The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 24

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Weekly Quiz Competition

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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

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The Triangle and Its Properties Test - 24

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The Triangle and Its Properties Test - 24

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Question 1

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

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

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

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

Solution

Question 2

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

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

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

x=30∘,∠A=30∘,∠B=41∘,∠C=109∘x={ 30 }^{ \circ },\angle A={ 30 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 109 }^{ \circ }x=30∘,∠A=30∘,∠B=41∘,∠C=109∘

Solution

Question 3

XYXYXY

YZYZYZ

XZXZXZ

None of the above

Solution

Question 4

25o,30o25^{o},30^{o}25o,30o

35o,31o35^{o},31^{o}35o,31o

50o,28o50^{o},28^{o}50o,28o

45o,33o45^{o},33^{o}45o,33o

Solution

Question 5

151515

131313

555

121212

Solution

Question 6

3.8 cm.

3.6 cm.

2.4 cm.

5.4 cm.

Solution

Question 7

8cm,15cm,17cm8 cm, 15 cm, 17 cm8cm,15cm,17cm

17cm,24cm,26cm17 cm, 24 cm, 26 cm17cm,24cm,26cm

9cm,16cm,18cm9 cm, 16 cm, 18 cm9cm,16cm,18cm

12cm,19cm,21cm12 cm, 19 cm, 21 cm12cm,19cm,21cm

Solution

Question 8

70∘70 ^ { \circ }70∘

80∘80 ^ { \circ }80∘

90∘90 ^ { \circ }90∘

none

Solution

Question 9

9 cm

10 cm

11 cm

12 cm

Solution

Question 10

135∘135^{\circ}135∘

84∘84^{\circ}84∘

105∘105^{\circ}105∘

none of these

Solution

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

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

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

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

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

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

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

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

$XY$

$YZ$

$XZ$

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

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

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

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

$15$

$13$

$5$

$12$

$8 cm, 15 cm, 17 cm$

$17 cm, 24 cm, 26 cm$

$9 cm, 16 cm, 18 cm$

$12 cm, 19 cm, 21 cm$

$70 ^ { \circ }$

$80 ^ { \circ }$

$90 ^ { \circ }$

$135^{\circ}$

$84^{\circ}$

$105^{\circ}$