M.C.Q - MATHS STD 9 Questions - Vidyadip

Questions

M.C.Q

🎯

Test yourself on this topic

30 questions · timed · auto-graded

Question 11 Mark

In Fig. AB and CD are parallel lines and transversal EF intersect them at P and Q respectively. If $\angle\text{APR}=25^\circ,\angle\text{RQC}=30^\circ$ and $\angle\text{CQF}=65^\circ,$ then:

x = 55º, y = 40º
x = 50º, y = 45º
x = 60º, y = 35º
x = 35º, y = 60º

Answer

x = 55º, y = 40º

Solution:

$\angle\text{OQP}=180^\circ-\angle\text{OQF}$
$=180^\circ-(30^\circ+65^\circ)$
$\Rightarrow\angle\text{OQP}=85^\circ\dots(1)$
$\angle\text{APQ}=\angle\text{CQF}$ (Corresponding angles)
$\Rightarrow25^\circ+\text{y}^\circ=65^\circ$
$\Rightarrow\text{y}^\circ=65^\circ-25^\circ$
$\Rightarrow\text{y}^\circ=40^\circ$
Now in $\triangle\text{OPQ}$
$\angle\text{O}+\angle\text{OPQ}+\angle\text{PQO}=180^\circ$
$\Rightarrow\text{x}^\circ+40^\circ+85^\circ=180^\circ$
$\text{x}^\circ=180^\circ-85^\circ-40^\circ=55^\circ$
$\Rightarrow\text{x}^\circ=55^\circ,\text{y}=40^\circ$

View full question & answer→

MCQ 21 Mark

In Fig. if $l_1 || l_2,$ the value of $x$ is:

A
$22\frac{1}{2}$
B
$30^\circ$
✓
$45^\circ$
D
$60^\circ$

Answer

Correct option: C.

$45^\circ$

From figure,
$\angle\text{ACS}=180^\circ-2\text{b}^\circ$
also $\angle\text{ACS}=\angle\text{PAC}=2\text{a}^\circ ($alternate angles$)$
$\Rightarrow2\text{a}^\circ=180^\circ-2\text{b}^\circ$
$\Rightarrow\text{a}^\circ+\text{b}^\circ=90^\circ$
Now, in $\triangle\text{ABC}$
$\text{a}^\circ+\text{b}^\circ+\angle\text{ABC}=180^\circ$
$\angle\text{ABC}=180^\circ-2\text{x}^\circ$
$\Rightarrow\text{a}^\circ+\text{b}^\circ+180^\circ-2\text{x}^\circ=180^\circ$
$\Rightarrow2\text{x}^\circ=\text{a}^\circ+\text{b}^\circ=90^\circ$
$\Rightarrow\text{x}^\circ=45^\circ$

View full question & answer→

Question 31 Mark

In a $\triangle\text{ABC},$ if $\angle\text{A}=60^\circ,\angle\text{B}=80^\circ$ and the bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at O, then $\angle\text{BOC}=$

60º
120º
150º
30º

Answer

120º

Solution:

O is point where bisectors of $\angle\text{C }\& \angle\text{B}$ meets.
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$60^\circ+80^\circ+\angle\text{C}=108^\circ$
$\angle\text{C}=40^\circ$
$\frac{\angle\text{C}}{2}=20^\circ$
$\frac{\angle\text{C}}{2}=20^\circ=\angle\text{BCO}\dots(1)$
$\frac{\angle\text{B}}{2}=\frac{80^\circ}{2}=40^\circ=\angle\text{OBC}\dots(2)$
In $\triangle\text{BOC}$
$\angle\text{BCO}+\angle\text{OBC}+\angle\text{BOC}=180^\circ$
from (1) and (2)
$20^\circ+40^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=180^\circ-60^\circ=120^\circ$

View full question & answer→

Question 41 Mark

In Fig. x + y =

270º
230º
210º
190º

Answer

230º

Solution:
$\triangle\text{ACO}$
$\angle\text{ACO}+\angle\text{COA}+\angle\text{COA}=180^\circ$
Now, $\angle\text{OAC}=180^\circ-\text{x}^\circ$
$\Rightarrow80^\circ+40^\circ+180^\circ-\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=120^\circ$
$\angle\text{BOD}=\angle\text{COA}=40^\circ$ (Opposite angles)
$\angle\text{BDO}=70^\circ$
In $\triangle\text{OBD},$
$\angle\text{OBD}=180^\circ-40^\circ-70^\circ=70^\circ$
Also, $\text{y}^\circ=180^\circ-\angle\text{OBD}=180^\circ-70^\circ=110^\circ$
$\Rightarrow\text{x}^\circ+\text{y}^\circ=120^\circ+110^\circ=230^\circ$

View full question & answer→

Question 51 Mark

In $\triangle\text{ABC},\angle\text{B}=\angle\text{C}$ and ray AX bisects the exterior angle $\angle\text{DAC}.$ If $\angle\text{DAX}=70^\circ$ then $\angle\text{ACB}=$

35º
90º
70º
55º

Answer

70º

Solution:

AX is bisector of $\angle\text{DAC}.$
$\Rightarrow\angle\text{DAX}=\angle\text{XAC}=70^\circ$
$\Rightarrow\angle\text{DAC}=2\times70^\circ=140^\circ$
Now, $\angle\text{A}=180^\circ-\angle\text{DAC}=180^\circ-140^\circ=40^\circ$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow40^\circ+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow40^\circ+\angle\text{C}+\angle\text{C}=180^\circ\dots(\angle\text{B}=\angle\text{C})$
$\Rightarrow2\angle\text{C}=140^\circ$
$\Rightarrow\angle\text{C}=70^\circ$
$\text{i.e}\angle\text{ACB}=70^\circ$

View full question & answer→

Question 61 Mark

The side BC of $\triangle\text{ABC}$ is produced to a poin D. The bisector of $\angle\text{A}$ meet side BC in L. If $\angle\text{ABC}=30^\circ$ and $\angle\text{ACD}=115^\circ,$ then $\angle\text{ALC}=$

$85^\circ$
$72\frac{1}{2}^\circ$
$145^\circ$
$\text{None of these}$

Answer

$72\frac{1}{2}^\circ$

Solution:

$\angle\text{C}=180^\circ-\angle\text{ACD}=180^\circ-115^\circ=65^\circ$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}=180^\circ-30^\circ-65^\circ$
$\Rightarrow\angle\text{A}=85^\circ$
Now in $\angle\text{ALC}$
$\angle\text{ALC}+\angle\text{LAC}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{ALC}=180^\circ-\angle\text{LAC}-\angle\text{C}$
$=180^\circ-\frac{\angle\text{A}}{2}-\angle\text{C}$
$=180^\circ-\frac{85^\circ}{2}-65^\circ$
$=\frac{145^\circ}{2}$
$=72\frac{1}{2}^\circ$

View full question & answer→

Question 71 Mark

An exterior angle of a triangle is 108º and its interior opposite angles are in the ratio 4 : 5 The angles of the triangle are:

48º, 60º, 72º
50º, 60º, 70º
52º, 56º, 72º
42º, 60º, 76º

Answer

48º, 60º, 72º

Solution:

From figure, we have
$\angle\text{A}+\angle\text{B}=\angle\text{ACD}$
$\Rightarrow4\text{x}^\circ+5\text{x}^\circ=108^\circ$
$\Rightarrow9\text{x}^\circ=108^\circ$
$\Rightarrow\text{x}=12^\circ$
So, $\angle\text{A}=48^\circ,\angle\text{B}=60^\circ$
$\Rightarrow\angle\text{C}=180^\circ-48^\circ-60^\circ=72^\circ$

View full question & answer→

Question 81 Mark

In Fig the value of x is:

65º
80º
95º
120º

Answer

120º

Solution:

In $\triangle\text{ABD}$
$\angle\text{A}+\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow55^\circ+\angle\text{DBA}+25^\circ=180^\circ$
$\Rightarrow\angle\text{DBA}=180^\circ-55^\circ-25^\circ$
$=180^\circ-80^\circ$
$\Rightarrow\angle\text{DBA}=100^\circ$
So, $\angle\text{DBC}=180^\circ-\angle\text{DBA}$
$=180^\circ-100^\circ$
$\Rightarrow\angle\text{DBC}=80^\circ$
Now, in $\triangle\text{EBC}$
$\angle\text{E}+\angle\text{EBC}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{E}+80^\circ+40^\circ=180^\circ$ $\big(\angle\text{DBC}=\angle\text{EBC}\big)$
$\Rightarrow\angle\text{E}=180^\circ-120^\circ=60^\circ$
Also, $\text{x}=180^\circ-\angle\text{E}=180^\circ-60^\circ$
$\Rightarrow\text{x}=120^\circ$

View full question & answer→

Question 91 Mark

In Fig. if BP || CQ and AC = BC, then the measure of x is:

20º
25º
30º
35º

Answer

30º

Solution:

$\angle\text{PBC}=\angle\text{QCD}$ (Corresponding angles, OP || CQ and BC is transverse)
$\Rightarrow\angle\text{PBC}=70^\circ$
Now, $\angle\text{PBA}+\angle\text{ABC}+\angle\text{PBC}$
$\Rightarrow20^\circ+\angle\text{ABC}=70^\circ$
$\Rightarrow\angle\text{ABC}=50^\circ$
In $\triangle\text{ABC},$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ\dots(1)$
Now, $\angle\text{ABC}=\angle\text{BAC}=50^\circ$ $($isosceles $\triangle)$
And, $\angle\text{ACB}=180^\circ-(70^\circ+\text{x})$
From (1)
$50^\circ+50^\circ+180^\circ-(70^\circ+\text{x})=180^\circ$
$\Rightarrow\text{x}=30^\circ$

View full question & answer→

Question 101 Mark

If the measure of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure of the smallest angle of the triangle?

25º
30º
45º
60º

Answer

45º

Solution:
The measures of angles of a triangle are in ratio 3 : 4 : 5.
Let the angles be 3x, 4x, and 5x.
in any triangle, sum of all angles = 180º
⇒ 3x + 4x + 5x = 180º
⇒ 12x = 180º
⇒ x = 15º
So, smallest angle = 3 × 15º = 45º

View full question & answer→

Question 111 Mark

In a $\triangle\text{ABC},\angle\text{A}=50^\circ$ and BC is produced to a point D. If the bisectors of $\angle\text{ABC}$ and $\angle\text{ACD}$ meet at E, then $\angle\text{E}=$

25º
50º
100º
75º

Answer

25º

Solution:

BE and CE are bisectors of $\angle\text{B}$ and $\angle\text{ACD}.$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}=180^\circ-50^\circ=130^\circ\dots(1)$
Now, in $\triangle\text{BEC}$
$\angle\text{CBE}+\angle\text{BEC}+\angle\text{ECB}=180^\circ\dots(2)$
$\angle\text{CBE}=\frac{\angle\text{B}}{2},\angle\text{BEC}=\angle\text{E},\angle\text{ECB}=\angle\text{C}+\angle\text{ACE}$
Now, $\angle\text{ACD}=180^\circ-\angle\text{C}$
$\angle\text{ACE}=\frac{\angle\text{ACD}}{2}=\frac{180^\circ-\angle\text{C}}{2}=90^\circ-\frac{\angle\text{C}}{2}$
So, $\angle\text{ECB}=\angle\text{C}+90^\circ-\frac{\angle\text{C}}{2}$
$\Rightarrow\angle\text{ECB}=90^\circ+\frac{\angle\text{C}}{2}$
Now putting all value in eq (2)
$\frac{\angle\text{B}}{2}+\angle\text{E}+90^\circ+\frac{\angle\text{C}}{2}=180^\circ$
$\Rightarrow\angle\text{E}=180^\circ-90^\circ-\Big(\frac{\angle\text{B}+\angle\text{C}}{2}\Big)$
$=90^\circ-\Big(\frac{\angle\text{B}+\angle\text{C}}{2}\Big)$
$=90^\circ-\frac{130^\circ}{2}$[From eq (1)]
$\Rightarrow\angle\text{E}=25^\circ$

View full question & answer→

Question 121 Mark

The bisects of exterior angles at B and C of $\triangle\text{ABC}$ meet at O. If $\angle\text{A}=\text{x}^\circ,$ then $\angle\text{BOC}=$

$90^\circ+\frac{\text{x}^\circ}{2}$
$90^\circ-\frac{\text{x}^\circ}{2}$
$180^\circ+\frac{\text{x}^\circ}{2}$
$180^\circ-\frac{\text{x}^\circ}{2}$

Answer

$90^\circ-\frac{\text{x}^\circ}{2}$

Solution:

In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}=180^\circ-\text{x}^\circ\dots(1)$
$\Rightarrow\angle\text{CBD}=180^\circ-\angle\text{B}\dots(2)$
$\Rightarrow\angle\text{ECB}=180^\circ-\angle\text{C}\dots(3)$
$\Rightarrow\frac{\angle\text{CBD}}{2}=\angle\text{OBC}=90^\circ-\frac{\angle\text{B}}{2}\dots(4)$ [frpom eq(2)]
$\frac{\angle\text{ECB}}{2}=\angle\text{OCB}=90^\circ-\frac{\angle\text{C}}{2}\dots(5)$
Now, in $\triangle\text{BOC}$
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=180^\circ-(\angle\text{OBC}+\angle\text{OCB})$
From eq (4) and (5),
$\angle\text{BOC}=180^\circ-\Big(90^\circ-\frac{\angle\text{B}}{2}+90^\circ-\frac{\angle\text{C}}{2}\Big)$
$=180^\circ-\Big(180^\circ-\frac{\angle\text{B}}{2}-\frac{\angle\text{C}}{2}\Big)$
$=\Big(\frac{\angle\text{B}+\angle\text{C}}{2}\Big)$ [from eq (1)]
$=\frac{180^\circ-\text{x}^\circ}{2}$
$\Rightarrow\angle\text{BOC}=90^\circ-\frac{\text{x}^\circ}{2}$

View full question & answer→

Question 131 Mark

In Fig. if $\text{EC }||\text{ AB},\angle\text{ECD}=70^\circ$ and $\angle\text{ECD}=70^\circ$ and $\angle\text{BDO}=20^\circ,$ then $\angle\text{OBD}$ is:

20º
50º
60º
70º

Answer

50º

Solution:
EC || AB And, CD is transverse to it.
Now $\angle\text{ECD}=\angle\text{AOD}=70^\circ$ (Corresponding angles)
In $\triangle\text{OBD}$
$\angle\text{OBD}+\angle\text{BOD}+\angle\text{ODB}=180^\circ$
$\angle\text{BOD}=180^\circ-\angle\text{AOD}=180^\circ-70^\circ=110^\circ$
So $\angle\text{OBD}=180^\circ-\angle\text{BOD}-\angle\text{ODB}$
$=180^\circ-110^\circ-20^\circ$
$=50^\circ$

View full question & answer→

Question 141 Mark

In Fig. what is the value of x?

35º
45º
50º
60º

Answer

60º

Solution:
In $\triangle\text{ABC},$
$\angle\text{BCA}+\angle\text{CAB}+\angle\text{ABC}=180^\circ$
$\Rightarrow3\text{y}^\circ+\text{x}^\circ+5\text{y}^\circ=180^\circ$
$\Rightarrow8\text{y}^\circ+\text{x}^\circ=180^\circ\dots(1)$
Also, $5\text{y}^\circ=180^\circ-7\text{y}^\circ$
$\Rightarrow12\text{y}^\circ=180^\circ$
$\Rightarrow\text{y}^\circ=15^\circ$
From (1), $\text{x}^\circ=180^\circ-8\text{y}^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-8\times15^\circ$
$\Rightarrow\text{x}^\circ=60^\circ$

View full question & answer→

Question 151 Mark

In a triangle, an exterior angle at a vertex is 95º and its one of the interior opposite angle is 55º, then the measure of the other interior angle is:

55º
85º
40º
9.0º

Answer

40º

Solution:
Let the other interior opposite angle be xº.
Then, we have
xº + 55º = 95º
⇒ xº = 95º - 55º = 40º

View full question & answer→

Question 161 Mark

In $\triangle\text{RST}$ what is the value of x?

40º
90º
80º
100º

Answer

100º

Solution:

In $\triangle\text{RST}$
$\angle\text{R}+\angle\text{S}+\angle\text{T}=180^\circ$
$\Rightarrow2\text{a}^\circ+\text{x}^\circ+2\text{b}^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-2(\text{a}+\text{b})^\circ\dots(1)$
Now in $\triangle\text{ROT}$
$\angle\text{ORT}+\angle\text{ROT}+\angle\text{OTR}=180^\circ$
$\Rightarrow\text{a}^\circ+140^\circ+\text{b}^\circ=180^\circ$
$\Rightarrow(\text{a}+\text{b})^\circ=180^\circ-140^\circ=40^\circ\dots(2)$
From (1) and (2)
$\text{x}^\circ=180^\circ-2(40^\circ)$
$\Rightarrow\text{x}=100^\circ$

View full question & answer→

Question 171 Mark

If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is:

90º
180º
270º
360º

Answer

360º

Solution:

In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
Now, $\angle\text{FAB}=180^\circ-\angle\text{A}\dots(1)$
$\angle\text{DCA}=180^\circ-\angle\text{C}\dots(2)$
$\angle\text{EBC}=180^\circ-\angle\text{B}\dots(3)$
Adding equation (1), (2) and (3)$\angle\text{FAB}+\angle\text{DCA}+\angle\text{EBC}=180^\circ-\angle\text{A}+180^\circ-\angle\text{C}+180^\circ-\angle\text{B}$
$=540^\circ-(\angle\text{A}+\angle\text{B}+\angle\text{C})$
$=540^\circ-180^\circ$
⇒ Sum of all exterior angles = 360º

View full question & answer→

Question 181 Mark

If the bisectors of the acute angles of a right triangle meet at O, then the angle at O between the two bisectors is:

45º
95º
135º
90º

Answer

135º

Solution:

In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+90^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{C}=90^\circ\dots(1)$
Now, in $\triangle\text{AOC},$
$\angle\text{COA}+\angle\text{OAC}+\angle\text{OCA}=180^\circ$
$\Rightarrow\angle\text{COA}+\frac{\angle\text{A}}{2}+\frac{\angle\text{C}}{2}=180^\circ$ {AO and CO bisects angle $\angle\text{A}$ and $\angle\text{C}$}
$\Rightarrow\angle\text{COA}=180-\Big(\frac{\angle\text{A}+\angle\text{C}}{2}\Big)$
$=180^\circ-\Big(\frac{90^\circ}{2}\Big)$ {From (1)}
$=100^\circ-45^\circ$
$=135^\circ$

View full question & answer→

MCQ 191 Mark

In Fig. for which value of $x$ is $l_1 || l_2$?

A
$37^\circ$
B
$43^\circ$
C
$45^\circ$
✓
$47^\circ$

Answer

Correct option: D.

$47^\circ$

Let if $l_1 || l_2$ and $AB$ is tranverse to it.
Then,
$\angle\text{PBA}$ should be equal $\angle\text{BAS} ($Alternate angles$)$
So if $l_1 || l_2$, then $\angle\text{BAS}=70^\circ$
$\Rightarrow\angle\text{BAC}=78^\circ-35^\circ=43^\circ\dots(1)$
Now, in $\triangle\text{ABC}$
$\text{x}^\circ+\angle\text{C}+\angle\text{BAC}=180^\circ$
$\Rightarrow\text{x}^\circ+90^\circ+43^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-90^\circ-43^\circ=47^\circ$
$\Rightarrow\text{x}^\circ=47^\circ$
So if $x^\circ = 47^\circ$ then $l_1 || l_2$

View full question & answer→

Question 201 Mark

An exterior angle of a triangle is equal to 100º and two interrior opposite angles are equal. Each of these angles is equal to:

75º
80º
40º
50º

Answer

50º

Solution:
Let the two interior opposite angles be xº each.
Now, the exterior angle is equal to the sum of the two interior opposite angles.
xº + xº = 180º
⇒ 2xº = 100º
⇒ xº = 50º

View full question & answer→

Question 211 Mark

Side BC of a triangle ABC has been produced to a point D such that $\angle\text{ACD}=120^\circ.$ If $\angle\text{B}=\frac{1}{2}\angle\text{A},$ then $\angle\text{A}$ is equal to :

80º
75º
60º
90º

Answer

80º

Solution:
$\angle\text{B}=\frac{1}{2}\angle\text{A}$
$\angle\text{ACD}$ is an exterior angle.
$\Rightarrow\angle\text{A}+\angle\text{B}=\angle\text{ACD}$
$\Rightarrow\angle\text{A}=\frac{1}{2}\angle\text{A}=120^\circ$
$\Rightarrow\frac{3\angle\text{A}}{2}=120^\circ$
$\Rightarrow3\angle\text{A}=240^\circ$
$\Rightarrow\angle\text{A}=80^\circ$

View full question & answer→

Question 221 Mark

Line segments AB and CD intersect at O such that AC || DB. If $\angle\text{CAB}=45^\circ$ and $\angle\text{CDB}=55^\circ,$ then $\angle\text{BOD}=$

100º
80º
90º
135º

Answer

80º

Solution:

AC || BD
And, AB is transverse to these parallal lines
So $\angle\text{CAB}=\angle\text{ABD}$ (Alternate angles)
$\Rightarrow\angle\text{ABD}=45^\circ$
Now In $\triangle\text{BOD}$
$\angle\text{BOD}+\angle\text{ODB}+\angle\text{DBA}=180^\circ$
$\angle\text{DBA}=\angle\text{ABD}=45^\circ,\angle\text{ODB}=55^\circ$
So $\angle\text{BOD}=180^\circ-45^\circ-55^\circ$
$=80^\circ$

View full question & answer→

Question 231 Mark

In Fig. what is y in terms of x?

$\frac{3}{2}\text{x}^\circ$
$\frac{4}{3}\text{x}^\circ$
$\text{x}^\circ$
$\frac{3}{4}\text{x}^\circ$

Answer

$\frac{3}{2}\text{x}^\circ$

Solution:

From figure,
$\angle\text{DOC}=180^\circ-\angle\text{AOD}$ (Both are Supplementary)
$\Rightarrow\angle\text{DOC}=180^\circ-3\text{y}^\circ$
Also, $\angle\text{ACB}=180^\circ-\angle\text{A}-\angle\text{B}$
$\Rightarrow\angle\text{ACB}=180^\circ-\text{x}^\circ-2\text{x}^\circ=180^\circ-3\text{x}^\circ$
And $\angle\text{ACD}=180^\circ-\angle\text{ACB}$
$=180^\circ-(180^\circ-3\text{x}^\circ)$
$\Rightarrow\angle\text{ACD}=3\text{x}^\circ$
Now, in $\triangle\text{OCD}$
$\angle\text{DOC}+\angle\text{OCD}+\angle\text{D}=180^\circ$
$180^\circ-3\text{y}^\circ+3\text{x}^\circ+\text{y}^\circ=180^\circ$ $\big[\angle\text{OCD}=\angle\text{ACD}\big]$
$\Rightarrow2\text{y}^\circ=3\text{x}^\circ$
$\Rightarrow\text{y}=\frac{3}{2}\text{x}^\circ$

View full question & answer→

Question 241 Mark

In $\triangle\text{ABC},$ if $\angle\text{A}=100^\circ,\text{AD}$ bisects $\angle\text{A}$ and $\text{AD}\perp\text{BC}.$ Then, $\angle\text{B}=$

50º
90º
40º
100º

Answer

40º

Solution:

$\text{AD}\perp\text{BC}$ and AD bisects $\angle\text{A}.$
$\Rightarrow\angle\text{BAD}=\angle\text{CAD}=50^\circ$
In Right $\triangle\text{ADB}$
$\angle\text{BAD}=50^\circ,\angle\text{ADB}=90^\circ$
Also sum of all interior angles = 180º
$\Rightarrow\angle\text{BAD}+\angle\text{ADB}+\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=180^\circ-50^\circ-90^\circ$
$\Rightarrow\angle\text{B}=40^\circ$

View full question & answer→

Question 251 Mark

If all the three angles of a triangle are equal, then each one of them is equal to:

90º
45º
60º
30º

Answer

60º

Solution:
Let the measure of each angle be x°.
Now, the sum of all angles of any triangle is 180°.
Thus, x° + x° + x° = 180°
i.e. 3x° = 180°
i.e. x° = 60°

View full question & answer→

Question 261 Mark

The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94º and 126º. Then, $\angle\text{BAC}=$

94º
54º
40º
44º

Answer

40º

Solution:

$\angle\text{ABC}=180^\circ-126^\circ=54^\circ$
$\angle\text{ACB}=180^\circ-94^\circ=86^\circ$
Now, in $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{BAC}=180^\circ-\angle\text{ABC}-\angle\text{ACB}$
$=180^\circ-54^\circ-86^\circ$
$\Rightarrow\text{BAC}=40^\circ$

View full question & answer→

Question 271 Mark

If two acute angles of a right triangle are equal, then each acute is equal to:

30º
45º
60º
90º

Answer

45º

Solution:
Let the measure of each acute angle of a triangle be x°.
Then, we have
x° + x° + 90° = 180°
i.e. 2x° = 90°
i.e. x° = 45°

View full question & answer→

Question 281 Mark

In Fig. if $\text{AB}\perp\text{BC},$ then x =

18º
22º
25º
32º

Answer

22º

Solution:
$\text{AB}\perp\text{BC}$
$\Rightarrow\angle\text{ABC}=90^\circ$
$\angle\text{CAB}=32^\circ$ (Opposite angles)
Now, in $\triangle\text{ABD}$
$\angle\text{DAB}=\text{x}^\circ+32^\circ$
$\angle\text{ABD}=90^\circ$
$\angle\text{BDA}=\text{x}^\circ+14^\circ$
In a $\triangle,$ sum of all angles = 180º
$\Rightarrow\angle\text{DAB}+\angle\text{ABD}+\angle\text{BDA}=180^\circ$
$\Rightarrow\text{x}^\circ+32^\circ+90^\circ+\text{x}^\circ+14^\circ=180^\circ$
$\Rightarrow2\text{x}^\circ=180^\circ-136^\circ$
$\Rightarrow2\text{x}^\circ=44^\circ$
$\Rightarrow\text{x}^\circ=22^\circ$

View full question & answer→

Question 291 Mark

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:

An isosceles triangle
An obtuse triangle
An equilateral triangle
A right triangle

Answer

A right triangle

Solution:
Let the three angles of a triangle be A, B and C.
Now, A + B + C = 180°
If A = B + C
Then A + (A) = 180°
i.e. 2A = 180°
i.e. A = 90°
Since, one of the angle is 90°, the triangle is a Right triangle.

View full question & answer→

Question 301 Mark

In Fig. what is z in terms of x and y?

x + y + 180º
x + y - 180º
180º - (x + y)
x + y + 360º

Answer

x + y - 180º

Solution:
From figure
$\angle\text{A}=\text{z}^\circ$
$\angle\text{ACB}=180^\circ-\text{x}^\circ$
$\angle\text{ABC}=180^\circ-\text{y}^\circ$
Now, in $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\text{z}^\circ+180^\circ-\text{y}^\circ+180^\circ-\text{x}^\circ=180^\circ$
$\Rightarrow\text{z}^\circ=\text{x}^\circ+\text{y}^\circ-180^\circ$

View full question & answer→

Generate Paper Free All Triangle And Its Angles topics

M.C.Q - MATHS STD 9 Questions - Vidyadip