The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, ABCD is a trapezium in which AB$$=7$$cm, AD$$=$$BC$$=5$$cm, DC$$=$$x cm and the distance between AB and DC is $$4$$cm. Then the value of x is ____________.

A
$$13$$

B
$$16$$

C
$$19$$

D
Cannot be determined

Solution

In $$\triangle ALD$$ and $$\triangle BMC$$:
$$DL^2=AD^2-AL^2=5^2-4^2=9\\ \Rightarrow LD=3\ cm$$
Similarly, $$MC=3\ cm$$
$$\therefore x=DL+LM+MC$$ (Length of $$LM=AB=7\ cm$$)
$$x=3+7+3=13\ cm$$
Question 2
1 / -0

In figure, $$ \angle C = 90^\circ$$ in isosceles triangle $$\Delta ABC$$, then $$AB^2 = $$ ................

A
$$\sqrt 2 BC^2$$

B
$$2BC^2$$

C
$$BC^2$$

D
$$4 BC^2$$

Solution

In $$\Delta ACB$$,

Given: $$\angle C = 90^\circ$$

Also, $$AC = BC$$ [$$\Delta ABC$$ is isosceles]

$$\therefore AB^2 = AC^2 + BC^2$$ [By Pythagoras theorem]

$$\therefore AB^2 = BC^2 + BC^2$$

$$\therefore AB^2 = 2BC^2$$
Question 3
1 / -0

In a triangle, the angles are in ratio $$1: 3: 2$$. Find the difference between the greatest and smallest angle of the triangle.

A
$$10^o$$

B
$$70^o$$

C
$$60^o$$

D
$$20^o$$

Solution

Given, the angles are in the ratio $$1:3:2$$.

Let the angles be $$x,3x$$ and $$2x$$.

Using angle sum property of triangle,
$$x+3x+2x={ 180 }^{ \circ }\\ \Rightarrow 6x={ 180 }^{ \circ }\\ \Rightarrow x={ 30 }^{ \circ }.$$

So the angles are :
$$x={ 30 }^{ \circ }$$,
$$ 3x=3\times { 30 }^{ \circ }={ 90 }^{ \circ }$$
and $$2x=2\times { 30 }^{ \circ }={ 60 }^{ \circ }$$.

Therefore, the difference between the greatest and the smallest angle $$={ 90 }^{ \circ }-{ 30 }^{ \circ }={ 60 }^{ \circ }$$.

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

Which of the following can be the sides of a right angled triangle ?

A
$$ 35, 17,18$$

B
$$ 39, 19,18$$

C
$$ 35, 27,18$$

D
$$ 41,40,9$$

Solution

We know that in a right angle triangle,
The sum of the squares on two sides of a triangle is equal to the square on the third side, then the triangle is a right angle.
$$Base^2 + Prependicular^2 =Hypotenuse^2$$
Now, we go to the options,
A) $$18^2+17^2 $$ is not equal to $$35^2$$ so leave this option.
B) $$19^2+18^2$$ is not equal to $$39^2$$ so leave this option.
C) $$27^2+18^2$$ is not equal to $$35^2$$ so leave this option.
D)$$40^2+9^2 $$ is equal to $$41^2$$ so right this option.
Hence, all option is incorrect.
Now , we take the triplet $$41$$,$$40$$ ,$$9$$
$$40^2+9^2=1681$$
$$40^2+9^2=41^2$$
Hence, $$41$$,$$40$$,$$9$$ are the sides of a right angle triangle
option $$D$$ is correct.
Question 5
1 / -0

If you place a ladder against a wall at an angle, what type of a triangle will you get?

A
Equilateral triangle

B
Right angle triangle

C
Isosceles triangle

D
Scalene triangle

Solution

Here $$AO$$ is wall and $$AB$$ is ladder.

Hence,irrespective of what $$\theta$$ is, $$\angle AOB$$ will always be $${ 90 }^{ \circ }$$.

$$\therefore$$ Triangle will always be right-angled.
Question 6
1 / -0

In $$\Delta ABC;\;m \angle B=90^{\circ};\;AB=5\;cm$$ and $$AC=13$$ then, $$BC=....$$

A
$$10$$

B
$$18$$

C
$$8$$

D
$$12$$

Solution

Since in a triangle $$ABC$$ angle B is given to be 90°.
So by pythagoras theorem we have
$$BC^2 = AC^2-AB^2$$
$$\Rightarrow$$ $$BC^2 = 13^2-5^2 $$
$$\Rightarrow$$ $$BC= 12$$
Question 7
1 / -0

In $$\triangle ABC$$ is $$\angle B$$ is right angle. If $$AB = a=16$$ and $$BC = c=12$$ then $$AC = b=$$ __________

A
$$8$$

B
$$18$$

C
$$20$$

D
$$28$$

Solution

Given:
$$\angle$$ $$B$$ is $$90$$$$ ^{\circ}$$
$$a = 16$$
$$c = 12$$

$$\therefore $$ using Pythagoras theorem
$$a^{2}+c^{2}=b^{2}$$

$$16^{2}+12^{2}=b^{2}$$

$$256+144=b^{2}$$

$$400=b^{2}$$

$$\therefore $$ $$b = 20.$$
Question 8
1 / -0

Find the hypotenuse of right angled triangle if the other sides are $$3,4$$ respectively.

A
5

B
3

C
2

D
1

Solution

Given sides of triangle are $$3,4$$
From Pytagorous theorm ,
$$AC^2=AB^2+BC^2\\AC^2=3^2+4^2\\AC^2=9+16\\AC^2=25\\AC=\sqrt{25}\\AC=5$$
Question 9
1 / -0

Which of the following are angles of Obtuse angled triangle?

A
$$30^{o},90^{o},60^{o}$$

B
$$30^{o},70^{o},80^{o}$$

C
$$120^{o},20^{o},40^{o}$$

D
$$90^{o},45^{o},45^{o}$$

Solution

Ans- From given angles obtuse angled triangle pair is
$$120^{\circ},20^{\circ},40^{\circ}$$
obtuse angle:- Angle greater than $$90^{\circ}$$ but smaller than $$180^{\circ}$$
Question 10
1 / -0

Find hypotenuse of right angled triangle if the sides are $$12,4\sqrt 3$$

A
$$8\sqrt 3$$

B
$$4\sqrt 3$$

C
$$6\sqrt 3$$

D
$$12\sqrt 3$$

Solution

The hypotenuse is given by
$$\sqrt {12^2+(4\sqrt 3)^2}\\\sqrt {144+48}\\\sqrt {192}=8\sqrt 3$$

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Re-Attempt Weekly Quiz Competition

The Triangle and Its Properties Test - 14

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The Triangle and Its Properties Test - 14

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SHARING IS CARING

Question 1

$$13$$

$$16$$

$$19$$

Cannot be determined

Solution

Question 2

$$\sqrt 2 BC^2$$

$$2BC^2$$

$$BC^2$$

$$4 BC^2$$

Solution

Question 3

$$10^o$$

$$70^o$$

$$60^o$$

$$20^o$$

Solution

Question 4

$$ 35, 17,18$$

$$ 39, 19,18$$

$$ 35, 27,18$$

$$ 41,40,9$$

Solution

Question 5

Equilateral triangle

Right angle triangle

Isosceles triangle

Scalene triangle

Solution

Question 6

$$10$$

$$18$$

$$8$$

$$12$$

Solution

Question 7

$$8$$

$$18$$

$$20$$

$$28$$

Solution

Question 8

5

3

2

1

Solution

Question 9

$$30^{o},90^{o},60^{o}$$

$$30^{o},70^{o},80^{o}$$

$$120^{o},20^{o},40^{o}$$

$$90^{o},45^{o},45^{o}$$

Solution

Question 10

$$8\sqrt 3$$

$$4\sqrt 3$$

$$6\sqrt 3$$

$$12\sqrt 3$$

Solution

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