common side)
$$\triangle ABC\cong \triangle CDA$$ (ASA criteria)
Question 8
1 / -0

If in two triangles $$PQR$$ and $$DEF$$, $$PR=\,EF$$, $$QR=\,DE$$ and $$PQ=\,FD$$, then $$\triangle PQR\cong$$ $$\triangle$$ ___.

A
$$FDE$$

B
$$DEF$$

C
$$FED$$

D
$$DFE$$

Solution

Given: $$PR=\,EF$$, $$QR=\,DE$$ and $$PQ=\,FD$$
Now, In $$\triangle PRQ$$ and $$\triangle DEF$$
$$PR = EF$$
$$QR = DE$$
$$PQ = FD$$
Thus, $$\triangle PQR \cong \triangle FDE$$ (SSS rule).
Question 9
1 / -0

D is a point on the side BC of a $$\triangle ABC$$ such that AD bisects $$\angle BAC$$. Then

A
$$BD=CD$$

B
$$BA>BD$$

C
$$BD>BA$$

D
$$CD>CA$$

Solution

Given-
AD is the angular bisector of a triangle ABC.
AD meets BC at D.
To find out-
which of the given options is true.
Solution-
AD is the bisector of $$\quad \angle BAC.$$

$$ \therefore \quad \angle BAD=\angle CAD.\quad $$

Again , in $$\quad \Delta ACD,\quad \quad \angle ADC\quad $$is the external angle .

We know that the external angle of a triangle is greater than each intrenal opposite angle of the same triangle.

$$\quad \therefore \quad \angle ADB=\angle DAC+DCA.$$

$$=\angle BAD+\angle DCA\quad \left( \because \quad \angle DAC=\angle BAD \right) \\ i.e\quad \angle ADC>\angle BAD.$$

$$\therefore \quad BA>BD.\quad $$

So option B is true.

Option A is true only when the triangle is equlateral or isosceles.

Option C is false since Option B is true.

option D is false since

$$ \angle ADC>\angle BAD\quad or\ \angle CAD\quad $$

$$ \therefore \quad CD\ngtr CA\quad $$\

Ans- Option B.
Question 10
1 / -0

In $$\Delta ABC$$, if $$\angle A = 50^0$$ and $$\angle B = 60^0$$, then the shortest and largest sides of the triangle, respectively are:

A
$$AC$$ and $$AB$$

B
$$BC$$ and $$AB$$

C
$$AC$$ and $$BC$$

D
$$BC$$ and $$AC$$

Solution

We have,
$$\angle A = 50$$ and $$\angle B = 60$$
$$\therefore \angle A + \angle B + \angle C = 180$$ $$[$$ Angle sum property of a $$\Delta$$ $$]$$
$$\Rightarrow 50 + 60 + \angle C = 180$$
$$\Rightarrow 110 + \angle C = 180$$
$$\Rightarrow \angle C = 180 - 110$$
$$\Rightarrow \angle C = 70$$
Since $$\angle A$$ and $$\angle C$$ are the smallest and largest angles respectively, therefore corresponding sides $$BC$$ and $$AB$$ are the smallest and largest sides of the triangle respectively.
Hence, option $$B$$ is the correct answer.