Triangles Test

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Triangles Test - 33

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Question 1
1 / -0

In the following fig. if AB = AC and BD = DC then $$\angle ADC\, =$$

A
$$60^{\circ}$$

B
$$120^{\circ}$$

C
$$90^{\circ}$$

D
None

Solution

Given that, in $$\triangle ABC$$,
$$AB = AC, \ BD = DC$$
To find out: $$\angle ADC$$

In $$\triangle ADB$$ and $$\triangle ADC$$,
$$AB = AC\quad \quad [given]$$
$$BD = DC\quad \quad [given]$$
Also, $$AD = AD\quad \quad [common]$$
$$\therefore \ $$ By $$SSS$$ congruence criterion, $$\Delta ADB\, \cong\, \Delta ADC$$.

Hence, $$\angle ADB=\angle ADB$$
Also, $$\angle ADB\, +\, \angle ADC\, =\, 180^{\circ}\quad [linear\ pair]$$
$$\therefore \ \angle ADB\, =\, \angle ADC\, =\, 90^{\circ}$$

Hence, $$\angle ADC=90^\circ$$
Question 2
1 / -0

In the given figure, $$OA=OB, OC=OD, \angle AOB=\angle COD.$$ Which of the following statements is true$$\: ?$$

A
$$AC=CD$$

B
$$OA=OD$$

C
$$AC=BD$$

D
$$\angle OCA=\angle ODC$$

Solution

In $$\triangle $$s $$AOC$$ and $$BOD$$
$$OA=OB$$          (Given)
$$OC=OD$$          (Given)
$$\angle AOB-\angle COB=\angle COD-\angle COB,$$
i.e., $$\angle AOC=\angle BOD$$
$$\therefore $$ $$\triangle AOC\cong \triangle BOD$$          (SAS)
$$\Rightarrow $$ $$AC=BD$$          (cpct)
Question 3
1 / -0

In a $$\triangle PQR$$, PS is bisector of $$\angle P$$ and $$\angle Q =70^o, \angle R = 30^o$$, then:

A
QS > PQ > PR

B
QS < PQ < PR

C
PQ > QS > SR

D
PQ < QS < SR

Solution

According to question,
$$\angle QPR=180^o-(70^o+30^o)=80^o$$

So,$$\angle QPS=\angle SPR=40^o$$
$$\angle PSQ=180^o-(40^o+70^o)=70^o$$

As, $$\angle PRQ<\angle PQR$$
So,
$$PQ<PR$$ ......(1)

Similarly,
$$\angle QPS<\angle PQS$$
So,
$$QS<PR$$ ........(2)

And,
$$\angle QPS<\angle PSQ$$
So,
$$QS<PQ$$ .......(3)

Therefore, from eq. (1), (2), (3)
$$QS<PQ<PR$$.
Question 4
1 / -0

In a $$\triangle$$PQR, PS is bisector of $$\angle P$$ and $$\angle Q = 70^o \angle R = 30^o$$, then

A
$$QS > PQ > PR$$

B
$$QS < PQ < PR$$

C
$$QS < PQ > PR$$

D
None of these

Solution

Given: $$\angle Q = 70^o, \angle R = 30^o$$
$$\angle QPR =180^o - (70^o + 30^o ) =80^o$$ {angle sum property of a triangle}
$$\therefore \angle QPS = \angle SPR =40^o$$
$$\angle PRQ < \angle PQR$$
$$\therefore PQ <PR$$ ....... (i) {Side opposite to the larger angle is larger}
$$\angle QPS < PQS$$
$$ \therefore QS < PR$$ ...... (ii)   {Side opposite to the larger angle is larger}

In $$\Delta PQS$$,
$$\angle PSQ = 180^o -(70^o + 40^o) =70^o$$   {angle sum property of a triangle}
$$\angle QPS < \angle PSQ$$
$$\therefore QS < PQ$$   ........ (iii)   {Side opposite to the larger angle is larger}
From (i), (ii) & (iii)
$$QS < PQ < PR$$.
Question 5
1 / -0

If a $$\triangle PQR$$ is constructed taking $$QR = 5\text{ cm},$$ $$PQ = 3\text{ cm},$$ and $$PR = 4\text{ cm}$$ then the correct order of the angles of the triangle is:

A
$$\angle P < \angle Q < \angle R$$

B
$$\angle P>\angle Q<\angle R$$

C
$$\angle P > \angle Q >\angle R$$

D
$$\angle P < \angle Q>\angle R$$

Solution

In a triangle, the angle is determined by its sides if it is given. The largest side will have the largest angle opposite to it. The smallest side will have the smallest angle opposite to it.

So, $$QR=5\text{ cm},$$ is the largest side. Hence the angle opposite to it will also be largest that is$$\angle P.$$

Then the side$$PR=4 \text{ cm},$$, smaller than $$QR$$. Hence the $$\angle Q$$ will be smaller than $$\angle P$$

Finally, the smallest side is $$PQ=3\text{ cm}$$ with its corresponding angle $$\angle R$$ is smallest.

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

Which of the following statement is false?

A
The sum of two sides of a $$\Delta$$ is greater than the third side

B
In a right angled $$\Delta$$ hypotenuse is the longest side

C
A, B, C are collinear if AB + BC = AC

D
None of these

Solution

$${\textbf{Step 1: Consider, option A}}{\textbf{.}}$$

$${\text{The sum of two sides of a}}$$ $$\vartriangle $$ $${\text{is greater than the third side}}{\text{.}}$$

$${\text{From triangle inequality theorem we have,}}$$
$${\text{The sum of the lengths of any two sides of a triangle is greater than the length of the third side}}{\text{.}}$$

$${\text{Thus, option A}}{\text{. is true}}{\text{.}}$$

$${\textbf{Step 2: Consider, option B}}{\textbf{.}}$$

$${\text{In a right angled}}$$ $$\vartriangle $$ $${\text{hypotenuse is the longest side}}{\text{.}}$$

$${\text{By Pythagoras theorem we have,}}$$

$${\text{In a right angled triangle, the square of the hypotenuse side is equal to the sum of}}$$
$${\text{square of the two sides}}{\text{.}}$$

$${\text{It implies that hypotenuse is the longest side}}{\text{.}}$$

$${\text{Thus, option B}}{\text{. is true}}{\text{.}}$$

$${\textbf{Step 3: Consider, option C}}{\textbf{.}}$$

$${\text{A, B, C are collinear if }}AB + BC = AC.$$

$${\text{As we know that, collinear points are the points that lie on the same line}}{\text{.}}$$
$${\text{These points lie on a line close to or far from each other}}{\text{.}}$$

$${\text{Thus, option C}}{\text{. is true}}{\text{.}}$$

$${\text{Hence, all of the above statements are true}}{\text{.}}$$

$${\text{So, none of these is false}}{\text{.}}$$

$${\textbf{Hence, option D is correct answer.}}$$
Question 7
1 / -0

In a triangle ABC. The relation which is true for its sides is-

A
AB + BC < AC

B
AB + BC > AC

C
AB +BC = CA

D
AB = BC + AC

Solution

In $$ABC$$,
Sum of the two sides is always greater than third side.
So, $$AB + BC > AC$$
Question 8
1 / -0

If a triangle $$PQR$$ has been constructed taking $$QR = 6 $$ cm, $$PQ = 3 $$ cm and $$PR = 4 $$ cm, then the correct order of the angle of triangle is

A
$$\displaystyle \angle P< \angle Q< \angle R $$

B
$$\displaystyle \angle P> \angle Q< \angle R $$

C
$$\displaystyle \angle P> \angle Q> \angle R $$

D
$$\displaystyle \angle P< \angle Q> \angle R $$

Solution

Given, in $$\triangle PQR$$, $$QR=6$$ cm, $$PQ=3$$ cm, $$PR=4$$ cm
We know,
(i) the shortest side is always opposite the smallest interior angle.
(ii) the longest side is always opposite the largest interior angle.
Here, $$QR=6$$ cm is the largest side, therefore $$\angle P$$ is the greatest.
And $$PQ=3$$ cm is the smallest side, therefore $$\angle R $$ is the smallest angle.
Therefore, the correct order is $$\angle P>\angle Q>\angle R$$.
Question 9
1 / -0

In the given figure, $$\triangle RTQ\cong \triangle PSQ$$ by ASA congruency condition. Which of the following pairs does not satisfy the condition?

A
$$PQ=QR$$

B
$$\angle P=\angle R$$

C
$$\angle TQP=\angle SQR$$

D
None of these

Solution

Given, $$\triangle RTQ \cong \triangle PSQ$$
Thus, $$PQ = QR$$ (by cpct)
$$\angle P = \angle R$$ (by cpct)

Also, $$\angle RQT = \angle PQS$$ (By CPCT)
$$\angle RQT - \angle TQS = \angle PQS - \angle TQS$$
Thus, $$\angle TQP = \angle SQR$$
Question 10
1 / -0

In the given figure, $$AD=BC, AC=BD.$$ Then $$\triangle PAB$$ is

A
equilateral

B
right angled

C
scalene

D
isosceles

Solution

In $$\triangle ADB$$ and $$\triangle ACB$$
$$AD=BC$$          (Given)
$$AC=BD$$
$$AB=BA$$          (Common)
$$\therefore $$ $$\triangle ADB\cong \triangle ACB$$          (SSS)
$$\Rightarrow $$ $$\angle ABD=\angle CAB$$          (cpct)
$$\Rightarrow $$ $$\angle ABP=\angle PAB$$
$$\Rightarrow $$ $$PA=PB$$          (Sides opp. equal angles are equal)
$$\Rightarrow $$ $$\triangle PAB$$ is isosceles.