MCQ 2011 Mark
In Fig. for which value of x is $I_1 \| I_2$?
- A
$37^{\circ}$
- B
$43^{\circ}$
- C
$45^{\circ}$
- ✓
$47^{\circ}$
AnswerCorrect option: D. $47^{\circ}$
Let if $I_1 \| I_2$ and $AB$ is tranverse to it.
Then,
$\angle\text{PBA}$ should be equal $\angle\text{BAS}$ (Alternate angles)
So if $I_1 \| I_2$, 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 $I_1 \| I_2$
View full question & answer→MCQ 2021 Mark
In $\triangle\text{ABC, AB = AC}$ and $\angle\text{B}=50^{\circ}.$ Then, $\angle\text{A = } ?$
- A
$40^\circ$
- B
$50^\circ$
- ✓
$80^\circ$
- D
$130^\circ$
AnswerCorrect option: C. $80^\circ$
In $\triangle\text{ABC,}$
$\text{AB = AC}$
$\Rightarrow\angle\text{C}=\angle\text{B}$ (angles opposite to equal sides are equal)
$\Rightarrow\angle\text{C}=50^{\circ}$
Now, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{A}+50^{\circ}+50^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{A}=80^{\circ}$
View full question & answer→MCQ 2031 Mark
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is:
- ✓
$360^\circ$
- B
$180^\circ$
- C
$90^\circ$
- D
$270^\circ$
AnswerCorrect option: A. $360^\circ$
In $\triangle\text{ABC}$
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
Now, $\angle\text{FAB} = 180^\circ- \angle\text{A}\ \cdots(\text{i})$
$\angle\text{DCA} = 180^\circ - \angle\text{C}\ \cdots(\text{ii})$
$\angle\text{EBC} = 180^\circ - \angle\text{B}\ \cdots(\text{iii})$
Adding equations $(i), (ii)$ and $(iii)$
$\angle\text{FAB} + \angle\text{DCA} + \angle\text{EBC} = 180^\circ$
$- \angle\text{A} + 180^\circ - \angle\text{C} + 180^\circ - \angle\text{B}$
$= 540^\circ - (\angle\text{A} + \angle\text{B} + \angle\text{C})$
$= 540^\circ - 180^\circ$
$\Rightarrow $ Sum of All exterior angles $= 360^\circ $ View full question & answer→MCQ 2041 Mark
In a triangle $ABC, \angle\text{B} = 35^\circ$ and $\angle\text{C} = 60^\circ,$ then the shortest side is:
- A
$BC$
- B
$AB$ and $BC$
- ✓
$AC$
- D
$AB$
AnswerSide opposite to smallest angle is shortest side angles of triangle are $35, 60$ and $85.$
View full question & answer→MCQ 2051 Mark
The sum of the interior angles of a triangle is:
- A
$270^\circ$
- B
$360^\circ$
- ✓
$180^\circ$
- D
$90^\circ$
AnswerCorrect option: C. $180^\circ$
For a triangle,
Number of sides$ (n) = 3$
Sum of interior angles$ = (n - 2) \times 180^\circ $
$= (3 - 2) \times 180^\circ $
$= 1 \times 180^\circ $
$= 180^\circ $
View full question & answer→MCQ 2061 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
$40^\circ$
- ✓
$50^\circ$
AnswerCorrect 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 $
View full question & answer→MCQ 2071 Mark
In the adjoining figure, $AB = AC$ and $AD$ is bisector of $\angle\text{A}.$ The rule by which $\triangle\text{ABD}\cong\triangle\text{ACD}.$
AnswerIn $\triangle\text{ABD}$ and $\triangle\text{ADC},$ we have$AB = AC$ (Given)
$\angle\text{BAD} = \angle\text{DAC}$ (Since AD, bisects $\angle\text{A}$)
$AD = AD$ (common in both)
Hence, $\triangle\text{ABD}\cong\triangle\text{ACD}$ by $SAS.$
View full question & answer→MCQ 2081 Mark
The perimeter of a triangle is $36\ cm$ and its sides are in the ratio $a : b : c = 3 : 4 : 5$ then $a, b, c$ are respectively:
- A
$9\ cm, 15\ cm, 12\ cm$
- ✓
$9\ cm, 12\ cm, 15\ cm$
- C
$12\ cm, 9\ cm, 15\ cm$
- D
$15\ cm, 12\ cm, 9\ cm$
AnswerCorrect option: B. $9\ cm, 12\ cm, 15\ cm$
Let the three sides $a, b, c$ be $3x, 4x$ and $5x$ respectively.
Then according to the conditions given in the question, we have
$3x + 4x + 5x = 36$
$12x = 36$
$x = 3cm$
Thus, the three sides are:
$a = 3 \times 3 = 9cm, b = 4 \times 3 = 12cm$ and $c = 5 \times 3 = 15cm$
View full question & answer→MCQ 2091 Mark
In the given figure, $ABC$ is an equilateral triangle. The value of $x + y$ is:
- A
$120^{\circ}$
- B
$180^{\circ}$
- ✓
$240^{\circ}$
- D
$200^{\circ}$
AnswerCorrect option: C. $240^{\circ}$
As triangle $ABC$ is an equilateral traingle, therefore all the three angles are equal, that is,$ 60^{\circ}$ each.$x = 180 - 60 = 120^{\circ}$
$y = 180 - 60 = 120^{\circ}$
$x + y = 120 + 120 = 240^{\circ}$
View full question & answer→MCQ 2101 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
$90^\circ$
AnswerCorrect option: A. $80^\circ$
$\angle\text{B}=\frac{1}{2}\angle\text{A}$
$\angle\text{ACD}$ is an exterior angle.
$\Rightarrow\angle\text{A}+\angle\text{B}=\angle\text{ACD}$
$\Rightarrow\angle\text{A}=\frac{1}{2}\angle\text{A}=120^\circ$
$\Rightarrow\frac{3\angle\text{A}}{2}=120^\circ$
$\Rightarrow3\angle\text{A}=240^\circ$
$\Rightarrow\angle\text{A}=80^\circ$
View full question & answer→MCQ 2111 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
$110^\circ$
- B
$100^\circ$
- C
$130^\circ$
- ✓
$120^\circ$
AnswerCorrect option: D. $120^\circ$
Let the base angles be $x$ each,
Vertex angle $= 2(x + x) = 4x$
Now, since the sum of all the angles of a triangle is $180^\circ $
$x + x + 4x = 180^\circ $
$6x = 180^\circ $
$x = 30^\circ $
Therefore, vertex angle $= 4x = 120^\circ $
View full question & answer→MCQ 2121 Mark
In $\triangle\text{PQR},\ \angle\text{P} = 60^\circ,\ \angle\text{Q} = 50^\circ.$ Which side of the triangle is the longest?
AnswerIn $\triangle\text{PQR},\ \angle\text{P} = 60^\circ,\ \angle\text{Q} = 50^\circ.$
Now, by angle sum property, $\angle\text{P} +\angle\text{Q} +\angle\text{R} = 180^\circ$
$60^\circ + 50^\circ + \angle\text{R} = 180^\circ$
or, $\angle\text{R} = 180^\circ - 110^\circ = 70^\circ$
So, $\angle\text{R}$ is the largest angle and the side opposite to it, i.e, $PQ$ will be the longest side.
View full question & answer→MCQ 2131 Mark
In the given figure, $AB = AC$ and $OB = OC$. Then, $\angle\text{ABO}:\angle\text{ACO} = ?$ 
- ✓
$1 : 1$
- B
$2 : 1$
- C
$1 : 2$
- D
AnswerCorrect option: A. $1 : 1$
In $\triangle\text{ABC},$
$\text{AB = AC}\Rightarrow\angle\text{ABC}=\angle\text{ACB}...(\text{i})$
In $\triangle\text{OBC},$
$\text{OB = OC} \Rightarrow\angle\text{OBC}=\angle\text{OCB}...(\text{ii})$
Subtraction (ii) from (i), we get
$\Rightarrow\angle\text{ABO}=\angle\text{ACO}$
So, $\angle\text{ABO}:\angle\text{ACO}=1:1$
View full question & answer→MCQ 2141 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
$135^\circ$
AnswerCorrect 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 $\angle\text{BOD}=180^\circ-45^\circ-55^\circ$
$=80^\circ$ View full question & answer→MCQ 2151 Mark
An angle is $14^\circ $ more than its complement. Find its measure.
- ✓
$52^\circ$
- B
$62^\circ$
- C
$32^\circ$
- D
$42^\circ$
AnswerCorrect option: A. $52^\circ$
Let the angle $= x$
Its complement $= 90^\circ - x$
According to the question, $x$ is $14^\circ $ more than its complement,
$\Rightarrow x = (90^\circ - x) + 14^\circ $
$\Rightarrow x + x = 104^\circ $
$\Rightarrow 2x = 104^\circ $
$\Rightarrow\ \text{X}=\frac{104^\circ}{2}=52^\circ$
View full question & answer→MCQ 2161 Mark
In the following, write the correct answer. In $\triangle\text{ABC}$ if $AB = AC$ and $\angle\text{B}=50^{\circ}$ then is equal to:
- A
$40^\circ$
- ✓
$50^\circ$
- C
$80^\circ$
- D
$130^\circ$
AnswerCorrect option: B. $50^\circ$
Given $\triangle\text{ABC}$ such that $AB = AC$ and $\angle\text{B}=50^{\circ}$
View full question & answer→MCQ 2171 Mark
In triangles $ABC$ and $PQR$, if $\angle\text{A}=\angle\text{R},\angle\text{B}=\angle\text{P}$ and $\text{AB}=\text{RP},$ then which one of the following congruence conditioins applies:
AnswerFrom given conditions,
$\angle\text{B}=\angle\text{P}$
$\angle\text{A}=\angle\text{R}$
And the side containing then is also equal
i.e $\text{AB}=\text{PR}$
So ASA property.
Hence, correct option is $(b).$
View full question & answer→MCQ 2181 Mark
In figure, $PS = QR$ and $\angle\text{SPQ} = \angle\text{RQP}.$ Then
- ✓
$PR = QS$
- B
$PR > QS$
- C
$PR ≤ QS$
- D
$PR < QS$
AnswerCorrect option: A. $PR = QS$
In $\triangle\text{QPR}$ and $\triangle\text{PQS},$
$QR = PS$ (Given)
$\angle\text{RQP} = \angle\text{SPQ}$ (Given)
$PQ = PQ$ (Common)
$\therefore \triangle\text{QPR} ≅ \triangle\text{PQS}$ (SAS Axio)
$\therefore PR = QS (C.P.C.T.)$
View full question & answer→MCQ 2191 Mark
In Figure, $AB$ and $CD$ are parallel lines and transversal $EF$ intersects 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 $
AnswerCorrect option: A. $x = 55^\circ , y = 40^\circ $
$\angle\text{OQP} = 180^\circ - \angle\text{OQF}$
$= 180^\circ - (30^\circ + 65^\circ )$
$\Rightarrow \angle\text{OQP} = 85^\circ\ ...\ (\text{i})$
$\angle\text{APQ}=\angle\text{CQF}$ (Corresponding angles)
$\Rightarrow 25^\circ + y^\circ = 65^\circ $
$\Rightarrow y^\circ = 65^\circ - 25^\circ $
$\Rightarrow y^\circ = 40^\circ $
Now in $\triangle\text{OPQ}$
$\angle\text{O}+\angle\text{OPQ}+\angle\text{PQO} = 180^\circ$
$\Rightarrow x^\circ + 40^\circ + 85^\circ = 180^\circ $
$x^\circ = 180^\circ - 85^\circ - 40^\circ = 55^\circ $
$\Rightarrow x = 55^\circ , y = 40^\circ $
View full question & answer→MCQ 2201 Mark
It is given that$\triangle\text{ABC}\cong\triangle\text{FDE}$ in which $AB = 5\ cm,$ $\angle\text{B}=40^{\circ},\angle\text{A}=80^{\circ}$and $FD = 5\ cm$. Then, which of the following is true?
- A
$\angle\text{D}=60^{\circ}$
- ✓
$\angle\text{E}=60^{\circ}$
- C
$\angle\text{F}=60^{\circ}$
- D
$\angle\text{D}=80^{\circ}$
AnswerCorrect option: B. $\angle\text{E}=60^{\circ}$
Given $\triangle\text{ABC}\cong\triangle\text{FDE}$
$\angle\text{A}=\angle\text{F}=80^{\circ} ...(C.P.C.T.)$
$\angle\text{B}=\angle\text{D}=40^{\circ}...(C.P.C.T.)$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow80^{\circ}+40^{\circ}+\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{C}=60^{\circ}$
$\therefore\angle\text{E}=\angle\text{C}=60^{\circ}$
$\therefore\angle\text{E}=60^{\circ}$
View full question & answer→MCQ 2211 Mark
If all the three angles of a triangle are equal, then each one of them is equal to:
- ✓
$60^\circ$
- B
$45^\circ$
- C
$90^\circ$
- D
$30^\circ$
AnswerCorrect option: A. $60^\circ$
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 $
View full question & answer→MCQ 2221 Mark
In the adjoining figure, $PQ > PR$. If $OQ$ and $OR$ are bisectors of $\angle\text{Q}$ and $\angle\text{R}$ respectively, then 
- ✓
$OQ > OR$
- B
$OQ < OR$
- C
$OQ ≤ OR$
- D
$OQ = OR$
AnswerCorrect option: A. $OQ > OR$
Since $PQ > PR$ then $\angle\text{R}>\angle\text{Q}$ and hence their bisectors follow the same.
I.e $\frac{\text{R}}{2}>\frac{\text{Q}}{2}$ and hence $OQ > OR.$
View full question & answer→MCQ 2231 Mark
In the given figure, side $BC$ of $\triangle\text{ABC}$ has been produced to a point $D$. If $\angle\text{A}=3\text{y},\angle\text{B}=\text{x}^\circ,\angle\text{C}=5\text{y}^\circ$ and $\angle\text{CBD}=7\text{y}^\circ.$ Then, the value of $x$ is:
Answer$\angle\text{ACB}=\angle\text{ACD}=180^\circ$ (linear pair)
$\Rightarrow5\text{y}+7\text{y}=180^\circ$
$\Rightarrow12\text{y}=180^\circ$
$\Rightarrow\text{y}=15^\circ$
Now, $\angle\text{ACD}=\angle\text{ABC}+\angle\text{BAC}$ (Exterior angle property)
$\Rightarrow7\text{y}=\text{x}+3\text{y}$
$\Rightarrow7(15^\circ)=\text{x}+3(15^\circ)$
$\Rightarrow105^\circ=\text{x}+45^\circ$
$\Rightarrow\text{x}=60^\circ$
View full question & answer→MCQ 2241 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 $O$, the $\angle\text{BOC} =$
- ✓
$120^\circ $
- B
$150^\circ$
- C
$30^\circ$
- D
$60^\circ$
AnswerCorrect option: A. $120^\circ $
$O$ is point where bisectors of $\angle\text{C}$ & $\angle\text{B}$ meets.
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$60^\circ + 80^\circ + \angle\text{C} = 180^\circ$
$\angle\text{C} = 40^\circ$
$\angle\text{C}2 = 20^\circ$
$\angle\text{C}2 = 20^\circ = \angle\text{BCO}\ ...\ (\text{i})$
$\angle\text{B}2 = 80^\circ = 40^\circ = \angle\text{OBC}\ ...\ (\text{ii})$
In $\triangle\text{BOC}$
$\angle\text{BCO} + \angle\text{OBC} + \angle\text{BOC} = 180^\circ$
From $(i)$ and $(ii)$
$20^\circ + 40^\circ + \angle\text{BOC} = 180^\circ$
$\angle\text{BOC} = 180^\circ - 60^\circ = 120^\circ$ View full question & answer→MCQ 2251 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
$\frac{3}{4}\text{x}^\circ$
AnswerCorrect 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 $\triangle\text{OCD}$
$\angle\text{DOC}+\angle\text{OCD}+\angle\text{D}=180^\circ$
$180^\circ-3\text{y}^\circ+3\text{x}^\circ+\text{y}^\circ=180^\circ$ $\big[\angle\text{OCD}=\angle\text{ACD}\big]$
$\Rightarrow2\text{y}^\circ=3\text{x}^\circ$
$\Rightarrow\text{y}=\frac{3}{2}\text{x}^\circ$ View full question & answer→MCQ 2261 Mark
In figure, $X$ is a point in the interior of square $ABCD. AXYZ$ is also a square. If $DY = 3\ cm$ and $AZ = 2\ cm$, then $BY =$
- A
$6\ cm$
- ✓
$7\ cm$
- C
$8\ cm$
- D
$5\ cm$
AnswerCorrect option: B. $7\ cm$
$\angle\text{Z} = 90^\circ$ (Angle of square)
Therefore, $AZD$ is a right angle triangle,
By Pythagoras theorem,
$\mathrm{AD}^2=\mathrm{AZ}^2+\mathrm{ZD}^2$
$\mathrm{AD}^2=22+(2+3)^2$
$\mathrm{AD}^2=4+25$
$\text{AD} = \sqrt{29}$
In $\triangle\text{AXB},$ with $X$ as right angle,
By Pythagoras theorem,
$\mathrm{AB}^2=A X^2+X B^2$
$X B^2=29-4$
$X B=5$
$B Y=X B+X Y$
$=5+2$
$=7 \mathrm{~cm}$
View full question & answer→MCQ 2271 Mark
Find the measure of each exterior angle of an equilateral triangle.
- A
$100^\circ$
- B
$150^\circ$
- C
$110^\circ$
- ✓
$120^\circ$
AnswerCorrect option: D. $120^\circ$
We know that in equilateral triangle each angle is $60^\circ $
and we know sum of interior angle and exterior angle is $180^\circ $
let exterior angle be $x$
$60^\circ + x = 180^\circ $
$x = 180^\circ - 60^\circ $
$x = 120^\circ $
View full question & answer→MCQ 2281 Mark
In $\triangle\text{ABC}, \ \angle\text{C} = \angle\text{A}$ and $BC = 6\ cm$ and $AC = 5\ cm$. Then the length of $AB$ is:
- ✓
$6\ cm$
- B
$3\ cm$
- C
$2.5\ cm$
- D
$5\ cm$
AnswerCorrect option: A. $6\ cm$
Sides opposite to equal angles are equal. Since, $\angle\text{C} = \angle\text{A},$ hence, $AB = BC = 6\ cm.$
View full question & answer→MCQ 2291 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
$30^\circ$
AnswerCorrect 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 $
View full question & answer→MCQ 2301 Mark
Which of the following is not possible in case of triangle $ABC?$
- A
$AB = 5\ cm, BC = 8\ cm, CA = 7\ cm.$
- ✓
$AB = 2\ cm, BC = 4\ cm, CA = 7\ cm.$
- C
$\angle\text{A} = 50^\circ, \ \angle\text{B} = 60^\circ,\ \angle\text{C} = 70^\circ.$
- D
$AB = 3\ cm, BC = 4\ cm, CA = 5\ cm.$
AnswerCorrect option: B. $AB = 2\ cm, BC = 4\ cm, CA = 7\ cm.$
Sum of any two sides is greater than third side, but here $2 + 4 < 7.$
View full question & answer→MCQ 2311 Mark
In right-angled $\triangle\text{DEF}$ if $\angle\text{E} = 90^\circ$ then:
- A
$DE$ is the longest side.
- B
$DF$ is the shortest side.
- C
$EF$ is the longest side.
- ✓
$DF$ is the longest side.
AnswerCorrect option: D. $DF$ is the longest side.
In a triangle, only one right angle is possible, hence it is the greatest angle and the side opposite to it is the longest side. Here, the side opposite to $\angle\text{E}$ is $DF$, hence, it is the longest side.
View full question & answer→MCQ 2321 Mark
In Fig. $\text{AB}\perp\text{BE}$ and $\text{FE}\perp\text{BE},$ If $\text{BC}=\text{DE}$ and $\text{AB}=\text{EF},$ then $\triangle\text{ABD}$ is congruent to:
- A
$\triangle\text{EFC}$
- B
$\triangle\text{ECF}$
- C
$\triangle\text{CEF}$
- ✓
$\triangle\text{FEC}$
AnswerCorrect option: D. $\triangle\text{FEC}$
$AB = EF$
$BC = DE$
$BC + CD = DE + CD$ (adding $CD$ both sides)
$BC + CD = BD, DE + CD = CE$
So $BD = CE$
Now Consider $\triangle\text{ABD},$ & $\triangle\text{FEC}$
$\text{AB}=\text{FE}$
$\text{BD}=\text{EC}$
$\angle\text{ABD}=\angle\text{FEC}=90^\circ$
So $\triangle\text{ABD}\cong\triangle\text{FEC}$ by SAS creterion.
Hence, correct option is $(d).$ View full question & answer→MCQ 2331 Mark
In a $\triangle\text{ABC},$ side $BC$ is produced to $D$. If $\text{ABC}=50^\circ$ and $\angle\text{ACD}=110^\circ$ then $\angle\text{A} = ?$
- A
$160^\circ$
- ✓
$60^\circ$
- C
$30^\circ$
- D
$80^\circ$
AnswerCorrect option: B. $60^\circ$
$∴\ \angle\text{A}+\angle\text{B}=\angle\text{ACD}$
$⇒\angle\text{A}+50^\circ=110^\circ$
$⇒\angle\text{A}=60^\circ$
View full question & answer→MCQ 2341 Mark
Side $BC$ of a triangle $ABC$ has been produced to a point $D$ such that $\angle\text{ACD} = 120^\circ$ If $\angle\text{B} = 12\angle\text{A},$ then $\angle\text{A}$ is equal to:
- ✓
$80^\circ$
- B
$90^\circ$
- C
$75^\circ$
- D
$60^\circ$
AnswerCorrect option: A. $80^\circ$
$\angle\text{B}=\frac{1}{2}\angle\text{A}$
$\angle\text{ACD}$ is an exterior angle
$\Rightarrow \angle\text{A}+\angle\text{B}=\angle\text{ACD}$
$\Rightarrow \angle\text{A}+\frac{1}{2}\angle\text{A}=120^\circ$
$\Rightarrow \frac{3\angle\text{A}}{2}=120^\circ$
$\Rightarrow\ 3\angle\text{A}=240^\circ$
$\Rightarrow\ \angle\text{A}=80^\circ$
View full question & answer→MCQ 2351 Mark
In triangles $ABC$ and $DEF, AB = FD$ and $\angle\text{A} = \angle\text{D}.$ The two triangles will be congruent by $SAS$ axiom if:
- A
$AC = EF$
- B
$BC = EF$
- ✓
$AC = DE$
- D
$BC = DE$
AnswerCorrect option: C. $AC = DE$
Given: $\triangle\text{ABC}$ and $\triangle\text{DEF}, \ \text{AB = FD}$ and $\angle\text{A} = \angle\text{D}$
The two triangles will be congruent by $SAS$ axiom if two sides including one angle of one triangle is equal to the another triangle.
Hence, $AC = DE$ (As $\angle\text{A}$ is between $AB$ and $AC)$ View full question & answer→MCQ 2361 Mark
In a $\triangle\text{ABC},$ side BC is produced to $D$. If $\angle\text{ABC}=50^\circ$ and $\angle\text{ACD}=110^\circ$ then $\angle\text{A}=?$
- A
$160^\circ$
- ✓
$60^\circ$
- C
$80^\circ$
- D
$30^\circ$
AnswerCorrect option: B. $60^\circ$
$\angle\text{ACD}=\angle\text{B}+\angle\text{A}$ (Exterior angle property)
$\Rightarrow110^\circ=50^\circ+\angle\text{A}$
$\Rightarrow\angle\text{A}=60^\circ$
View full question & answer→MCQ 2371 Mark
in $\triangle\text{ABC}$ and $\triangle\text{DEF}$ it is given that $AB = DE$ and $BC = EF$ in order that $\triangle\text{ABC}≅\triangle\text{DEF},$ we must have
- A
$\text{None of these}$
- ✓
$\angle\text{B} = \angle\text{E}$
- C
$\angle\text{A} = \angle\text{D}$
- D
$\angle\text{C} = \angle\text{F}$
AnswerCorrect option: B. $\angle\text{B} = \angle\text{E}$
Given, $AB = DE$ and $BC = EF$
for, $\angle\text{B} = \angle\text{E}$
$\triangle\text{ABC}≅\triangle\text{DEF},$ [SSA] View full question & answer→MCQ 2381 Mark
If triangle $ABC$ is obtuse angled and $\angle\text{C}$ is obtuse, then
- ✓
$AB > BC$
- B
$AB = BC$
- C
$AB < BC$
- D
$AC > AB$
AnswerCorrect option: A. $AB > BC$
Side opposite to the largest angle is always greatest and a triangle can have only $1$ angle as obtuse.
View full question & answer→MCQ 2391 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
$44^\circ$
AnswerCorrect option: C. $40^\circ$
$\angle\text{ABC}=180^\circ-126^\circ=54^\circ$
$\angle\text{ACB}=180^\circ-94^\circ=86^\circ$
Now, in $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{BAC}=180^\circ-\angle\text{ABC}-\angle\text{ACB}$
$=180^\circ-54^\circ-86^\circ$
$\Rightarrow\text{BAC}=40^\circ$ View full question & answer→MCQ 2401 Mark
P is a point on side $BC$ of a $\triangle\text{ABC}$ such that AP bisects $\angle\text{BAC}.$ Then,
- A
$BP = CP$
- B
$CP < CA$
- ✓
$BA > BP$
- D
$BP > BA$
AnswerCorrect option: C. $BA > BP$
Since, $AP$ bisects $\angle\text{BAC}.$
Hence the point $P$ on $BC$ also bisects it as it the opposite side of $\angle\text{BAC}.$
Hence, we can conclude that, $BA > BP.$
View full question & answer→MCQ 2411 Mark
In the given figure, two rays $BD$ and $CE$ intersect at a point A. The side $BC$ of $\triangle\text{ABC}$ have been produced on both sides to points $F$ and $G$ respectively. If $\angle\text{ABF}=\text{x}^\circ,\angle\text{ACG}=\text{y}^\circ$ and $\angle\text{DAE}=\text{z}^\circ$ then $z =?$
- ✓
$x + y - 180$
- B
$x + y + 180$
- C
$180 - (x + y)$
- D
$x + y + 360^\circ $
AnswerCorrect option: A. $x + y - 180$
$\angle\text{ABF}+\angle\text{ABC}=180^\circ$ (linear pair)
$\Rightarrow\text{x}+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-\text{x}$
$\angle\text{ACG}+\angle\text{ACB}=180^\circ$ (linear pair)
$\Rightarrow\text{y}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-\text{y}$
In $\triangle\text{ABC},$ by angle sum property
$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$
$\Rightarrow(180^\circ-\text{x})+(180^\circ-\text{y})+\angle\text{BAC}=180^\circ$
$\Rightarrow\angle\text{BAC}-\text{x}-\text{y}+180^\circ=0$
$\Rightarrow\angle\text{BAC}=\text{x}+\text{y}-180^\circ$
Now, $\angle\text{EAD}=\angle\text{BAC}$ (vertically opposite angles)
$\Rightarrow\text{z}=\text{x}+\text{y}-180^\circ$
View full question & answer→MCQ 2421 Mark
In figure, the value of $x$ is:
- A
$95^\circ$
- B
$65^\circ$
- ✓
$120^\circ$
- D
$80^\circ$
AnswerCorrect option: C. $120^\circ$
In $\triangle\text{ABD}$
$\angle\text{A} + \angle\text{B} + \angle\text{D} = 180^\circ$
$\Rightarrow 55^\circ + \angle\text{DBA} + 25^\circ = 180^\circ$
$\Rightarrow \angle\text{DBA} = 180^\circ - 55^\circ - 25^\circ$
$= 180^\circ - 80^\circ$
$\Rightarrow \angle\text{DBA} = 100^\circ$
So, $\angle\text{DBC} = 180^\circ - \angle\text{DBA}$
$= 180^\circ - 100^\circ$
$ \angle\text{DBC} = 80^\circ$
Now, $\triangle\text{EBC}$
$\angle\text{A} + \angle\text{EBC} + \angle\text{C} = 180^\circ$
$\Rightarrow \angle\text{E} + 80^\circ + 40^\circ = 180^\circ ( \angle\text{DBC} = \angle\text{EBC})$
$\Rightarrow \angle\text{E} = 180^\circ - 120^\circ = 60^\circ$
Also, $\text{x} = 180^\circ - \angle\text{E} = 180^\circ - 60^\circ$
$\Rightarrow \text{x} = 120^\circ$
View full question & answer→MCQ 2431 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 $
AnswerCorrect 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 $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\text{z}^\circ+180^\circ-\text{y}^\circ+180^\circ-\text{x}^\circ=180^\circ$
$\Rightarrow\text{z}^\circ=\text{x}^\circ+\text{y}^\circ-180^\circ$
View full question & answer→MCQ 2441 Mark
In a $\triangle\text{ABC},$ it is given that $\angle\text{A}:\angle\text{B}:\angle\text{C}=3:2:1$ and $\angle\text{ACD}=90^\circ.$ If BC produced to E then $\angle\text{ECD}=?$
- ✓
$60^\circ$
- B
$50^\circ$
- C
$40^\circ$
- D
$25^\circ$
AnswerCorrect option: A. $60^\circ$
we know that
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ .....(Angle sum property)
$\therefore\angle\text{A}=\Big(180\times\frac{3}{6}\Big)=90^\circ$
$\angle\text{B}=\Big(180\times\frac{2}{6}\Big)=60^\circ\ \text{and}$
$\angle\text{C}=\Big(180\times\frac{1}{6}\Big)=30^\circ$
Now,
$\angle\text{ACE}=\angle\text{A}+\angle\text{B}$ .....(Exterior angle is equal to sum of the remote interior angles)
$=90^\circ+60^\circ$
$=150^\circ$
$\angle\text{ACE}=\angle\text{ECD}+\angle\text{ACD}$
$\therefore150^\circ=\angle\text{ECD}+90^\circ$
$\therefore\angle\text{ECD}=60^\circ$
View full question & answer→MCQ 2451 Mark
In the given figure, lines $AB$ and $CD$ intersect at a point $O$. The sides $CA$ and $OB$ have been produced to $E$ and repectively such that $\angle\text{OAE}=\text{x}^\circ$ and $\angle\text{DBF}=\text{y}^\circ.$
If $\angle\text{OCA}=80^\circ,\angle\text{COA}=40^\circ$ and $\angle\text{BDO}=70^\circ$ then $\text{x} ^\circ+\text{y}^\circ=?$ - A
$190^\circ$
- ✓
$230^\circ$
- C
$210^\circ$
- D
$270^\circ$
AnswerCorrect option: B. $230^\circ$
In $\triangle\text{OAC},$ by angle sum property
$\angle\text{OCA}+\angle\text{COA}+\angle\text{CAO}=180^\circ $
$\Rightarrow80^\circ+40^\circ+\angle\text{CAO}=180^\circ$
$\Rightarrow\angle\text{CAO}=60^\circ$
$\angle\text{CAO}+\angle\text{OAE}=180^\circ$ (linear pair)
$\Rightarrow60^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=120^\circ$
$\angle\text{COA}=\angle\text{BOD}$ (vertically opposite angles)
$\Rightarrow\angle\text{BOD}=40^\circ$
In $\triangle\text{OBD},$ by angle sum property
$\angle\text{OBD}+\angle\text{BOD}+\angle\text{ODB}=180^\circ$
$\Rightarrow\angle\text{OBD}+40^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{OBD}=70^\circ$
$\angle\text{OBD}+\angle\text{DBF}=180^\circ$ (linear pair)
$\Rightarrow70^\circ+\text{y}=180^\circ$
$\Rightarrow\text{y}=110^\circ$
$\therefore\text{x}+\text{y}=120^\circ+110 ^\circ=230^\circ$
View full question & answer→MCQ 2461 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
$40^\circ$
- B
$80^\circ$
- C
$75^\circ$
- ✓
$50^\circ$
AnswerCorrect 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 $
View full question & answer→MCQ 2471 Mark
In the following, write the correct answer. In $\triangle\text{ABC}$ if $BC = AB$ and $\angle\text{B}=80^{\circ}$ then is equal to:
- A
$80^\circ$
- B
$40^\circ$
- ✓
$50^\circ$
- D
$100^\circ$
AnswerCorrect option: C. $50^\circ$
In $\triangle\text{ABC}$ we have
$BC = AB$ But $\angle\text{B}=80^{\circ}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$
$=\angle\text{A}+80^{\circ}\angle\text{A}=180^{\circ}$
$=2\angle\text{A}=100^{\circ}$
$=\angle\text{A}=100^{\circ}\div2=50^{\circ}$
View full question & answer→MCQ 2481 Mark
If $\triangle\text{PQR}\cong\triangle\text{EFD},$ then $\angle\text{E}=$
- ✓
$\angle\text{P}$
- B
$\angle\text{Q}$
- C
$\angle\text{R}$
- D
AnswerCorrect option: A. $\angle\text{P}$
$\triangle\text{PQR}\cong\triangle\text{EFD},$$\Rightarrow\angle\text{E}=\angle\text{P}$ (congruent angles of congruent triangles)
Hence, correct option is $(a).$
View full question & answer→MCQ 2491 Mark
In the following, write the correct answer. In $\triangle\text{PQR},$ if $\angle\text{R}>\angle\text{Q}$ then:
- A
$QR > PR$
- ✓
$PQ > PR$
- C
$PQ < PR$
- D
$QR < PR$
AnswerCorrect option: B. $PQ > PR$
Given, $\angle\text{R}>\angle\text{Q}$
$PQ > PR.$ View full question & answer→MCQ 2501 Mark
In the adjoining figure, $AB = AC$ and $AD$ is bisector of $\angle\text{A}.$ The rule by which $\triangle\text{ABD} \cong\triangle\text{ACD}$ is:
AnswerIn $\triangle\text{ABD}$ and $\triangle\text{ADC},$ we have
$AB = AC$ (Given)
$\angle\text{BAD} = \angle\text{DAC}$ ( Since $AD$, bisects $\angle\text{A}$)
$AD = AD$ ( conunon in both)
Hence, $\triangle\text{ABD}\cong\triangle\text{ACD}$ by $SAS.$
View full question & answer→