MCQ 511 Mark
IF $\text{AB}⊥\text{BC}$ and $\angle\text{BAC} = \angle\text{BCA},$ then which of the following statements is true?
Answer- AB = BC
Solution:
ABC is a right-angled isosceles triangle with $\angle\text{BAC} = \angle\text{BCA},$ and hence sides opposite to equal angles must be equal. Hence, we can say that AB = BC.
View full question & answer→MCQ 521 Mark
In $\triangle\text{ABC}$ and $\triangle\text{DEF}$ its is given that $\angle\text{B}=\angle\text{E}$ and $\angle\text{C}=\angle\text{F}$ in order that $\triangle\text{ABC}≅\triangle\text{DEF}$ we must have,
Answer- BC = EF
Solution:
In $\triangle\text{ABC}$ and $\triangle\text{DEF}$
$\angle\text{B}=\angle\text{E}$ and $\angle\text{C}=\angle\text{F}$
For congruence, BC = EF
Therefore by AAS axiom
$\triangle\text{ABC}≅\triangle\text{DEF}$
View full question & answer→MCQ 531 Mark
Answer- AB + BC + AC > 2AD
Solution:
In triangle ADB
AB + BD > AD
In triangle ADC
AC + DC > AD
Adding both
AB + AC + BD + DC > 2AD
Now BD + DC = BC
So, AB + AC + BC > 2AD
View full question & answer→MCQ 541 Mark
In triangles ABC and PQR three equality relations between some parts are as follows: AB = QP, $\angle\text{B} = \angle\text{P},$ BC = PR. State which of the congruence conditions applies:
Answer- SAS
Solution:
Two sides and included angle are equal and is SAS axiom.
View full question & answer→MCQ 551 Mark
In a $\triangle\text{ABC},$ if $\angle\text{B} = \angle\text{C} = 45^\circ,$ which is the longest side?
Answer- BC
Solution:
We know that sum of all angles of a triangle is 180°
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$\angle\text{B} = \angle\text{C} = 45^\circ$
$\angle\text{A} + 45^\circ + 45^\circ = 180^\circ$
$\angle\text{A} + 90^\circ = 180^\circ$
$\angle\text{A} = 180^\circ - 90^\circ$
$\angle\text{A} = 90^\circ$
So, angle A is the largest and the side opposite to the greatest angle is the longest so, side BC is the longest.
View full question & answer→MCQ 561 Mark
Answer- RHS
Solution:
In $\triangle\text{ABC}$ and $\angle\text{BAC}=\angle\text{ABD}$
BAD, we have (Right angles)
BC = AD (Hypotentuses and Given)
AB = AB (common in both)
Hence, $\triangle\text{ABC}\cong\triangle\text{BAD}$ by RHS criterion.
View full question & answer→MCQ 571 Mark
Answer- 75º
Solution:
We have,
$∴\ \angle\text{ABD}+\angle\text{ABC}=180^\circ$ [$∵$ E is a straight line]
$⇒125^\circ+\angle\text{ABC}=180^\circ$
$⇒\angle\text{ABC}=55^\circ$
Also,
$\angle\text{ACE}+\angle\text{ACB}=180^\circ$
$⇒130^\circ+\angle\text{ACB}=180^\circ$
$⇒\angle\text{ACB}=50^\circ$
$∴\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ [Sum of the angles of a triangle]
$⇒\angle\text{BAC}+55^\circ+50^\circ=180^\circ$
$⇒\angle\text{BAC}=75^\circ$
$⇒\angle\text{A}=75^\circ.$
View full question & answer→MCQ 581 Mark
Answer- AAS
Solution:
In $\triangle\text{ABD}$ and $\triangle\text{ADC},$ we have
$\angle\text{ABD} = \angle\text{ACD} $ (given)
$\angle\text{BDA} = \angle\text{CDA} $ (90o)
AD = AD (common in both)
Hence, $\triangle\text{ABD}\cong\triangle\text{ADC}$ by AAS criterion.
View full question & answer→MCQ 591 Mark
Answer- 65º
Solution:
Since BAE is a straight line, we have
$\angle\text{BAC}+\angle\text{CAE}=180^\circ$ .....(Supplementary angles)
$\Rightarrow\angle\text{BAC}+135^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=45^\circ$
Since CBD is a straight line, we have
$\angle\text{ABD}+\angle\text{ABC}=180^\circ$ ....(Supplmenetary angles)
$\Rightarrow110^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=70^\circ$
In $\triangle\text{ABC},$ we have
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ .....(Angle sum property)
$\Rightarrow45^\circ+70^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=65^\circ$
View full question & answer→MCQ 601 Mark
Two sides of a triangle are of length 4cm and 2.5cm. The length of the third side of the triangle cannot be:
Answer- 6.5cm
Solution:
The sum of any two sides of a triangle is greater than the third side.
Since, 4cm + 2.5cm = 6.5cm
The length of third side of a triangle cannot be 6.5cm.
View full question & answer→MCQ 611 Mark
In $\triangle\text{AOC}$ and $\triangle\text{XYZ},\ \angle\text{A}=\angle\text{X},\ \angle\text{AO} = \angle\text{XY},\ \angle\text{AC} = \angle\text{XZ},$ then by which congruence rule $\triangle\text{AOC}\cong\triangle\text{XYZ}?$
Answer- SAS
Solution:
According to SAS criterion, if the corresponding sides and their included angles are equal, then the triangles are congruent. Here, in $\triangle\text{AOC}$ and $\triangle\text{XYZ},$ AO = XY, and AC = XZ are the corresponding sides and $\angle\text{A}=\angle\text{X}$ are included angles, Hence, $\triangle\text{AOC}\cong\triangle\text{XYZ},$ by SAS.
View full question & answer→MCQ 621 Mark
The triangle is not possible with the given measurements:
Answer- 5.4cm, 2.3cm, 3.1cm
Solution:
In a triangle, the sum of any two sides must be greater than the third side and here 2.3 + 3.1 = 5.4 and hence, the triangle is not possible with the given measurements.
View full question & answer→MCQ 631 Mark
In a $\triangle\text{ABC},$ if $\angle\text{A}−\angle\text{B}=42^\circ$ and $\angle\text{B}−\angle\text{C}=21^\circ$ then $\angle\text{B} = ?$
Answer- 53º
Solution:
Let,
$\angle\text{A}−\angle\text{B}=42°...\ \text{(i)}$ and
$\angle\text{B}−\angle\text{C}=21^\circ ...\ \text{(ii)}$
Adding (i) and (ii),we get
$\angle\text{A}−\angle\text{C}=63° ...\ \text{(iii)}$
$\angle\text{B}=\angle\text{A}−42^\circ$ [using (i)]
$\angle\text{C}=\angle\text{A}−63^\circ$ [Using (iii)]
$∴\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ [Sum of the angles of a triangle]
$⇒\angle\text{A}+\angle\text{A}−42^\circ+\angle\text{A}−63^\circ=180^\circ$
$⇒3\angle\text{A}−105^\circ=180^\circ$
$⇒3\angle\text{A}=285^\circ$
$∴\angle\text{B}=(95−42)^\circ$
$⇒\angle\text{B}=53^\circ$
View full question & answer→MCQ 641 Mark
Answer- SSS
Solution:
In $\triangle\text{ABC} $ and $\triangle\text{ADC}, $ we have
AB = AD (4cm)
BC = DC (2.7cm)
AC = AC (commom in both)
Hence, $\triangle\text{ABC}\cong\triangle\text{ADC}$ by SSS criterion.
View full question & answer→MCQ 651 Mark
If $\triangle\text{ABC}≅\triangle\text{PQR}$ and $\triangle\text{ABC}$ is not congruent to $\triangle\text{RPQ},$ then which of the following is not true:
Answer- BC = PQ
Solution:
According to the condition given in the question,
If $\triangle\text{ABC}≅\triangle\text{PQR}$ and $\triangle\text{ABC}$ is not congruent to $\triangle\text{RPQ}.$
Then, clearly BC ≠ PQ
$\therefore$ It is false
View full question & answer→MCQ 661 Mark
Answer- 60º
Solution:
Let $ \angle\text{A}=(3\text{x})^\circ,\ \angle\text{B}=(2\text{x})^\circ$ and $\angle\text{C}=\text{x}^\circ$
Then,
3x + 2x + x = 180º [Sum of the angles of a triangle]
⇒ 6x = 180º
⇒ x = 30º
Hence, the angles are
$\angle\text{A}=3×30^\circ=90^\circ,\ \angle\text{B}=2×30^\circ=60^\circ$ and $\angle\text{C}=30^\circ$
Side BC of triangle ABC is produced to E.
$\therefore\ \angle\text{ACE}=\angle\text{A}+\angle\text{B}$
$⇒\angle\text{ACD}+\angle\text{ECD}=90^\circ+60^\circ$
$⇒90^\circ+\angle\text{ECD}=150^\circ$
$⇒\angle\text{ECD}=60^\circ$
View full question & answer→MCQ 671 Mark
What is the sum of the angles of a quadrilateral?
Answer- 360º
Solution:
For a quadrilateral
Number of sides (n) = 4
Sum of interior angles = (n - 2) × 180º
p = (4 - 2) × 180º
p = 2 × 180º
p = 360º
View full question & answer→MCQ 681 Mark
Answer- 50º
Soluton:
C = 180º - 115º = 65º AB = AC And hence $\angle\text{B} =\angle\text{C} = 65^\circ.$
Then A = 180º - 2 × 65º = 50º
View full question & answer→MCQ 691 Mark
In the following, write the correct answer.
If AB = QR, BC = PR and CA = PQ, then:
- A
$\triangle\text{ABC}\cong\triangle\text{QRP}$
- B
$\triangle\text{CBA}\cong\triangle\text{PRQ}$
- C
$\triangle\text{BAC}\cong\triangle\text{RQP}$
- D
$\triangle\text{PQR}\cong\triangle\text{BCA}$
Answer- $\triangle\text{CBA}\cong\triangle\text{PRQ}$
Solution:
We know that, if is congruent to, then sides of $\triangle\text{RST}$ fall on corresponding equal sides of angles of fall on corresponding equal angles.
Here, given AB = QR, BC = PR and CA = PQ, which shows that AB covers QR, BC covers PR and CA covers PQ i.e., A correspond to Q, B correspond to R and C correspond to P.
View full question & answer→MCQ 701 Mark
Answer- AB > AD
Solution:
Given $\text{AB > AC}$
$\therefore\angle\text{ACB}>\angle\text{ABC}$ ...(We know that, if two sides of a triangle unequal, then the longer side has the greater angle opposite to it.)
$\angle\text{ADB}>\angle\text{ACD}$ ...(Exterior angle of a triangle is greater that the interior opposite angles)
$\therefore\angle\text{ADB}>\angle\text{ACB}>\angle\text{ABC}$
$\therefore\angle\text{ADB}>\angle\text{ABD}$
$\therefore\text{AB > AD}$
View full question & answer→MCQ 711 Mark
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→MCQ 721 Mark
In $\angle\text{ABC}$ and $\angle\text{DEF},$ AB = DE and $\angle\text{A} = \angle\text{D}.$ Then, the two triangles will be congruent by SAS axiom if:
Answer- AC = DF
Solution:
The SAS rule states that:
If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.
Here, in $\angle\text{ABC},$ the two sides are AB and AC and the included angle is $\angle\text{A}.$ For $\angle\text{DEF},$ the two corresponding sides are DE and DF and the included angle is $\angle\text{D}.$
Hence, the two triangles will be congruent by SAS axiom if AC = DF.
View full question & answer→MCQ 731 Mark
Answer- 55º
Solution:
Since, It is given that AB = AC, then $\angle\text{B}=\angle\text{C}$ (Isosceles triangle property)
Given $\angle\text{A}=70^\circ$ Let angle B and C be xº.
Sum of all the three angles of triangle = 180º, therefore $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
70 + x + x = 180º
x = 55º
$\angle\text{C}=55^\circ$
View full question & answer→MCQ 741 Mark
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→MCQ 751 Mark
In triangles ABC and PQR, if $\angle\text{A}=\angle\text{R},\ \angle\text{B}=\angle\text{P}$ and AB = RP, then which one of the following congruence conditions applies:
Answer- ASA
Solution:
Since, the two adjacent angles and the contained side are shown equal while proving the two triangles congruent. Hence, by A.S.A. theorem the two triangles can be proved congruent.
View full question & answer→MCQ 761 Mark
Answer- 50º
Solution:
Since, $\triangle\text{ABC}$ and $\triangle\text{PQR}$ are similar triangles.
then, $\angle\text{B} = \angle\text{Q} = 83^\circ$
Thus, in $\triangle\text{ABC},$
$\angle\text{C} = 180^\circ - (\angle\text{A} + \angle\text{B})$
or, $\angle\text{C} = 180^\circ - (47^\circ + 83^\circ)$
$\angle\text{C}=50^\circ$
View full question & answer→MCQ 771 Mark
Answer- 57.5º
Solution:
As BC = AC, therefore triangle ABC is an isoscelestriangle.
Given $\angle\text{ACD} = 115^\circ,\ \angle\text{ACB} = 180 - 115 = 65^\circ$ (Linear Pair)
As AC = BC, therefore $\angle\text{A} =\angle\text{B}$
As sum of all the three angles of atriangle is 180º
Therefore. $\angle\text{A}+\angle\text{B} +\angle\text{ACB} = 180^\circ$
$\angle\text{A} =\angle\text{B}=57.5^\circ$
View full question & answer→MCQ 781 Mark
Answer- 50º
Solution:
$\angle\text{ACD} = 115^\circ,\ \angle\text{ACB} = 180-115=65^\circ$ (Linear Pair)
Since, It is given that AB = AC, then $\angle\text{ABC} = \angle\text{ACB}$ (Isosceles trangle property)
As $\angle\text{ACB} = 65^\circ,$ therefore $\angle\text{ACB} = 65^\circ$
Sum of all the three angles of triangle = 180º, therefore $\angle\text{ABC} + \angle\text{ACB} + \angle\text{A} = 180^\circ$
$\angle\text{A} = 180 - 65 - 65 = 50^\circ$
View full question & answer→MCQ 791 Mark
Answer- 115°
Solution:
In $\triangle\text{ABC}$ we have:
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ [Sum of the angles of a trianlge]
$⇒50^\circ+\angle\text{B}+\angle\text{C}=180^\circ$
$⇒\angle\text{B}+\angle\text{C}=130^\circ$
$\Rightarrow\ \frac{1}{2}\angle\text{B}\ +\ \frac{1}{2}\angle\text{C}=65^\circ\ ...\ (\text{i})$
In $\triangle\text{OBC},$ we have:
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ$
$\Rightarrow\ \frac{1}{2}\angle\text{B}\ +\ \frac{1}{2}\angle\text{C}\ +\ \angle\text{BOC}=180^\circ$
$⇒65^\circ+\angle\text{BOC}=180^\circ$
$⇒\angle\text{BOC}=115^\circ.$
View full question & answer→MCQ 801 Mark
Answer- AB + BC + AC > 2AD
Solution:
In triangle ADB
AB + BD > AD
In triangle ADC
AC + DC > AD
Adding both
AB + AC + BD + DC > 2AD
Now BD + DC = BC
So, AB + AC + BC > 2AD
View full question & answer→MCQ 811 Mark
- A
$\angle\text{CAD}$
- B
$\angle\text{BAD}$
- C
$\angle\text{BAC}$
- D
$\angle\text{ACD}$
Answer- $\angle\text{CAD}$
Solution:
As AB = CD,
so, $\angle\text{ACB}=\angle\text{CAD}$ (alternate angles).
View full question & answer→MCQ 821 Mark
If AB = QR, BC = RP and CA = PQ, then which of the following holds?
- A
$\triangle\text{ABC}\cong\triangle\text{PQR}$
- B
$\triangle\text{CBA}\cong\triangle\text{PQR}$
- C
$\triangle\text{CAB}\cong\triangle\text{PQR}$
- D
$\triangle\text{BCA}\cong\triangle\text{PQR}$
View full question & answer→MCQ 831 Mark
Answer- 22
Solution:
$\text{AB} ⊥ \text{BC}$
$⇒\ \angle\text{ABC}=90^\circ$
$\angle\text{CAB}=32^\circ$ (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 = 180°
$⇒ \angle\text{DAB} + \angle\text{ABD} + \angle\text{BDA} = 180^\circ$
$⇒\text{x}^\circ + ^\circ32°+90^\circ + \text{x}^\circ+14^\circ = 180^\circ$
$⇒\ 2\text{x}^\circ = 180^\circ - 136^\circ$
$⇒\ 2\text{x}^\circ = 44$
$⇒\ \text{x}^\circ= 22$
View full question & answer→MCQ 841 Mark
Answer- It is 1 : 1
Solution:
In $\triangle\text{ABC}$
$\text{AB = AC}$
$\therefore \angle\text{ABC} = \angle\text{ACB}$ (angles opposite to equal sides of a triangle are equal) ...(1)
In $\triangle\text{DBC},$
$\text{DB = DC},$
$\therefore \angle\text{DBC} = \angle\text{DCB}$ (angles opposite to equal sides of a triangle are equal) ...(2)
subtract 2 from 1
$\angle\text{ABC}-\angle\text{DBC} = \angle\text{ACB}-\angle\text{DCB}$ (equals subtracted from equals gives equal)
$= \angle\text{ABD} = \angle\text{ACD}$
Divide both the sides by $\triangle\text{ACD}$
$\Rightarrow\frac{\angle\text{ABD}}{\angle\text{ACD}}=1$
$\therefore \angle\text{ABD} : \angle\text{ACD}=1:1$
View full question & answer→MCQ 851 Mark
Answer- 75º
Solution:
Since DE is a straights line,
$\angle\text{ACB}+\angle\text{ACE}=180^\circ$
$\Rightarrow\angle\text{ACB}+130^\circ=180^\circ$
$\Rightarrow\angle\text{ACB}=50^\circ$
In $\triangle\text{ABC},$
$\angle\text{ABD}=\angle\text{BAC}+\angle\text{ACB}$ ......(Exterior angle is equal to sum of the remote interior angles)
$\Rightarrow125^\circ=\angle\text{BAC}+50^\circ$
$\Rightarrow\angle\text{BAC}=75^\circ$
that is, $\angle\text{A}=75^\circ.$
View full question & answer→MCQ 861 Mark
The length of two sides of a triangle are 7 units and 10 units. Which of the following length can be the length of the third side?
Answer- 13cm
Solution:
As per the rule in a triangle, sum of any 2 sides should be greater than the third side. So, the lenght of the third side should be 13, since with 7, 10 and 13 we have 7 + 10 > 13, 7 + 13 > 10 and 13 + 10 > 7.
View full question & answer→MCQ 871 Mark
Answer- 100º
Solution:
In $\triangle\text{RST}$
$\angle\text{R} + \angle\text{S} + \angle\text{T} = 180^\circ$
⇒ 2aº + xº + 2bº = 180º
⇒ xº = 180º - 2(a + b)º ... (i)
Now, in $\triangle\text{ROT}$
$\angle\text{ORT} + \angle\text{ROT} + \angle\text{OTR} = 180^\circ$
⇒ aº + 140º + bº = 180º
⇒ (a + b)º = 180º - 140º = 40º ...(ii)
From eq. (i) and (ii)
xº = 180º − 2(40º)
⇒ x = 100º
View full question & answer→MCQ 881 Mark
In $\triangle\text{ABC}$ and 6PQR, AB = PR and $\angle\text{A}=\angle\text{P}.$ Then, the two triangles will be congruent by SAS axiom if:
Answer- AC = PQ
Solution:
$\angle\text{A}$ is included between AB and AC and LP is included between PQ and PR and corresponding sides must be equal. Since AB = PR, hence AC = PQ for the given triangles to be congruent by SAS axiom.
View full question & answer→MCQ 891 Mark
Find the measure of each exterior angle of an equilateral triangle.
Answer- 120º
Solution:
We know that in equilateral triangle each angle is 60º
And we know sum of interior angle and exterior angle is 180º
Let exterior angle be x
60º + x = 180º
x = 180º - 60º
x = 120º
View full question & answer→MCQ 901 Mark
The sum of the interior angles of a triangle is:
Answer- 180º
Solution:
For a triangle,
Number of sides (n) = 3
Sum of interior angles = (n - 2) × 180º
= (3 - 2) × 180º
= 1 × 180º
= 180º
View full question & answer→MCQ 911 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:
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→MCQ 921 Mark
In a triangle ABC, $\angle\text{B} = 35^\circ$ and $\angle\text{C} = 60^\circ,$ then the shortest side is:
Answer- AC
Solution:
Side opposite to smallest angle is shortest side angles of triangle are 35, 60 and 85.
View full question & answer→MCQ 931 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 :
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→MCQ 941 Mark
In $\triangle\text{PQR},\ \angle\text{P} = 60^\circ,\ \angle\text{Q} = 50^\circ.$ Which side of the triangle is the longest?
Answer- PQ
Solution:
In $\triangle\text{PQR},\ \angle\text{P} = 60^\circ,\ \angle\text{Q} = 50^\circ.$
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.
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View full question & answer→MCQ 951 Mark
Answer- PR = QS
Solution:
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 961 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:
Answer- ASA
Solution:
From 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 971 Mark
Answer- OQ > OR
Solution:
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 981 Mark
It is given that$\triangle\text{ABC}\cong\triangle\text{FDE}$ in which AB = 5cm, $\angle\text{B}=40^{\circ},\angle\text{A}=80^{\circ}$and FD = 5cm. Then, which of the following is true?
- A
$\angle\text{D}=60^{\circ}$
- B
$\angle\text{E}=60^{\circ}$
- C
$\angle\text{F}=60^{\circ}$
- D
$\angle\text{D}=80^{\circ}$
View full question & answer→MCQ 991 Mark
Answer- 1 : 1
Solution:
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 1001 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 $\angle\text{BOC} =$
View full question & answer→
