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Maths M.C.Q

Triangle And Its Angles · Jharkhand Board (JAC) STD 9 (Jharkhand - English Medium). 110 questions.

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M.C.Q

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110 questions · 79 auto-graded MCQ + 31 self-marked written.

MCQ 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 = 55^\circ , y = 40^\circ $
  • B
    $x = 50^\circ , y = 45^\circ $
  • C
    $x = 60^\circ , y = 35^\circ$
  • D
    $x = 35^\circ , y = 60^\circ$
Answer
Correct option: A.
$x = 55^\circ , y = 40^\circ $

$\angle\text{OQP}=180^\circ-\angle\text{OQF}$
$=180^\circ-(30^\circ+65^\circ)$
$\Rightarrow\angle\text{OQP}=85^\circ\dots(1)$
$\angle\text{APQ}=\angle\text{CQF}$ (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 $\triangle\text{OPQ}$
$\angle\text{O}+\angle\text{OPQ}+\angle\text{PQO}=180^\circ$
$\Rightarrow\text{x}^\circ+40^\circ+85^\circ=180^\circ$
$\text{x}^\circ=180^\circ-85^\circ-40^\circ=55^\circ$
$\Rightarrow\text{x}^\circ=55^\circ,\text{y}=40^\circ$
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MCQ 21 Mark
In Fig. if $l_1$ || $l_2$, the value of $x$ is:
  • A
    $22\frac{1}{2}$
  • B
    $30$
  • ✓
    $45$
  • D
    $60$
Answer
Correct option: C.
$45$

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 $\triangle\text{ABC}$
$\text{a}^\circ+\text{b}^\circ+\angle\text{ABC}=180^\circ$
$\angle\text{ABC}=180^\circ-2\text{x}^\circ$
$\Rightarrow\text{a}^\circ+\text{b}^\circ+180^\circ-2\text{x}^\circ=180^\circ$
$\Rightarrow2\text{x}^\circ=\text{a}^\circ+\text{b}^\circ=90^\circ$
$\Rightarrow\text{x}^\circ=45^\circ$
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MCQ 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 $\angle\text{BOC}=$
  • A
    $60^\circ$
  • ✓
    $120^\circ$
  • C
    $150^\circ$
  • D
    $30^\circ$
Answer
Correct option: B.
$120^\circ$

$O$ 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 $(2)$
$20^\circ+40^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=180^\circ-60^\circ=120^\circ$
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MCQ 41 Mark
In Fig. $x + y =$
  • A
    $270^\circ$
  • ✓
    $230^\circ$
  • C
    $210^\circ$
  • D
    $190^\circ$
Answer
Correct option: B.
$230^\circ$
$\triangle\text{ACO}$
$\angle\text{ACO}+\angle\text{COA}+\angle\text{COA}=180^\circ$
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, $\text{y}^\circ=180^\circ-\angle\text{OBD}=180^\circ-70^\circ=110^\circ$
$\Rightarrow\text{x}^\circ+\text{y}^\circ=120^\circ+110^\circ=230^\circ$
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MCQ 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 $\angle\text{ACB}=$
  • A
    $35^\circ$
  • B
    $90^\circ$
  • ✓
    $70^\circ$
  • D
    $55^\circ$
Answer
Correct option: C.
$70^\circ$

$AX$ 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 $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow40^\circ+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow40^\circ+\angle\text{C}+\angle\text{C}=180^\circ\dots(\angle\text{B}=\angle\text{C})$
$\Rightarrow2\angle\text{C}=140^\circ$
$\Rightarrow\angle\text{C}=70^\circ$
$\text{i.e}\angle\text{ACB}=70^\circ$
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MCQ 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 $\angle\text{ALC}=$
  • A
    $85^\circ$
  • ✓
    $72\frac{1}{2}^\circ$
  • C
    $145^\circ$
  • D
    $\text{None of these}$
Answer
Correct option: B.
$72\frac{1}{2}^\circ$

$\angle\text{C}=180^\circ-\angle\text{ACD}=180^\circ-115^\circ=65^\circ$
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 $\angle\text{ALC}$
$\angle\text{ALC}+\angle\text{LAC}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{ALC}=180^\circ-\angle\text{LAC}-\angle\text{C}$
$=180^\circ-\frac{\angle\text{A}}{2}-\angle\text{C}$
$=180^\circ-\frac{85^\circ}{2}-65^\circ$
$=\frac{145^\circ}{2}$
$=​​72\frac{1}{2}^\circ$
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MCQ 71 Mark
An exterior angle of a triangle is $108^\circ$ and its interior opposite angles are in the ratio $4 : 5$ The angles of the triangle are:
  • ✓
    $48^\circ , 60^\circ , 72^\circ$
  • B
    $50^\circ , 60^\circ , 70^\circ$
  • C
    $52^\circ , 56^\circ , 72^\circ$
  • D
    $42^\circ , 60^\circ , 76^\circ $
Answer
Correct option: A.
$48^\circ , 60^\circ , 72^\circ$

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, $\angle\text{A}=48^\circ,\angle\text{B}=60^\circ$
$\Rightarrow\angle\text{C}=180^\circ-48^\circ-60^\circ=72^\circ$
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MCQ 81 Mark
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:
  • A
    An isosceles triangle
  • B
    An obtuse triangle
  • C
    An equilateral triangle
  • ✓
    A right triangle
Answer
Correct option: D.
A right triangle
Let the three angles of a triangle be $A, B$ and $C$.
Now, $A + B + C = 180^\circ$
If $A = B + C$
Then $A + (A) = 180^\circ$
i.e. $2A = 180^\circ $
i.e. $A = 90^\circ$
Since, one of the angle is $90^\circ$, the triangle is a Right triangle.
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MCQ 91 Mark
In Fig the value of $x$ is:
  • A
    $65^\circ$
  • B
    $80^\circ$
  • C
    $95^\circ$
  • ✓
    $120^\circ$
Answer
Correct option: D.
$120^\circ$


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, $\text{x}=180^\circ-\angle\text{E}=180^\circ-60^\circ$
$\Rightarrow\text{x}=120^\circ$
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MCQ 101 Mark
In Fig. if $BP || CQ$ and $AC = BC$, then the measure of $x$ is:
  • A
    $20^\circ$
  • B
    $25^\circ$
  • ✓
    $30^\circ$
  • D
    $35^\circ$
Answer
Correct option: C.
$30^\circ$

$\angle\text{PBC}=\angle\text{QCD}$ (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 $(1)$
$50^\circ+50^\circ+180^\circ-(70^\circ+\text{x})=180^\circ$
$\Rightarrow\text{x}=30^\circ$
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MCQ 111 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?
  • A
    $25^\circ$
  • B
    $30^\circ$
  • ✓
    $45^\circ$
  • D
    $60^\circ$
Answer
Correct option: C.
$45^\circ$
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 $= 180^\circ$
$\Rightarrow 3x + 4x + 5x = 180^\circ$
$\Rightarrow 12x = 180^\circ $
$\Rightarrow x = 15^\circ $
So, smallest angle $= 3 \times 15^\circ = 45^\circ$
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MCQ 121 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 $\angle\text{E}=$
  • ✓
    $25^\circ$
  • B
    $50^\circ$
  • C
    $100^\circ$
  • D
    $75^\circ$
Answer
Correct option: A.
$25^\circ$

$BE$ 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 $(1)$]
$\Rightarrow\angle\text{E}=25^\circ$
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MCQ 131 Mark
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is:
  • A
    $90^\circ$
  • B
    $180^\circ$
  • C
    $270^\circ$
  • ✓
    $360^\circ$
Answer
Correct option: D.
$360^\circ$

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}$
$=540^\circ-(\angle\text{A}+\angle\text{B}+\angle\text{C})$
$=540^\circ-180^\circ$
$\Rightarrow $ Sum of all exterior angles $= 360^\circ $
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MCQ 141 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 $\angle\text{BOC}=$
  • A
    $90^\circ+\frac{\text{x}^\circ}{2}$
  • ✓
    $90^\circ-\frac{\text{x}^\circ}{2}$
  • C
    $180^\circ+\frac{\text{x}^\circ}{2}$
  • D
    $180^\circ-\frac{\text{x}^\circ}{2}$
Answer
Correct option: B.
$90^\circ-\frac{\text{x}^\circ}{2}$

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 $(1)$]
$=\frac{180^\circ-\text{x}^\circ}{2}$
$\Rightarrow\angle\text{BOC}=90^\circ-\frac{\text{x}^\circ}{2}$
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MCQ 151 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:
  • A
    $20^\circ$
  • ✓
    $50^\circ$
  • C
    $60^\circ$
  • D
    $70^\circ$
Answer
Correct option: B.
$50^\circ$
$EC || AB$ 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 $\angle\text{OBD}=180^\circ-\angle\text{BOD}-\angle\text{ODB}$
$=180^\circ-110^\circ-20^\circ$
$=50^\circ$
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MCQ 161 Mark
In Fig. what is the value of $x$?
  • A
    $35^\circ$
  • B
    $45^\circ$
  • C
    $50^\circ$
  • ✓
    $60^\circ$
Answer
Correct option: D.
$60^\circ$
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 $(1)$, $\text{x}^\circ=180^\circ-8\text{y}^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-8\times15^\circ$
$\Rightarrow\text{x}^\circ=60^\circ$
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MCQ 171 Mark
In $\triangle\text{ABC},$ if $\angle\text{A}=100^\circ,\text{AD}$ bisects $\angle\text{A}$ and $\text{AD}\perp\text{BC}.$ Then, $\angle\text{B}=$
  • A
    $50^\circ$
  • B
    $90^\circ$
  • ✓
    $40^\circ$
  • D
    $100^\circ$
Answer
Correct option: C.
$40^\circ$

$\text{AD}\perp\text{BC}$ 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 $= 180^\circ$
$\Rightarrow\angle\text{BAD}+\angle\text{ADB}+\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=180^\circ-50^\circ-90^\circ$
$\Rightarrow\angle\text{B}=40^\circ$
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MCQ 181 Mark
In a triangle, an exterior angle at a vertex is $95^\circ$ and its one of the interior opposite angle is $55^\circ$ , then the measure of the other interior angle is:
  • A
    $55^\circ$
  • B
    $85^\circ$
  • ✓
    $40^\circ$
  • D
    $9.0^\circ$
Answer
Correct option: C.
$40^\circ$
Let the other interior opposite angle be $x^\circ$.
Then, we have
$x^\circ + 55^\circ = 95^\circ $
$\Rightarrow x^\circ = 95^\circ - 55^\circ = 40^\circ $
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MCQ 191 Mark
In $\triangle\text{RST}$ what is the value of $x$?
  • A
    $40^\circ$
  • B
    $90^\circ$
  • C
    $80^\circ$
  • ✓
    $100^\circ$
Answer
Correct option: D.
$100^\circ$

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 $(2)$
$\text{x}^\circ=180^\circ-2(40^\circ)$
$\Rightarrow\text{x}=100^\circ$
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MCQ 201 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:
  • A
    $45^\circ$
  • B
    $95^\circ$
  • ✓
    $135^\circ$
  • D
    $90^\circ$
Answer
Correct option: C.
$135^\circ$

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 $(1)$}
$=100^\circ-45^\circ$
$=135^\circ$
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MCQ 211 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 $l_1$ || $l_2$
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MCQ 221 Mark
An exterior angle of a triangle is equal to $100^\circ$ and two interrior opposite angles are equal. Each of these angles is equal to:
  • A
    $75^\circ$
  • B
    $80^\circ$
  • C
    $40^\circ$
  • ✓
    $50^\circ$
Answer
Correct option: D.
$50^\circ$
Let the two interior opposite angles be $x^\circ $ each.
Now, the exterior angle is equal to the sum of the two interior opposite angles.
$x^\circ + x^\circ = 180^\circ $
$\Rightarrow 2x^\circ = 100^\circ $
$\Rightarrow x^\circ = 50^\circ $
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MCQ 231 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 :
  • ✓
    $80^\circ$
  • B
    $75^\circ$
  • C
    $60^\circ$
  • D
    $90^\circ$
Answer
Correct option: A.
$80^\circ$
$\angle\text{B}=\frac{1}{2}\angle\text{A}$
$\angle\text{ACD}$ is an exterior angle.
$\Rightarrow\angle\text{A}+\angle\text{B}=\angle\text{ACD}$
$\Rightarrow\angle\text{A}=\frac{1}{2}\angle\text{A}=120^\circ$
$\Rightarrow\frac{3\angle\text{A}}{2}=120^\circ$
$\Rightarrow3\angle\text{A}=240^\circ$
$\Rightarrow\angle\text{A}=80^\circ$
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MCQ 241 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 $\angle\text{BOD}=$
  • A
    $100^\circ$
  • ✓
    $80^\circ$
  • C
    $90^\circ$
  • D
    $135^\circ$
Answer
Correct option: B.
$80^\circ$

$AC || BD$
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 $\angle\text{BOD}=180^\circ-45^\circ-55^\circ$
$=80^\circ$
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MCQ 251 Mark
In Fig. what is $y$ in terms of $x$?
  • ✓
    $\frac{3}{2}\text{x}^\circ$
  • B
    $\frac{4}{3}\text{x}^\circ$
  • C
    $\text{x}^\circ$
  • D
    $\frac{3}{4}\text{x}^\circ$
Answer
Correct option: A.
$\frac{3}{2}\text{x}^\circ$

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 $\triangle\text{OCD}$
$\angle\text{DOC}+\angle\text{OCD}+\angle\text{D}=180^\circ$
$180^\circ-3\text{y}^\circ+3\text{x}^\circ+\text{y}^\circ=180^\circ$ $\big[\angle\text{OCD}=\angle\text{ACD}\big]$
$\Rightarrow2\text{y}^\circ=3\text{x}^\circ$
$\Rightarrow\text{y}=\frac{3}{2}\text{x}^\circ$
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MCQ 261 Mark
If all the three angles of a triangle are equal, then each one of them is equal to:
  • A
    $90^\circ$
  • B
    $45^\circ$
  • ✓
    $60^\circ$
  • D
    $30^\circ$
Answer
Correct option: C.
$60^\circ$
Let the measure of each angle be $x^\circ $.
Now, the sum of all angles of any triangle is $180^\circ $.
Thus, $x^\circ + x^\circ + x^\circ = 180^\circ $
i.e. $3x^\circ = 180^\circ $
i.e. $x^\circ = 60^\circ$
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MCQ 271 Mark
The base $BC$ of triangle $ABC$ is produced both ways and the measure of exterior angles formed are $94^\circ$ and $126^\circ$. Then, $\angle\text{BAC}=$
  • A
    $94^\circ$
  • B
    $54^\circ$
  • ✓
    $40^\circ$
  • D
    $44^\circ$
Answer
Correct option: C.
$40^\circ$

$\angle\text{ABC}=180^\circ-126^\circ=54^\circ$
$\angle\text{ACB}=180^\circ-94^\circ=86^\circ$
Now, in $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{BAC}=180^\circ-\angle\text{ABC}-\angle\text{ACB}$
$=180^\circ-54^\circ-86^\circ$
$\Rightarrow\text{BAC}=40^\circ$
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MCQ 281 Mark
In Fig. what is $z$ in terms of $x$ and $y$?
  • A
    $x + y + 180^\circ $
  • ✓
    $x + y - 180^\circ$
  • C
    $180^\circ - (x + y)$
  • D
    $x + y + 360^\circ$
Answer
Correct option: B.
$x + y - 180^\circ$
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 $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\text{z}^\circ+180^\circ-\text{y}^\circ+180^\circ-\text{x}^\circ=180^\circ$
$\Rightarrow\text{z}^\circ=\text{x}^\circ+\text{y}^\circ-180^\circ$
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MCQ 291 Mark
If two acute angles of a right triangle are equal, then each acute is equal to:
  • A
    $30^\circ $
  • ✓
    $45^\circ$
  • C
    $60^\circ$
  • D
    $90^\circ$
Answer
Correct option: B.
$45^\circ$
Let the measure of each acute angle of a triangle be $x^\circ $.
Then, we have
$x^\circ + x^\circ + 90^\circ = 180^\circ $
i.e. $2x^\circ = 90^\circ$
i.e. $x^\circ = 45^\circ $
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MCQ 301 Mark
In Fig. if $\text{AB}\perp\text{BC},$ then x =
  • A
    18º
  • ✓
    22º
  • C
    25º
  • D
    32º
Answer
Correct option: B.
22º
$\text{AB}\perp\text{BC}$
$\Rightarrow\angle\text{ABC}=90^\circ$
$\angle\text{CAB}=32^\circ$ (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
$\Rightarrow\angle\text{DAB}+\angle\text{ABD}+\angle\text{BDA}=180^\circ$
$\Rightarrow\text{x}^\circ+32^\circ+90^\circ+\text{x}^\circ+14^\circ=180^\circ$
$\Rightarrow2\text{x}^\circ=180^\circ-136^\circ$
$\Rightarrow2\text{x}^\circ=44^\circ$
$\Rightarrow\text{x}^\circ=22^\circ$
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MCQ 311 Mark
In Fig , if $A C=B D$ and $\angle C A B=\angle D B A$, then $\angle A C B=$
Image
  • A
    $\angle B A D$
  • B
    $\angle A B C$
  • C
    $\angle A B D$
  • D
    $\angle B D A$
Answer
D. $\angle B D A$
In triangles $A B C$ and $A B D$, we have
$
A B=A B
$$\quad$[Common]
$\angle C A B=\angle D B A$$\quad$[Given]
and,$\quad$$
A C=B D
$$\quad$[Given]
So, by $S A S$ congruence criterion,
we obtain$
\triangle A B C \cong \triangle B A D \Rightarrow \angle A C B=\angle B D A
$
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MCQ 321 Mark
In Fig , $A B C D$ is a quadrilateral in which $A D=C B$ and $A B=C D$, then $\angle A C B$ is equal to
Image
  • A
    $\angle A C D$
  • B
    $\angle B A C$
  • C
    $\angle C A D$
  • D
    $\angle B A D$
Answer
C. $\angle C A D$
solution In triangles $A C D$ and $C A B$, we have
$
A D=C B
$$\quad$[Given]
$
A B=C D
$$\quad$$\quad$[Given]
and,$\quad$$
A C=A C
$$\quad$[Common]
So, by $S S S$ congruence criterion, we obtain$
\triangle A C D \cong \triangle C A B \Rightarrow \angle A C B=\angle C A D
$
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MCQ 331 Mark
In Fig , ABCD is a quadrilateral in which $B N$ and DM are perpendiculars drawn to $A C$ such that $B N=D M$. If $O B=4 cm$, then $B D=$
Image
  • A
    6 cm
  • B
    8 cm
  • C
    10 cm
  • D
    12 cm
Answer
B. 8 cm
 In triangles $O N B$ and $O M D$, we have
$
\angle O N B=\angle O M D\quad$[Each equal to $90^{\circ}$ ]
$\angle B O N=\angle D O M \quad$ [Vertically opposite angles]
and,$\quad$$ B N=D M$
So, by using $A A S$ congruence criterion, we obtain$
\begin{array}{ll}
& \triangle O N B \cong \triangle O M D \Rightarrow O B=O D \Rightarrow O D=O B=4 cm \\
\therefore & B D=O B+O D=4 cm+4 cm=8 cm
\end{array}
$
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MCQ 341 Mark
In Fig , if $\triangle A B C$ is an isosceles triangle with $A B=A C$ and $B D=C D$, then the congruence criterion by which $\triangle A D B \cong \triangle A D C$ is
  • A
    AAS
  • B
    RHS
  • C
    ASA
  • D
    SSS
Answer
B. RHS
In $\triangle A B C$, we have, $A B=A C$ and $D$ is the mid-point of $B C$.
Therefore, $A D \perp B C$. Thus, in $\triangle A B D$ and $\triangle A C D$, we have$
A B=A C, B D=C D \text { and } \angle A D B=\angle A D C=90^{\circ}
$
So, by RHS congruence criterion, we obtain $\triangle A D B \cong \triangle A D C$.
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MCQ 351 Mark
In Fig , $\triangle A B C$ is an isosceles triangle with $A B=A C$. If $A D$ is the bisector of $\angle A$, which one of the following options has words that correctly complete the following statement? By congruence criterion, $\triangle A B D \cong \triangle A C D$ and using c.p.c.t, we obtain $\angle A B C=\ldots \ldots$
  • A
    SAS; $\angle A C D$
  • B
    $SAS ; \angle A D C$
  • C
    $A S A ; \angle A C D$
  • D
    $A S A ; \angle A D C$
Answer
A. SAS; $\angle A C D$
In $\triangle^{\prime} s A B D$ and $A C D$, we have,
$
A B=A C, \angle B A D=\angle C A D \text { and } A D=A D
$
So, by SAS congruence criterion, we obtain$
\angle A B D \cong \triangle A C D \Rightarrow \angle A B D=\angle A C D \Rightarrow \angle A B C=\angle A C D
$
Hence, option (a) is correct.
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MCQ 361 Mark
In Fig. $A B C$ is an isosceles triangle with $A B=A C$ and $L M$ is parallel to $B C$. If $\angle A=50^{\circ}$, then $\angle L M C=$
  • A
    $65^{\circ}$
  • B
    $115^{\circ}$
  • C
    $130^{\circ}$
  • D
    $50^{\circ}$
Answer
B. $115^{\circ}$
In $\triangle A B C$, we have
Ab = AC $\Rightarrow$ $\angle C=\angle B$
$\therefore$ $\angle A+\angle B+\angle C=180^{\circ}$ $\Rightarrow$ $50^{\circ}+\angle B+\angle B=180^{\circ}$ $\Rightarrow$ $2 \angle B=130^{\circ}$ $\Rightarrow$ $\angle B=65^{\circ}$
Thus, we have $\angle B=\angle C=65^{\circ}$.
It is given that $L M \| B C$.
$\therefore$ $\angle A L M=\angle B$ and $\angle A M L=C$ $\Rightarrow$ $\angle A L M=65^{\circ}=\angle A M L$
$\therefore$ $\angle L M C=180^{\circ}-\angle A M L=180^{\circ}-65=115^{\circ}$
Image
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MCQ 371 Mark
In Fig , if $\triangle A B C \cong \triangle A D C, \angle B A C=30^{\circ}$ and $\angle A B C=100^{\circ}$, then $\angle A C D=$
Image
  • A
    $30^{\circ}$
  • B
    $80^{\circ}$
  • C
    $50^{\circ}$
  • D
    $70^{\circ}$
Answer
C. $50^{\circ}$
In $\triangle A B C$, we have
$
\begin{array}{ll}
& \angle B A C=30^{\circ} \text { and } \angle A B C=100^{\circ} \\
\therefore & \angle B A C+\angle A B C+\angle A C B=180^{\circ} \\
\Rightarrow & 30^{\circ}+100^{\circ}+\angle A C B=180^{\circ} \Rightarrow \angle A C B=50^{\circ} \\
\text { Now, } & \triangle A B C \cong \triangle A D C \Rightarrow \angle A C D=\angle A C B \Rightarrow \angle A C D=50^{\circ}
\end{array}
$
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MCQ 381 Mark
In Fig , if $A B=A C$ and $B D=C D$, then $\angle A B D: \angle A C D=$
Image
  • A
    $1: 1$
  • B
    $1: 2$
  • C
    $2: 1$
  • D
    $2: 3$
Answer
A. $1: 1$
  In triangles $A B C$ and $D B C$, we have
$
\begin{array}{ll}
& A B=A C \text { and } D B=D C \\
\Rightarrow & \angle C=\angle B \text { and } \angle D C B=\angle D B C \\
\Rightarrow & \angle C-\angle D C B=\angle B-\angle D B C \\
\Rightarrow & \angle A B D=\angle A C D \Rightarrow \angle A B D: \angle A C D=1: 1
\end{array}
$
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MCQ 391 Mark
In Fig , if $A B=F C, E F=B D$ and $\angle A F E=\angle C B D$. Then, the rule by which $\triangle A F E \cong \triangle C B D$ is
Image
  • A
    SAS
  • B
    ASA
  • C
    SSS
  • D
    AAS
Answer
A. SAS
We have,
$
A B=F C \Rightarrow A B+B F=B F+F C \Rightarrow A F=B C
$
Thus, in triangles $A F E$ and $C B D$, we obtain
$
\begin{aligned}
A F & =B C \\
\angle A F E & =\angle C B D \\
and, F E & =B D
\end{aligned}
$
So, by $S A S$ congruence criterion, we obtain $\triangle A F E \cong \triangle C B D$.
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MCQ 401 Mark
In triangles $A B C$ and $P Q R$, if $A B=A C, \angle C=\angle P$ and $\angle B=\angle Q$. Then, the two triangles
  • A
    isosceles but not cogruent
  • B
    isosceles and congruent
  • C
    congruent but not isosceles
  • D
    neither congruent nor isosceles
Answer
A. isosceles but not cogruent
 In $\triangle A B C$,
we have$
A B=A C \Rightarrow \angle C=\angle B \Rightarrow \angle P=\angle Q \Rightarrow Q R=P R
$
Thus, $\triangle P Q R$ is isosceles.
$\triangle A B C \cong \triangle R Q P$ only if $B C=Q P$, otherwise not.
Hence, two triangles are isosceles but not congruent.
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MCQ 411 Mark
In a $\triangle P Q R$, if $\angle P=\angle R, P R=5 cm$ and $Q R=4 cm$, then $P Q=$
  • A
    4 cm
  • B
    5 cm
  • C
    2 cm
  • D
    2.5 cm
Answer
A. 4 cm
 In $\triangle P Q R$, it is given that
$
\angle R=\angle P \Rightarrow P Q=Q R \Rightarrow P Q=4 cm
$[Sides opposite to equal angles]
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MCQ 421 Mark
In a $\triangle A B C, B C=A B$ and $\angle B=80^{\circ}$, then $\angle A=$
  • A
    $80^{\circ}$
  • B
    $50^{\circ}$
  • C
    $40^{\circ}$
  • D
    $100^{\circ}$
Answer
B. $50^{\circ}$
We have,
$B C=A B$ and $\angle B=80^{\circ}$$
\begin{array}{ll}
\therefore & \angle A=\angle C \text { and } \angle B=80^{\circ} \\
\text { Now } & \angle A+\angle B+\angle C=180^{\circ} \Rightarrow \angle A+80^{\circ}+\angle A=180^{\circ} \Rightarrow 2 \angle A=100 \Rightarrow \angle A=50^{\circ}
\end{array}
$
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MCQ 431 Mark
In a $\triangle A B C$, if $A B=A C$ and $\angle B=50^{\circ}$, then $\angle A=$
  • A
    $40^{\circ}$
  • B
    $50^{\circ}$
  • C
    $80^{\circ}$
  • D
    $130^{\circ}$
Answer
C. $80^{\circ}$ 
We have,
$
A B=A C \Rightarrow \angle C=\angle B \Rightarrow \angle C=50^{\circ} \quad \text { [Angles opposite to equal sides are equal] }
$
Thus, we obtain $\angle B=\angle C=50^{\circ}$.$
\therefore \quad \angle A+\angle B+\angle C=180^{\circ} \Rightarrow \angle A+50^{\circ}+50^{\circ}=180^{\circ} \Rightarrow \angle A=80^{\circ}
$
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MCQ 441 Mark
In Fig , by which criterion triangles $O A C$ and $O B D$ are congruent?
Image
  • A
    SAS
  • B
    ASA
  • C
    AAS
  • D
    SSS
Answer
A. SAS
 In triangles $O A C$ and $O B D$,
we find that$
O A=O B, \angle A O C=\angle B O D \text { and } O C=O D
$
So, by using $S A S$ congruence criterion, we obtain $\triangle O A C \cong \triangle O B D$.
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MCQ 451 Mark
In Fig , $P Q R S$ is a parallelogram. Can it be concluded that $\triangle R P S \cong \triangle Q S P$ ? Why or why not?
  • A
    Yes, because $P S=P S, \angle S P R=\angle P S Q, \angle R S P=\angle S P Q$
  • B
    Yes becouse $P Q=R S, P S=Q R, \angle S P R=\angle R P Q, \angle S R P=\angle P R Q, \angle P S R=\angle P Q R$ and $P R=P R$
  • C
    No, because angle measures are not equal
  • D
    No, because side lengths are not equal.
Answer
A. Yes, because $P S=P S, \angle S P R=\angle P S Q, \angle R S P=\angle S P Q$
Image
In triangles RPS and QSP, we find that
$P S=P S, \angle S P R=\angle P S Q, \angle R S P=\angle S P Q$ $[\because P Q \| R S$ and $P S$ is a transversal $]$
So, by $S A S$ congruence criterion, we obtain $\triangle R P S \cong \triangle Q S P$.
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MCQ 461 Mark
In triangles $A B C$ and $P Q R, A B=P Q$ and $\angle B=\angle Q$. The two triangles are congruent by SAS criterion, if
  • A
    AC = PR
  • B
    BC = PQ
  • C
    AC = QR
  • D
    BC = QR
Answer
D. BC = QR
In triangles $A B C$ and $P Q R$, we have $A B=P Q$ and $\angle B=\angle Q$. Therefore, two triangles will be congruent by $S A S$ criterion, if $B C=Q R$.
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MCQ 471 Mark
By which congruence criterion the following triangles are congruent?
Image
  • A
    SAS
  • B
    ASS
  • C
    AAS
  • D
    SSS
Answer
D. SSS
In triangles $A B C$ and $D E F$, we find that $A B=D E, B C=E F$ and $A C=D F$.
Hence by SSS congruence criterion, we obtain $\triangle A B C \cong \triangle D E F$.
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MCQ 481 Mark
If $\triangle P Q R \cong \triangle A C B$, then $A B$ is equal to
  • A
    QR
  • B
    PR
  • C
    PQ
  • D
    none of these
Answer
B. PR
We have,
$\triangle P Q R \cong \triangle A C B$$
\Rightarrow \quad P Q=A C, Q R=C B \text { and } P R=A B
$
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Question 491 Mark
Answer
B. In option (a), we find that in triangles $A B C$ and $D E F$, we have
$A B=D F, A C=D E$ and $B C=E F$
So, by using $S S S$ criterion of congruence, we obtain $\triangle A B C \cong \triangle D F E$
In option (c), we find that in triangles $A B C$ and $D E F$, we have
$\angle A=\angle D, \angle B=\angle F$ and $B C=F E$
So, by using $A A S$ criterion of congruence, we obtain $\triangle A B C \cong \triangle D F E$.
In option (d), we observe that $\angle A=\angle D, \angle B=\angle F$ and $A B=D F$. So, by $S A S$ congruence criterion, we obtain $\triangle A B C \cong \triangle D F E$.
In option (b) we observe that $\angle A=\angle D, \angle B=\angle E$ and $\angle C=\angle F$. But, $\triangle A B C$ need not be congruent to $\triangle D E F$ as the corresponding sides of two triangles may not be same.
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MCQ 501 Mark
In Fig , two triangles $A B C$ and $P Q R$ shown. Which congruence criterion can be used to show that the triangles are congruent?

Image
  • A
    SAS
  • B
    SSS
  • C
    ASA
  • D
    AAA
Answer
C. ASA
In triangles $A B C$ and $P Q R$, we find that
$\Rightarrow \quad \angle A=\angle R, A C=R Q$ and $\angle C=\angle Q$
Therefore, by ASA congruence criterion, we obtain $\triangle A B C \cong \triangle R P Q$
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MCQ 511 Mark
In two triangles $A B C$ and $P Q R$, if $A B=Q R, B C=R P$ and $C A=P Q$, then
  • A
    $\triangle A B C \cong \triangle P Q R$
  • B
    $\triangle C B A \cong \triangle P R Q$
  • C
    $\triangle B A C \cong \triangle R P Q$
  • D
    $\triangle P Q R \cong \triangle B C$
Answer
B. $\triangle C B A \cong \triangle P R Q$
We have, $A B=Q R, B C=R P$ and $C A=P Q$
$\Rightarrow \quad A \leftrightarrow Q, B \leftrightarrow R$ and $C \leftrightarrow P$
$\Rightarrow \quad \triangle A B C \cong \triangle Q R P, \triangle C B A \cong \triangle P R Q, \triangle B A C \cong \triangle R Q P$ and $\triangle B C A \cong \triangle R P Q$
Clearly, option (b) is correct.
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MCQ 521 Mark
Which of the following is not a criterion for congruence of triangles?
  • A
    SAS
  • B
    ASA
  • C
    SSA
  • D
    SSS
Answer
C. SSA
 We have, $S S S, S A S, A S A, A A S$ and RHS as criteria for congruence of triangles. Hence SSA is not a criterion for congruence of triangles.
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MCQ 531 Mark
In Fig , $a+b=$
Image
  • A
    $117^{\circ}$
  • B
    $130^{\circ}$
  • C
    $127^{\circ}$
  • D
    $158^{\circ}$
Answer
C. $127^{\circ}$
 Since $A R B$ is a straight line.
$
\begin{array}{ll}
\therefore & \frac{x}{2}+5\left(\frac{x}{2}-1^{\circ}\right)+x+90^{\circ}=180^{\circ} \\
\Rightarrow & 4 x+4^{\circ}=180^{\circ} \Rightarrow 4 x=176^{\circ} \Rightarrow x=44^{\circ}
\end{array}
$
Using exterior angle property in $P Q R$, we obtain$
\begin{aligned}
& \angle Q R C=a+b \\
\Rightarrow \quad & \frac{x}{2}+5\left(\frac{x}{2}-1^{\circ}\right)=a+b \Rightarrow a+b=3 x-5^{\circ} \Rightarrow a+b=3 \times 44^{\circ}-5=127^{\circ}
\end{aligned}
$
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MCQ 541 Mark
In Fig , if $P T$ is the bisector of $\angle Q P R$ in $\triangle P Q R, \angle P Q R=50^{\circ}, \angle P R Q=30^{\circ}$ and $P S \perp Q R$, then $x=$
Image
  • A
    $40^{\circ}$
  • B
    $20^{\circ}$
  • C
    $30^{\circ}$
  • D
    $10^{\circ}$
Answer
D. $10^{\circ}$
Using angle sum property in $\triangle P Q R$, we obtain$
\begin{array}{ll}
& \angle P Q R+\angle P R Q+\angle R P Q=180^{\circ} \\
\Rightarrow & 50^{\circ}+30^{\circ}+\angle R P Q=180^{\circ} \Rightarrow \angle R P Q=100^{\circ} \\
\therefore \quad & \angle Q P T=\frac{1}{2} \angle R P Q=50^{\circ} \quad[\because P T \text { is bisector of } \angle Q P R]
\end{array}
$
Using exterior angle property in $\triangle P Q S$, we obtain$
\begin{array}{ll}
& \angle P S T=\angle P Q S+\angle Q P S \Rightarrow 90^{\circ}=50^{\circ}+\angle Q P S \Rightarrow \angle Q P S=40^{\circ} \\
\therefore & x=\angle Q P T-\angle Q P S=50^{\circ}-40^{\circ}=10^{\circ}
\end{array}
$
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MCQ 551 Mark
In Fig , if the sides $B C, C A$ and $A B$ of $\triangle A B C$ have been produced to $D, E$ and $f$ respectively, then $\angle B A E+\angle C B F+\angle A C D=$
Image
  • A
    $180^{\circ}$
  • B
    $360^{\circ}$
  • C
    $240^{\circ}$
  • D
    $300^{\circ}$
Answer
B. $360^{\circ}$
Using exterior angle property in $\triangle A B C$, we obtain
$\angle A C D=\angle A+\angle B, \angle C B F=\angle A+\angle C$ and, $\angle B A E=\angle B+\angle C$
$\therefore$ $\angle B A E+\angle C B F+\angle A C D=(\angle B+\angle C)+(\angle A+\angle C)+(\angle A+\angle B)$
$=2(\angle A+\angle B+\angle C)=2 \times 180^{\circ}=360^{\circ}$
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MCQ 561 Mark
In a $\triangle A B C$, it is given that $\angle A: \angle B: C=3: 2: 1$ and $\angle A C D=90^{\circ}$. If $B C$ is produced to $E$, then $\angle E C D=$
  • A
    $60^{\circ}$
  • B
    $30^{\circ}$
  • C
    $50^{\circ}$
  • D
    $40^{\circ}$
Answer
A. $60^{\circ}$
Image
We have, $\angle A: \angle B: \angle C=3: 2: 1$
So, let $\angle A=3 x, \angle B=2 x$ and $\angle C=x$.
$\therefore$ $\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow$ $3 x+2 x+x=180^{\circ}$ $\Rightarrow$ $6 x=180^{\circ}$ $\Rightarrow$ $x=30^{\circ}$
$\therefore$ $\angle A=90^{\circ}, \angle B=60^{\circ}$ and $\angle C=30^{\circ}$
Using exterior angle property in $\triangle A C E$, we obtain
$\angle A C E=\angle A+\angle B$ $\Rightarrow$ $\angle A C D+\angle E C D=90^{\circ}+60^{\circ}$$\Rightarrow$$90^{\circ}+\angle E C D=150^{\circ}$$\Rightarrow$$\angle E C D=60^{\circ}$
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MCQ 571 Mark
In Fig , sides $C B$ and $B A$ of $\triangle A B C$ are produced to $D$ and $E$ respectively. If $\angle A B D=105^{\circ}$ and $\angle C A E=130^{\circ}$, then $\angle A C B=$
Image
  • A
    $50^{\circ}$
  • B
    $55^{\circ}$
  • C
    $75^{\circ}$
  • D
    $130^{\circ}$
Answer
B. $55^{\circ}$
We have, $\angle C A E=130^{\circ}$$
\therefore \quad \angle B A C=180^{\circ}-130^{\circ}=50^{\circ} \quad\left[\because \angle B A C+\angle C A E=180^{\circ}\right]
$
Using exterior angle property in $\triangle A B C$, we obtain$
\begin{array}{ll}
& \angle A B D=\angle B A C+\angle A C B \\
\Rightarrow \quad & 105^{\circ}=50^{\circ}+\angle A C B \Rightarrow \angle A C B=105^{\circ}-50^{\circ}=55^{\circ}
\end{array}
$
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MCQ 581 Mark
In Fig , $\angle A C D=120^{\circ}$ and $\angle A B C=40^{\circ}$, then $\angle B A C=$
Image
  • A
    $80^{\circ}$
  • B
    $60^{\circ}$
  • C
    $50^{\circ}$
  • D
    $40^{\circ}$
Answer
A. $80^{\circ}$
 In $\triangle A B C$, side $B C$ is produced to $D$. Using exterior angle property, we obtain$
\begin{array}{ll}
& \angle A C D=\angle A B C+\angle B A C \\
\Rightarrow \quad & 120^{\circ}=40^{\circ}+\angle B A C \Rightarrow \angle B A C=80^{\circ}
\end{array}
$
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MCQ 591 Mark
In Fig , $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=$
Image
  • A
    $180^{\circ}$
  • B
    $360^{\circ}$
  • C
    $540^{\circ}$
  • D
    $90^{\circ}$
Answer
B. $360^{\circ}$
Using angle sum property in $\triangle^{\prime} s A B C$ and $D E F$, we obtain$
\begin{array}{ll}
& \angle A+\angle B+\angle C=180^{\circ} \text { and } \angle D+\angle E+\angle F=180^{\circ} \\
\Rightarrow \quad & \angle A+\angle B+\angle C+\angle D+\angle E+\angle F=180^{\circ}+180^{\circ}=360^{\circ}
\end{array}
$
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MCQ 601 Mark
In Fig , $B C \| P Q, B P$ and $C Q$ intersect at $O$. If $x+y=80^{\circ}$ and $x-y=55^{\circ}$, then $z=$
Image
  • A
    $80^{\circ}$
  • B
    $55^{\circ}$
  • C
    $90^{\circ}$
  • D
    $100^{\circ}$
Answer
D. $100^{\circ}$
It is given that $B C \| P Q$ and transversal $B P$ cuts them at $B$ and $P$ respectively.$
\begin{array}{ll}
\therefore & \angle C B P=\angle B P Q \\
\Rightarrow & \angle B P Q=x
\end{array}
$
Using angle sum property in $\triangle O P Q$, we obtain$
\angle P+\angle Q+\angle P O Q=180^{\circ} \Rightarrow x+y+z=180^{\circ} \Rightarrow 80^{\circ}+z=180^{\circ} \Rightarrow z=100^{\circ} \quad\left[\because x+y=80^{\circ} \text { (given) }\right]
$
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MCQ 611 Mark
In a $\triangle A B C$, if $\angle A=\angle B+\angle C$, then $\triangle A B C$ is
  • A
    isosceles triangle
  • B
    equilateral triangle
  • C
    right triangle
  • D
    none of these
Answer
C. right triangle
We have, \[ \angle A=\angle B+\angle C \Rightarrow \angle A+\angle A=\angle A+\angle B+\angle C \Rightarrow 2 \angle A=180^{\circ} \Rightarrow \angle A=90^{\circ} \]
 Hence, $\triangle A B C$ is a right triangle.
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MCQ 621 Mark
In Fig. ABC is a triangle in which $\angle B=2 \angle C$. D is a point on side BC such that AD bisects $\angle B A C$ and $A B=C D$. BE is the bisector of $\angle B$. The measure of $\angle B A C$ is
Image
  • ✓
    $72^{\circ}$
  • B
    $73^{\circ}$
  • C
    $74^{\circ}$
  • D
    $95^{\circ}$
Answer
Correct option: A.
$72^{\circ}$
(a) $72^{\circ}$
[Hint: $\triangle A B E \cong \triangle B C E$ ]
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MCQ 631 Mark
In fig. ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, $\angle B A D=$
Image
  • ✓
    $55^{\circ}$
  • B
    $70^{\circ}$
  • C
    $35^{\circ}$
  • D
    $110^{\circ}$
Answer
Correct option: A.
$55^{\circ}$
a
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MCQ 641 Mark
$D, E, F$ are the mid-point of the sides $B C, C A$ and $A B$ respectively of $\triangle A B C$. Then $\triangle D E F$ is congruent to triangle
  • A
    ABC
  • B
    AEF
  • C
    BFD, CDE
  • ✓
    AFE, BFD, CDE
Answer
Correct option: D.
AFE, BFD, CDE
d
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MCQ 651 Mark
In Fig. if AC is bisector of $\angle B A D$ such that $A B=3 cm$ and $A C=5 cm$, then CD =
Image
  • A
    2 cm
  • B
    3 cm
  • ✓
    4 cm
  • D
    5 cm
Answer
Correct option: C.
4 cm
c
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MCQ 661 Mark
In Fig. ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x, is
Image
  • A
    $52^{\circ}$
  • B
    $76^{\circ}$
  • C
    $156^{\circ}$
  • ✓
    $104^{\circ}$
Answer
Correct option: D.
$104^{\circ}$
d
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MCQ 671 Mark
In Fig. if $A E \| D C$ and $A B=A C$, the value of $\angle A B D$ is
Image
  • A
    $70^{\circ}$
  • ✓
    $110^{\circ}$
  • C
    $120^{\circ}$
  • D
    $130^{\circ}$
Answer
Correct option: B.
$110^{\circ}$
b
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MCQ 681 Mark
In Fig. $A B \perp B E$ and $F E \perp B E$. If $B C=D E$ and $A B=E F$, then $\triangle A B D$ is congruent to
Image
  • A
    $\triangle E F C$
  • B
    $\triangle E C F$
  • C
    $\triangle C E F$
  • ✓
    $\triangle F E C$
Answer
Correct option: D.
$\triangle F E C$
d
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MCQ 691 Mark
In Fig. the measure of $\angle B^{\prime} A^{\prime} C^{\prime}$ is
Image
  • A
    $50^{\circ}$
  • ✓
    $60^{\circ}$
  • C
    $70^{\circ}$
  • D
    $80^{\circ}$
Answer
Correct option: B.
$60^{\circ}$
b
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MCQ 701 Mark
If $A B C$ and $D E F$ are two triangles such that $\triangle A B C \cong \triangle F D E$ and $A B=5 cm, \angle B=40^{\circ}$ and $\angle A=80^{\circ}$. Then, which of the following is true?
  • A
    $D F=5 cm, \angle F=60^{\circ}$
  • B
    $D E=5 cm, \angle E=60^{\circ}$
  • ✓
    $D F=5 cm, \angle E=60^{\circ}$
  • D
    $D E=5 cm, \angle D=40^{\circ}$
Answer
Correct option: C.
$D F=5 cm, \angle E=60^{\circ}$
c
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MCQ 711 Mark
Which of the following is not a criterion for congruence of triangles?
  • A
    SAS
  • ✓
    SSA
  • C
    ASA
  • D
    SSS
Answer
Correct option: B.
SSA
b
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MCQ 721 Mark
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is
  • A
    $100^{\circ}$
  • ✓
    $120^{\circ}$
  • C
    $110^{\circ}$
  • D
    $130^{\circ}$
Answer
Correct option: B.
$120^{\circ}$
b
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MCQ 731 Mark
In a $\triangle A B C$, if AB = AC and BC is produced to D such that $\angle A C D=100^{\circ}$ then $\angle A=$
  • ✓
    $20^{\circ}$
  • B
    $40^{\circ}$
  • C
    $60^{\circ}$
  • D
    $80^{\circ}$
Answer
Correct option: A.
$20^{\circ}$
a
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MCQ 741 Mark
If $\triangle P Q R \cong \triangle E F D$, then $\angle E=$
  • ✓
    $\angle P$
  • B
    $\angle Q$
  • C
    $\angle R$
  • D
    None of these
Answer
Correct option: A.
$\angle P$
a
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MCQ 751 Mark
If $\triangle P Q R \cong \triangle E F D$, then $E D=$
  • A
    PQ
  • B
    QR
  • ✓
    PR
  • D
    None of these
Answer
Correct option: C.
PR
c
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MCQ 761 Mark
In triangles ABC and PQR, if $\angle A=\angle R, \angle B=\angle P$ and $A B=R P$, then which one of the following congruence conditions applies:
  • A
    SAS
  • ✓
    ASA
  • C
    SSS
  • D
    RHS
Answer
Correct option: B.
ASA
b
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MCQ 771 Mark
In triangles ABC and PQR three equality relations between some parts are as follows:
$A B=Q P, \angle B=\angle P$ and $B C=P R$
State which of the congruence conditions applies:
  • ✓
    SAS
  • B
    ASA
  • C
    SSS
  • D
    RHS
Answer
Correct option: A.
SAS
a
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MCQ 781 Mark
If $\triangle A B C \cong \triangle P Q R$ and $\triangle A B C$ is not congruent to $\triangle R P Q$, then which of the following is not true:
  • ✓
    BC = PQ
  • B
    AC = PR
  • C
    AB = PQ
  • D
    QR = BC
Answer
Correct option: A.
BC = PQ
a
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MCQ 791 Mark
If $\triangle A B C \cong \triangle A C B$, then $\triangle A B C$ is isosceles with
  • ✓
    AB = AC
  • B
    AB = BC
  • C
    AC = BC
  • D
    None of these
Answer
Correct option: A.
AB = AC
a
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MCQ 801 Mark
If $\triangle A B C \cong \triangle L K M$, then side of $\triangle L K M$ equal to side AC of $\triangle A B C$ is
  • A
    LK
  • B
    KM
  • ✓
    LM
  • D
    None of these
Answer
Correct option: C.
LM
c
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MCQ 811 Mark
In $\triangle R S T$ (See Fig.) what is the value of x?
Image
  • A
    40
  • B
    $90^{\circ}$
  • C
    $80^{\circ}$
  • ✓
    100
Answer
Correct option: D.
100
d
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MCQ 831 Mark
The side BC of $\triangle A B C$ is produced to a point D. The bisector of $\angle A$ meets side BC in L. If $\angle A B C=30^{\circ}$ and $\angle A C D=115^{\circ}$, then $\angle A L C=$
  • A
    $85^{\circ}$
  • ✓
    $72 \frac{1}{2}^{\circ}$
  • C
    $145^{\circ}$
  • D
    none of these
Answer
Correct option: B.
$72 \frac{1}{2}^{\circ}$
b
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MCQ 841 Mark
In a $\triangle A B C, \angle A=50^{\circ}$ and BC is produced to a point D. If the bisectors of $\angle A B C$ and $\angle A C D$ meet at $E$, then $\angle E=$
  • ✓
    $25^{\circ}$
  • B
    $50^{\circ}$
  • C
    $100^{\circ}$
  • D
    $75^{\circ}$
Answer
Correct option: A.
$25^{\circ}$
a
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MCQ 851 Mark
The bisects of exterior angles at B and C of $\triangle A B C$ meet at O. If $\angle A=x^{\circ}$, then $\angle B O C=$
  • A
    $90^{\circ}+\frac{x^{\circ}}{2}$
  • ✓
    $90^{\circ}-\frac{x^{\circ}}{2}$
  • C
    $180^{\circ}+\frac{x^{\circ}}{2}$
  • D
    $180^{\circ}-\frac{x^{\circ}}{2}$
Answer
Correct option: B.
$90^{\circ}-\frac{x^{\circ}}{2}$
b
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MCQ 861 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
  • A
    $45^{\circ}$
  • B
    $95^{\circ}$
  • ✓
    $135^{\circ}$
  • D
    $90^{\circ}$
Answer
Correct option: C.
$135^{\circ}$
c
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MCQ 871 Mark
In Fig., AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If $\angle A P R=25^{\circ}, \angle R Q C=30^{\circ}$ and $\angle C Q F=65^{\circ}$, then
Image
  • ✓
    $x=55^{\circ}, y=40^{\circ}$
  • B
    $x=50^{\circ}, y=45^{\circ}$
  • C
    $x=60^{\circ}, y=35^{\circ}$
  • D
    $x=35^{\circ}, y=60^{\circ}$
Answer
Correct option: A.
$x=55^{\circ}, y=40^{\circ}$
a
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MCQ 891 Mark
In Fig. what is y in terms of x?
Image
  • ✓
    $\frac{3}{2} x$
  • B
    $\frac{4}{3} x$
  • C
    x
  • D
    $\frac{3}{4} x$
Answer
Correct option: A.
$\frac{3}{2} x$
a
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MCQ 901 Mark
In Fig. what is z in terms of x and y?
Image
  • A
    x + y + 180
  • ✓
    x + y - 180
  • C
    $180^{\circ}-(x+y)$
  • D
    $x+y+360^{\circ}$
Answer
Correct option: B.
x + y - 180
b
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MCQ 921 Mark
The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are $94^{\circ}$ and $126^{\circ}$. Then, $\angle B A C=$
  • A
    $94^{\circ}$
  • B
    $54^{\circ}$
  • ✓
    $40^{\circ}$
  • D
    $44^{\circ}$
Answer
Correct option: C.
$40^{\circ}$
c
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MCQ 931 Mark
In Fig. if BP ||CQ and AC = BC then the measure of x is
Image
  • A
    $20^{\circ}$
  • B
    $25^{\circ}$
  • ✓
    $30^{\circ}$
  • D
    $35^{\circ}$
Answer
Correct option: C.
$30^{\circ}$
c
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MCQ 941 Mark
In Fig. the value of x is
Image
  • A
    $65^{\circ}$
  • B
    $80^{\circ}$
  • C
    $95^{\circ}$
  • ✓
    $120^{\circ}$
Answer
Correct option: D.
$120^{\circ}$
d
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MCQ 961 Mark
If the measures of angles of a triangle are in the ratio of 3 : 4 : 5 what is the measure of the smallest angle of the triangle?
  • A
    $25^{\circ}$
  • B
    $30^{\circ}$
  • ✓
    $45^{\circ}$
  • D
    $60^{\circ}$
Answer
Correct option: C.
$45^{\circ}$
c
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MCQ 981 Mark
In Fig. if EC ||AB, $\angle E C D=70^{\circ}$ and $\angle B D O=20^{\circ}$, then $\angle O B D$ is
Image
  • A
    $20^{\circ}$
  • ✓
    $50^{\circ}$
  • C
    $60^{\circ}$
  • D
    $70^{\circ}$
Answer
Correct option: B.
$50^{\circ}$
b
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MCQ 991 Mark
Line segments AB and CD intersect at O such that  AC|| DB . If $\angle C A B=45^{\circ}$ and $\angle C D B=55^{\circ}$, then $\angle B O D=$
  • A
    $100^{\circ}$
  • ✓
    $80^{\circ}$
  • C
    $90^{\circ}$
  • D
    $135^{\circ}$
Answer
Correct option: B.
$80^{\circ}$
b
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MCQ 1001 Mark
In a $\triangle A B C$, if $\angle A=60^{\circ}, \angle B=80^{\circ}$ and the bisectors of $\angle B$ and $\angle C$ meet at $O$, then $\angle B O C=$
  • A
    $60^{\circ}$
  • ✓
    $120^{\circ}$
  • C
    $150^{\circ}$
  • D
    $30^{\circ}$
Answer
Correct option: B.
$120^{\circ}$
b
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MCQ 1011 Mark
An exterior angle of a triangle is $108^{\circ}$ and its interior opposite angles are in the ratio 4:5. The angles of the triangle are
  • ✓
    $48^{\circ}, 60^{\circ}, 72^{\circ}$
  • B
    $50^{\circ}, 60^{\circ}, 70^{\circ}$
  • C
    $52^{\circ}, 56^{\circ}, 72^{\circ}$
  • D
    $42^{\circ}, 60^{\circ}, 76^{\circ}$
Answer
Correct option: A.
$48^{\circ}, 60^{\circ}, 72^{\circ}$
a
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MCQ 1021 Mark
In $\triangle A B C$, if $\angle A=100^{\circ}, A D$ bisects $\angle A$ and $A D \perp B C$. Then, $\angle B=$
  • A
    $50^{\circ}$
  • B
    $90^{\circ}$
  • ✓
    $40^{\circ}$
  • D
    $100^{\circ}$
Answer
Correct option: C.
$40^{\circ}$
c
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MCQ 1031 Mark
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is
  • A
    $90^{\circ}$
  • B
    $180^{\circ}$
  • C
    $270^{\circ}$
  • ✓
    $360^{\circ}$
Answer
Correct option: D.
$360^{\circ}$
d
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MCQ 1041 Mark
In a triangle, an exterior angle at a vertex is $95^{\circ}$ and its one of the interior opposite angle is $55^{\circ}$ then the measure of the other interior angle is
  • A
    $55^{\circ}$
  • B
    $85^{\circ}$
  • ✓
    $40^{\circ}$
  • D
    $9.0^{\circ}$
Answer
Correct option: C.
$40^{\circ}$
c
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MCQ 1051 Mark
In $\triangle A B C, \angle B=\angle C$ and ray $A X$ bisects the exterior angle $\angle D A C$. If $\angle D A X=70^{\circ}$, then $\angle A C B=$
  • A
    $35^{\circ}$
  • B
    $90^{\circ}$
  • ✓
    $70^{\circ}$
  • D
    $55^{\circ}$
Answer
Correct option: C.
$70^{\circ}$
c
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MCQ 1061 Mark
Side BC of a triangle ABC has been produced to a point D such that $\angle A C D=120^{\circ}$. If $\angle B=\frac{1}{2} \angle A$, then $\angle A$ is equal to
  • ✓
    $80^{\circ}$
  • B
    $75^{\circ}$
  • C
    $60^{\circ}$
  • D
    $90^{\circ}$
Answer
Correct option: A.
$80^{\circ}$
a
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MCQ 1071 Mark
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
  • A
    an isosceles triangle
  • B
    an obtuse triangle
  • C
    an equilateral triangle
  • ✓
    a right triangle
Answer
Correct option: D.
a right triangle
d
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MCQ 1081 Mark
An exterior angle of a triangle is equal to $100^{\circ}$ and two interior opposite angles are equal. Each of these angles is equal to
  • A
    $75^{\circ}$
  • B
    $80^{\circ}$
  • C
    $40^{\circ}$
  • ✓
    $50^{\circ}$
Answer
Correct option: D.
$50^{\circ}$
d
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MCQ 1091 Mark
If two acute angles of a right triangle are equal, then each acute is equal to
  • A
    $30^{\circ}$
  • ✓
    $45^{\circ}$
  • C
    $60^{\circ}$
  • D
    $90^{\circ}$
Answer
Correct option: B.
$45^{\circ}$
b
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MCQ 1101 Mark
If all the three angles of a triangle are equal, then each one of them is equal to
  • A
    $90^{\circ}$
  • B
    $45^{\circ}$
  • ✓
    $60^{\circ}$
  • D
    $30^{\circ}$
Answer
Correct option: C.
$60^{\circ}$
c
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