The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 24

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Question 1
1 / -0

Find the value of x, y and z in the adjoining figure.

A
$$x=160$$, $$y=60$$, $$z=80$$

B
$$x=60$$, $$y=80$$, $$z=40$$

C
$$x=80$$, $$y=60$$, $$z=140$$

D
$$x=140$$, $$y=80$$, $$z=60$$

Solution

$$In\triangle BCE$$
$$\implies\quad \angle BEC+\angle BCE+\angle CBE={ 180 }^{ \circ }$$
$$\implies\quad { 90 }^{ \circ }+{ 30 }^{ \circ }+\angle CBE={ 180 }^{ \circ }$$
$$\implies\quad \angle CBE={ 60 }^{ \circ }$$
$$ y={ 60 }^{ \circ }$$
$$ In\triangle APC:$$
$$\implies\quad { 50 }^{ \circ }+{ 30 }^{ \circ }+\angle APC={ 180 }^{ \circ }$$
$$\implies\quad \angle APC={ 100 }^{ \circ }$$
$$ \therefore x=180-\angle APC$$
$$ (\because sum\quad of\quad angles\quad on\quad a\quad straight\quad line={ 180 }^{ \circ })$$
$$\implies\quad x={ 180 }^{ \circ }-{ 100 }^{ \circ }={ 80 }^{ \circ }$$
$$ In\triangle BOP:$$
$$ z=x+y\quad (exterior\quad angle\quad of\quad a\quad triangle=sum\quad of\quad two\quad opposite\quad interior\quad angles)$$
$$\implies\quad z=80+60={ 140 }^{ \circ }$$
$$ \therefore x={ 80 }^{ \circ }\quad ,\quad y={ 60 }^{ \circ }\quad ,z={ 140 }^{ \circ }$$
Question 2
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In $$\triangle ABC,\angle A={ x }^{ \circ },\angle B={ \left( 2x-15 \right) }^{ \circ}$$ and $$\angle C ={ \left( 3x+21 \right) }^{ \circ }$$. Find the value of $$x$$ and the measure of each angle of the triangle.

A
$$x={ 27 }^{ \circ },\angle A={ 27 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 101 }^{ \circ }$$

B
$$x={ 29 }^{ \circ },\angle A={ 29 }^{ \circ },\angle B={ 43 }^{ \circ },\angle C={ 108 }^{ \circ }$$

C
$$x={ 27 }^{ \circ },\angle A={ 27 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 108 }^{ \circ }$$

D
$$x={ 30 }^{ \circ },\angle A={ 30 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 109 }^{ \circ }$$

Solution

Given, $$\angle A=x^\circ$$, $$\angle B=(2x-15)^\circ$$ and $$\angle C=(3x+21)^\circ$$.
We know, by angle sum property, the sum of angles of a triangle is $$180^\circ$$.
Then, $$\angle A+ \angle B+ \angle C=180^\circ$$
$$\implies$$ $$x^\circ+(2x-15)^\circ+ (3x+21)^\circ=180^\circ$$
$$\implies$$ $$6x^\circ+6^\circ=180^\circ$$
$$\implies$$ $$6x^\circ=180^\circ-6^\circ$$
$$\implies$$ $$6x^\circ=174^\circ$$
$$\implies$$ $$x^\circ=29^\circ$$.

Therefore,
$$\angle A=x^\circ=29^\circ$$,
$$\angle B=(2x-15)^\circ=2\times29^\circ-15^o=58^\circ-15^\circ=43^\circ$$
and $$\angle C=(3x+21)^\circ=3\times29^\circ+21^\circ=87^\circ+21^\circ=108^\circ$$.

Hence, option $$B$$ is correct.
Question 3
1 / -0

In $$\triangle XYZ$$ ______________is the base.

A
$$XY$$

B
$$YZ$$

C
$$XZ$$

D
None of the above

Solution
Question 4
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From the figure, find values of $$x$$ and $$y$$.

A
$$25^{o},30^{o}$$

B
$$35^{o},31^{o}$$

C
$$50^{o},28^{o}$$

D
$$45^{o},33^{o}$$

Solution

In the given triangle,
by angle sum property,
$$40^o+95^o+x^o=180^o.......(i)$$
and $$x^o+y^o+102^o=180^o.......(ii)$$.

From $$(i)$$,
$$40^o+95^o+x^o=180^o$$
$$\implies$$ $$135^o+x^o=180^o$$
$$\implies$$ $$x^o=180^o-135^o$$
$$\implies$$ $$x^o=45^o.......(iii)$$.

Substitute $$(iii)$$ in $$(ii)$$,
$$45^o+y^o+102^o=180^o$$
$$\implies$$ $$y^o+147^o=180^o$$
$$\implies$$ $$y^o=180^o-147^o$$
$$\implies$$ $$y^o=33^o$$.

Therefore, $$x^o=45^o$$ and $$y^o=33^o$$.
Hence, option $$D$$ is correct.
Question 5
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Find the length of the hypotenuse in a right angled triangle if the sum of the squares of the sides making right angle is 169.

A
$$15$$

B
$$13$$

C
$$5$$

D
$$12$$

Solution

According to the Pythagoras theorem, the sum of the squares of the sides making the right angle is equal to the square of the third side (hypotenuse).

$$\therefore\ $$ Square of the hypotenuse $$=169$$

$$\Rightarrow$$ Hypotenuse $$=\sqrt{169}$$

$$=13\space\mathrm{units}$$

Hence, option B is correct.
Question 6
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In the fig. then , L (DN) = ?

A
3.8 cm.

B
3.6 cm.

C
2.4 cm.

D
5.4 cm.

Solution
Question 7
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In $$\triangle ABC,\angle B={ 90 }^{ \circ }$$ find the sides of the triangle if $$AB=(x-3)~cm,BC=(x+4)~cm$$ and $$AC=(x+6)~cm$$.

A
$$8 cm, 15 cm, 17 cm$$

B
$$17 cm, 24 cm, 26 cm$$

C
$$9 cm, 16 cm, 18 cm$$

D
$$12 cm, 19 cm, 21 cm$$

Solution
Question 8
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In triangle, three angles are $$x , x + 10 ^ { \circ } + x + 20 ^ { \circ }$$ then the biggest is

A
$$70 ^ { \circ }$$

B
$$80 ^ { \circ }$$

C
$$90 ^ { \circ }$$

D
none

Solution
Question 9
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In a right-angled triangle the lengths of base and perpendicular are 6 cm and 8 cm.What is the length of the hypotenuse?

A
9 cm

B
10 cm

C
11 cm

D
12 cm

Solution

The sides of right angled triangle are $$6,8 cm$$

The hypotenuse is given as

$$a^2+b^2=c^2\\6^2+8^2=c^2\\36+64=c^2\\c^2= 100\\c=\sqrt {100}\\c=10cm$$
Question 10
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If the angles of triangle are in the ratio $$1:4:7$$, then the value of the largest angle is :

A
$$135^{\circ}$$

B
$$84^{\circ}$$

C
$$105^{\circ}$$

D
none of these

Solution