Triangles Test

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Triangles Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

$$Q$$ is a point on side RS of $$\Delta PSR$$ such that $$PQ =PR$$ then which of the following is correct:

A
$$PS = PQ$$

B
$$PS > PQ$$

C
$$PS < PQ$$

D
Cannot be determined

Solution

In $$\Delta PQR$$,
side $$PQ =$$ side $$PR$$ ....given
$$\angle PQR = \angle PRQ$$ ....isosceles triangle theorem
Now, $$\angle PQS + \angle PQR = 180^o$$ ....angles in linear pair
Since, $$\angle PQR$$ is an acute angle, $$\angle PQS$$ is obtuse.
i.e.$$m \angle PQS > 90^o$$
Now in $$\Delta PQS$$,
$$m \angle PQS > m \angle PSQ$$
$$\therefore PS > PQ$$
$$\therefore PS > PR$$ ...($$\because PQ=PR$$)
Question 2
1 / -0

If $$AB = 1$$, then radius of the inscribed circle

is:

A
$$ \frac{1}{3}$$

B
$$ \frac{1}{2}$$

C
$$ \frac{1}{\sqrt{3} + 1}$$

D
$$ \frac{1}{\sqrt{3} - 1}$$
Question 3
1 / -0

$$\Delta ABC$$ is congruent of $$\Delta DEF$$, if

A
$$AB \neq DE$$, $$AC = 6 = DF$$, $$\angle A = \angle D$$

B
$$AB = DE$$, $$AC = DF$$, $$\angle B =\angle E$$

C
$$AB = DE$$, $$AC \neq DF$$, $$\angle C =\angle F$$

D
$$AB = DE$$, $$AC = DF$$, $$\angle C \neq \angle F$$
Solution
We say that triangle $$\triangle ABC$$ is congruent to $$\triangle DEF$$ if -
AB = DF
BC = EF
CA = FD
$$\angle A = \angle D $$
$$\angle B = \angle E $$
$$\angle C = \angle F $$
For this equation all options satisfies the above conditions.
Therefore for this question, $$\triangle ABC$$ is congruent of $$\triangle DEF$$ if all the options are correct.
Question 4
1 / -0

If length of the largest side of a triangle is 12 cm then other two sides of triangle can be

A
4.8 cm, 8.2 cm

B
3.2 cm, 7.8 cm

C
6.4 cm, 2.8 cm

D
7.6 cm, 3.4 cm

Solution

Sum of any two sides of a triangle is greater than the third side.
Here the sum must be greater than $$12\ \ cm$$
In option $$A$$
$$4.8\ \ cm+8.2\ \ cm=13\ \ cm$$
$$\Rightarrow 13\ \ cm>12 \ \ cm$$
In rest of the options sum is less than $$12\ \ cm$$
So option $$A$$ is correct.
Question 5
1 / -0

In the figure given above, $$AP$$ $$\perp l$$ i.e., $$AP$$ is the shortest line segment that can be drawn from $$A$$ to the line $$l$$. If $$PR > PQ$$, which of the following is true.

A
$$AQ>AR$$

B
$$AR>AQ$$

C
$$AQ=2AR$$

D
$$AQ=\sqrt{2}AR$$

Solution

Cut off $$PS$$ $$=$$ $$PQ$$ from $$PR$$ join $$AS$$
In $$\Delta APQ$$ and $$\Delta APS$$
$$\angle APQ = \angle APS$$ (each $$90^{\circ}$$)
$$AP$$ $$=$$ $$AP$$ .... (common side)
$$PQ$$ $$=$$ $$PS$$
$$\therefore \Delta APQ \cong \Delta APS$$
$$AP$$ $$=$$ $$AS$$ and $$\angle 1 = \angle 3$$ ....(1)
In $$\Delta ARS$$, $$\angle 3 > \angle 2$$ ....(2)
[exterior angle property]
from (1) and (2)
$$\angle 1 > \angle 2$$
from $$\Delta AQR$$, $$\angle 1 > \angle 2$$
$$\therefore AR > QR$$
$$\therefore AR > AQ$$
Question 6
1 / -0

If $$E$$ is a point on side $$QR$$ of $$\Delta PQR$$ such that $$PE$$ bisects $$\angle QPR$$, then :

A
$$QE = ER$$

B
$$QP > QE$$

C
$$QE > QP$$

D
$$ER > RP$$

Solution

According to the given condition $$\angle QPE = \angle RPE$$
And from external angle theorem,
$$\angle PEQ= \angle RPE + \angle PRE $$
$$\Rightarrow \angle PEQ = \angle QPE + \angle PRE$$
$$\Rightarrow \angle PEQ > \angle QPE $$
In triangle, side opposite angle is greater
$$\Rightarrow QP > QE$$
Question 7
1 / -0

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

A
$$AB$$

B
$$BC$$

C
$$AC$$

D
Data insufficient

Solution

Since sum of angles in a triangle $$ = {180}^{o} $$
$$ \angle A = {180}^{o} - {30}^{o} - {70}^{o} = {80}^{o} $$
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 8
1 / -0

If $$AB = 7$$ cm, $$BP = 4$$ cm, $$AP = 5.4$$ cm, then compare the segments.

A
$$AP > BP > AB$$

B
$$BP < AP < AB$$

C
$$AP > BP < AB$$

D
None of these

Solution

$$AB=7cm$$, $$BP=4cm$$ and $$AP=5.4cm$$. Therefore $$AB>AP>BP$$.
Question 9
1 / -0

In the following figure; $$ AC= CD $$, $$ AD= BD $$ and $$ \angle C= 58^{\circ} $$. Find angle $$ CAB $$.

A
$$91.5^o$$

B
$$82^o$$

C
$$88.5^o$$

D
none of the above

Solution

In $$\triangle ACD$$,
by isosceles triangle property, angles opposite to equal sides are equal.

Then, let $$\angle CAD = \angle CDA = x$$.

We know, by angle sum property, the sum of angles $$= 180^o$$.
$$\angle CAD + \angle CDA + \angle ACD = 180^o$$
$$\implies$$ $$x + x + \angle ACD = 180^o$$
$$\implies$$ $$2x + 58^o = 180^o$$ ......(as $$\angle C=\angle ACD=58^o$$)
$$\implies$$ $$2x = 122^o$$
$$\implies$$ $$x = 61^{\circ}$$.
Therefore, $$\angle CAD = \angle CDA = 61^{\circ}$$.

Now, in $$\triangle ADB$$,
$$AD = BD$$ (Given)
$$\angle DAB = \angle ABD = y$$ ......(Isosceles triangle property).
Also, $$\angle CDA = \angle DBA + \angle DAB$$ ......(Exterior angle property)
$$\implies$$ $$61^o = y + y$$
$$\implies$$ $$2y = 61^o$$
$$\implies$$ $$y = 30.5^0$$

Then, $$\angle CAB = \angle CAD + \angle DAB$$
$$\implies$$ $$\angle CAB = 61^o + 30.5^o$$
$$\implies$$ $$\angle CAB = 91.5^{\circ}$$.

Therefore, option $$A$$ is correct.
Question 10
1 / -0

In the figure $$\angle ABC=135^{\circ},\angle ABX=90^{\circ} ,\angle XCD=55^{\circ}, \angle BCD=100^{\circ}$$,then determine whether $$\angle XBC$$ and $$\angle XCB$$ are congruent to each other.

A
$$\angle XBC = 45^{\circ}, \angle XCB = 46{\circ}$$

B
$$\angle XBC = 45^{\circ}, \angle XCB = 45^{\circ}$$

C
$$\angle XBC = 45^{\circ}, \angle XCB = 60^{\circ}$$

D
$$\angle XBC = 45^{\circ}, \angle XCB = 25^{\circ}$$

Solution

Given,
$$
\angle ABC={ 135 }^{ o },\quad ABX={ 90 }^{ o },\quad XCD={ 55 }^{ o },\quad BCD={ 100 }^{ o }$$.
Here, $$\angle ABC=\angle ABX+\angle XBC\Rightarrow XBC=\angle ABC-ABX={ 135 }^{ o }-{ 90 }^{ o }\Rightarrow XBC=45^o.$$
Again $$ \angle BCD=\angle XCB+\angle XCD\Rightarrow \angle XCB=\angle BCD-\angle XCD\Rightarrow \angle XCB={ 100 }^{ o }-{ 55 }^{ o }$$ $$=45^o$$.

Since, $$\angle XBC=\angle XCB=45^o$$, they are congruent.

Hence, option $$B$$ is correct.