MCQ 11 Mark
If $\triangle\text{ABC}\cong\triangle\text{ACB},$ then $\triangle\text{ABC}$ is isosceles with.
- AAB = AC
- BAB = BC
- CAC = BC
- DNone of these
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MCQ 21 Mark
If $\triangle\text{ABC}$ is an isosceles triangle with AB = AC and $\angle\text{B} = 65^\circ,$ find $\angle\text{A}.$
- A70º
- B50º
- C60º
- DNone of these
Answer
View full question & answer→- 50º
Solution:
In isosceles triangle ABC
AB = AC (Given)
Therefore $\angle\text{B} = \angle\text{C} = 65^\circ$ (angles opposite to equal side are equal).
So, by applying angle sum property i.e $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ,$
$\angle\text{A} + 65^\circ + 65^\circ = 180^\circ$
$\angle\text{A} = 180^\circ - 130^\circ$
$\angle\text{A} = 50^\circ$
MCQ 31 Mark
- A$\triangle\text{ABC}\cong\triangle\text{ABD}$
- B$\angle\text{C}=\angle\text{D}$
- C$\text{All are true}$
- D$\text{BC = BD}$
Answer
View full question & answer→- $\text{All are true}$
Solution:
$\text{AC = AD}$
$\angle\text{AB}= \angle\text{BAD}$
$\text{AB = AB}$
By SAS, we have
$\triangle\text{ABC}\cong\triangle\text{ABD}$
Hence, we have $\text{BC = BD}$ and $\angle\text{C}= \angle\text{D}.$
So, all the given options are true.
MCQ 41 Mark
If $\triangle\text{ABC}≅\triangle\text{LKM},$ then side of $\triangle\text{LKM}$ equal to side AC of $\triangle\text{ABC}$ is:
- ALM
- BKM
- CNone of these
- DLK
Answer
View full question & answer→- LM
Solution:
Since, by corresponding part of congruent triangle AC of $\triangle\text{ABC}$ is equal to the LM of $\triangle\text{LKM}.$
MCQ 51 Mark
If all the three angles of a triangle are equal, then each one of them is equal to:
- A90º
- B45º
- C60º
- D30º
Answer
View full question & answer→- 60º
Solution:
Let the measure of each angle be x°.
Now, the sum of all angles of any triangle is 180°.
Thus, x° + x° + x° = 180°
i.e. 3x° = 180°
i.e. x° = 60°
MCQ 61 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:
In $\triangle\text{ABC}$ if AB = AC and $\angle\text{B}=50^{\circ}$ then is equal to:
- A40º
- B50º
- C80º
- D130º
Answer
View full question & answer→- 50º
Solution: Given $\triangle\text{ABC}$ such that AB = AC and $\angle\text{B}=50^{\circ}$
MCQ 71 Mark
In the following, write the correct answer.
It is given that $\triangle\text{ABC}=\triangle\text{FDE}$ and AB = 5 cm, $∠\text{B} = 40°$and $∠\text{A} = 80°$then which of the following is true?
It is given that $\triangle\text{ABC}=\triangle\text{FDE}$ and AB = 5 cm, $∠\text{B} = 40°$and $∠\text{A} = 80°$then which of the following is true?
- A$\text{DF}=5\text{CM}, \angle\text{F}=60^{\circ}$
- B$\text{DF}=5\text{CM}, \angle\text{E}=60^{\circ}$
- C$\text{DE}=5\text{CM}, \angle\text{E}=60^{\circ}$
- D$\text{DE}=5\text{CM}, \angle\text{D}=40^{\circ}$
MCQ 81 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?
- A30º
- B25º
- C45º
- D60º
Answer
View full question & answer→- 45º
Solution:
The measures of angles of a triangle are in ratio 3: 4: 5.
Let the angles be 3x, 4x and 5x.
In any triangle, sum of all angles = 180º
⇒ 3x + 4x + 5x = 180º
⇒ 12x = 180º
⇒ x = 15º
So, smallest angle = 3 × 15º = 45º
MCQ 91 Mark
If ABC and DEF are two triangles such that $\triangle\text{ABC}\cong\triangle\text{FDE}$ and $\text{AB}=5\text{m},\angle\text{B}=40^\circ$ and $\angle\text{A}=80^\circ.$ Then, which of the following is true?
- A$\text{DF}=5\text{cm},\angle\text{F}=60^\circ$
- B$\text{DE}=5\text{cm},\angle\text{E}=60^\circ$
- C$\text{DF}=5\text{cm},\angle\text{E}=60^\circ$
- D$\text{DE}=5\text{cm},\angle\text{D}=40^\circ$
Answer
View full question & answer→- $\text{DF}=5\text{cm},\angle\text{E}=60^\circ$
Solution:
In $\triangle\text{ABC},$
$\angle\text{C}=180^\circ-\angle\text{A}+\angle\text{B}=180^\circ-80^\circ-40^\circ=60^\circ$
$\triangle\text{ABC}\cong\triangle\text{FDE}$
$\Rightarrow\text{AB}=\text{FD}=5\text{cm}$
$\Rightarrow\angle\text{B}=\angle\text{D}=40^\circ$
$\Rightarrow\angle\text{A}=\angle\text{F}=80^\circ$
$\Rightarrow\angle\text{C}=\angle\text{E}=60^\circ$
$\Rightarrow\text{DF}=\text{FD}=5\text{cm}$ and $\angle\text{E}=60^\circ$
Hence, correct option is (c).
MCQ 101 Mark
- A25º
- B20º
- C30º
- D35º
Answer
View full question & answer→- 30º
Solution:
$\angle\text{PBC} = \angle\text{QCD}$ (Corresponding angles, $\text{OP || CQ}$ and BC is transverse)
$⇒\angle\text{PBC} = 70^\circ$
Now, $\angle\text{PBA} + \angle\text{ABC} = \angle\text{PBC}$
$⇒20^\circ+\angle\text{ABC}=70^\circ$
$⇒\angle\text{ABC}=50^\circ$
In $\triangle\text{ABC},$
$\angle\text{ABC} + \angle\text{BAC} + \angle\text{ACB} = 180^\circ \ ...\ (\text{i})$
Now, $\angle\text{ABC} + \angle\text{BAC} = 50^\circ$ (isosceles $\triangle$)
And, $\angle\text{ACB} = 180^\circ - (70^\circ + \text{x})$
From (i),
50º + 50º + 180º - (70º + x) = 180º
Hence x = 30º
MCQ 111 Mark
If $\triangle\text{PQR}≡\triangle\text{EFD},$ then $\angle\text{E}=$
- A$\angle\text{Q}$
- B$\angle\text{R}$
- C$\text{None of these}$
- D$\angle\text{P}$
Answer
View full question & answer→- $\angle\text{P}$
Solution:
Since, by corresponding part of congruent $\angle\text{E}$ of $\triangle\text{EFD}$ is equal to the $\angle\text{P}$ of $\triangle\text{PQR}.$
MCQ 121 Mark
In Fig. ABC is a triangle in which $\angle\text{B}=2\angle\text{C}.$ D is a point on side BC such that AD bisects $\angle\text{BAC}$ and $\text{AB}=\text{CD}.$ Be is the bisector of $\angle\text{B}.$ The measure of $\angle\text{BAC}$ is:
- A72º
- B95º
- C73º
- D74º
MCQ 131 Mark
In $\triangle\text{ABC, BC = AB}$ and $\angle\text{B}=80^{\circ}.$ Then, $\angle\text{A = ?}$
- A50°
- B40°
- C100°
- D80°
Answer
View full question & answer→- 50°
Solution:
In $\triangle\text{ABC,}$
$\text{BC = AB}$
$\Rightarrow\angle\text{A}=\angle\text{C}$ (angles opposite to equal sides are equal)
Now, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{A}+80^{\circ}+\angle\text{A}=180^{\circ}$
$\Rightarrow2\angle\text{A}+100^{\circ}$
$\Rightarrow\angle\text{A}=50^{\circ}$
MCQ 141 Mark
An exterior angle of a triangle is 108º and its interior opposite angles are in the ratio 4 : 5. The angles of the triangle are:
- A42º, 60º, 76º
- B48º, 60º, 72º
- C50º, 60º, 70º
- D2º, 56º, 72º
MCQ 151 Mark
In $\triangle\text{ABC},\ \angle\text{A} = 50^\circ,\ \angle\text{B} = 60^\circ.$ Find the longest side of the triangle?
- ABC
- BAB
- CCannot be determined
- DCA
Answer
View full question & answer→- AB
Solution:
By angle sum property, we have,
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$\Rightarrow 50^\circ + 60^\circ + \angle\text{C} = 180^\circ$
$\Rightarrow\ \angle\text{C} = 180^\circ - (50^\circ+ 60^\circ) = 70^\circ$
Therefore, $\angle\text{C}$ is the largest angle in the triangle and the side opposite to it i.e. AB is the longest side.
MCQ 161 Mark
- A70°
- BNone of these
- C50°
- D60°
Answer
View full question & answer→- 60°
Solution:
In $\triangle\text{ABC}, \ \text{AB} = \text{AC}$
$⇒ \angle\text{ABC} = \angle\text{ACB}$
Also, $\angle\text{ACD}=120^\circ$
$⇒ \angle\text{ACB} = 180^\circ- \angle\text{ACD}$ (Linear pair)
$⇒ \angle\text{ACB} = 180^\circ- 120^\circ = 60^\circ$
$⇒ \angle\text{ABC} = 60^\circ$
By using angle sum property, we have
$\angle\text{ABC} + \angle\text{ACB} + \angle\text{BAC} = 180^\circ$
$60^\circ + 60^\circ+ \angle\text{A} = 180^\circ$
or, $\angle\text{A} = 180^\circ - 120^\circ= 60^\circ$
MCQ 171 Mark
- Ax + y + 180
- B180 - (x + y)
- Cx + y - 180
- Dx + y + 360º
Answer
View full question & answer→- x + y - 180
Solution:
In the given figure, $\angle\text{ABF}+\angle\text{ABC}=180^\circ$ (Linear pair of angles)
$∴\text{x}^\circ+\angle\text{ABC}=180^\circ$
$⇒\angle\text{ABC}=180^\circ−\text{x}°...\ (\text{i})$
Also, $\angle\text{ACG}+\angle\text{ACB}=180^\circ$ (Linear pair of angles)
$∴\text{y}′+\angle\text{ACB}=180^\circ$
$⇒\angle\text{ACB}=180^\circ−\text{y}′\ ...\ \text{(ii)}$
Also, $\angle\text{BAC}=\angle\text{DAE}=\text{z}° \ ....\ \text{(iii)}$ (Vertically opposite angles)
In $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$∴\text{z}^\circ+180−\text{x}^\circ+180^\circ−\text{y}^\circ=180^\circ$ [Using (1), (2) and (3)]
$⇒ \text{z} = \text{x} + \text{y} – 180$
MCQ 181 Mark
In $\triangle\text{ABC},$ if $\angle\text{C}>\angle\text{B},$ then
- ABC > AC
- BAB > AC
- CBC < AC
- DAB < AC
Answer
View full question & answer→- AB > AC
Solution:
Greater angle has greater side opposite to it.
MCQ 191 Mark
- A$\angle\text{BAD}$
- B$\angle\text{BAC}$
- C$\angle\text{BDA}$
- D$\angle\text{ABD}$
Answer
View full question & answer→- $\angle\text{BDA}$
Solution:
The two triangles are congruent according to (SAS CONGRUENCY) as AD = BC (given), $\angle\text{BAD} = \angle\text{ABC},$ (given) and AB = AB (common) and hence corresponding angles are equal (cpct).
MCQ 201 Mark
- A270º
- B230º
- C210º
- D190º
Answer
View full question & answer→- 230º
Solution:
$\triangle\text{ACO}$
$\angle\text{ACO}+\angle\text{COA}+\angle\text{COA}=180^\circ$
Now, $\angle\text{OAC}=180^\circ-\text{x}^\circ$
$\Rightarrow80^\circ+40^\circ+180^\circ-\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=120^\circ$
$\angle\text{BOD}=\angle\text{COA}=40^\circ$ (Opposite angles)
$\angle\text{BDO}=70^\circ$
In $\triangle\text{OBD},$
$\angle\text{OBD}=180^\circ-40^\circ-70^\circ=70^\circ$
Also, $\text{y}^\circ=180^\circ-\angle\text{OBD}=180^\circ-70^\circ=110^\circ$
$\Rightarrow\text{x}^\circ+\text{y}^\circ=120^\circ+110^\circ=230^\circ$
MCQ 211 Mark
If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is:
- AEquilateral
- BIsosceles
- CScalene
- DRight-angled
MCQ 221 Mark
- A60
- B45
- C50
- D35
MCQ 231 Mark
- A$90^\circ-\frac{\text{x}^\circ}{2}$
- B$180^\circ+\frac{\text{x}^\circ}{2}$
- C$90^\circ+\frac{\text{x}^\circ}{2}$
- D$180^\circ-\frac{\text{x}^\circ}{2}$
Answer
View full question & answer→- $90^\circ-\frac{\text{x}^\circ}{2}$
Solution:
In $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{B}+\angle\text{C}=180^\circ-\text{x}^\circ\ ...\ \text{(i)}$
$\angle\text{CBD}=180^\circ-\angle\text{B}\ ...\ \text{(ii)}$
$\angle\text{ECB}=180^\circ-\angle\text{C}\ ...\ \text{(iii)}$
$\Rightarrow\ \frac{\angle\text{CBD}}{2}=\angle\text{OBC}=90^\circ-\frac{\angle\text{B}}{2}\ ...\ \text{(iv)}$
$\Rightarrow\ \frac{\angle\text{ECB}}{2}=\angle\text{OCB}=90^\circ-\frac{\angle\text{C}}{2}\ ...\ \text{(v)}$
Now, in $\triangle\text{BOC}$
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ$
$\angle\text{BOC}=180^\circ-(\angle\text{OBC}\ +\ \angle\text{OCB})$
From eq. (iv) and (v)
$\angle\text{BOC}=180^\circ-\Big(90^\circ-\frac{\angle\text{B}}{2}+90^\circ-\frac{\angle\text{C}}{2}\Big)$
$=\ 180^\circ-\Big(180^\circ-\frac{\angle\text{B}}{2}-\frac{\angle\text{C}}{2}\Big)$
$=\frac{\angle\text{B}+\angle\text{C}}{2}$
$=\frac{180^\circ+\text{x}^\circ}{2}$
$\angle\text{BOC}=90^\circ-\frac{\text{x}^\circ}{2}$
MCQ 241 Mark
D is a point on the side BC of a $\triangle\text{ABC}$ such that AD bisects $\triangle\text{BAC}$ then:
- ACD > CA
- BBD = CD
- CBD > BA
- DBA > BD
Answer
View full question & answer→- BA > BD
Solution:
Since, $\triangle\text{BAC}$ is bisected by AD, then $\triangle\text{BAD}$ is less than $\triangle\text{ABC},$ hence the side opposite $\triangle\text{ABC},$ i.e., BA is greater than the side opposite to $\triangle\text{BAD}$ i.e., BD.
MCQ 251 Mark
The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94º and 126º. Then, $\angle\text{BAC} =$
- A54º
- B94º
- C44º
- D40º
MCQ 261 Mark
D, E, F are the mid-point of the sides BC, CA and AB respectively of $\triangle\text{ABC}.$ Then $\triangle\text{DEF}$ is congruent to triangle.
- AABC
- BAEF
- CBFD, CDE
- DAFE, BFD, CDE
Answer
View full question & answer→- AFE, BFD, CDE
Solution:
In anytriangle, a line joining the mid-points of any two sides is parallel to the third side.
⇒ EF || BC EF || DC and BD
Similiarly DF || EC
⇒DF || AE and EC
Also DE || AB.
⇒ DE || AF and BF
From this information it is clear that EFDC, EFBD, EAFD
are the parallelogram by property.
Now consider one parallelogram EFDC
Consider $\triangle\text{DEF}$ and $\triangle\text{EDC}$
$\text{DE}=\text{ED}$ (common)
$\angle\text{DEF}=\angle\text{EDC}$
$\angle\text{EDF}=\angle\text{DEC}$ (ASA property)
$\Rightarrow\triangle\text{DEF}\cong\triangle\text{EDC}$
Similiarly in parallelogram EAFD,
$\triangle\text{DEF}\cong\triangle\text{AFC}$
And in parallelogram EFBD
$\triangle\text{DEF}\cong\triangle\text{FBD}$
Hence, correct option is (d).
Note: Option (d) modified.
MCQ 271 Mark
An exterior angle of a triangle is 108º and its interior opposite angles are in the ratio 4 : 5 The angles of the triangle are:
- A48º, 60º, 72º
- B50º, 60º, 70º
- C52º, 56º, 72º
- D42º, 60º, 76º
MCQ 281 Mark
Find the measure of angles which is equal to its supplement.
- A45º
- B60º
- C120º
- D90º
Answer
View full question & answer→- 90º
Solution:
We know that the sum of supplementary angle is 180º
Let the angle be x the other angle is 180° - x and the two angles are equal,
⇒ x = 180º - x
⇒ 2x = 180º
$\Rightarrow\ \text{X}=\frac{180^\circ}{2}$
$\Rightarrow\ \text{X}=90^\circ$
MCQ 291 Mark
- A$\angle\text{BDA}$
- B$\angle\text{BAC}$
- C$\angle\text{ABD}$
- D$\angle\text{BAD}$
Answer
View full question & answer→- $\angle\text{BDA}$
Solution:
The two triangles are congruent according to (SAS CONGRUENCY) as AD = BC (given), $\angle\text{BAD} = \angle\text{ABC},$ (given) and AB = AB (common) and hence corresponding angles are equal (cpct).
MCQ 301 Mark
In $\triangle\text{ABC},$ BC = AB and $\angle\text{B} = 80^\circ.$ Then $\angle\text{A}$ is equal to:
- A100º
- B80º
- C40º
- D50º
MCQ 311 Mark
If the altitudes from two vertices of a triangle to the opposite sides are equal then the triangle is:
- ARight angled
- BEquilatera
- CScalene
- DIsosceles
MCQ 321 Mark
In $\triangle\text{ABC},$ if $\angle\text{A} = 45^\circ$ and $\angle\text{B} = 70^\circ,$ then the shortest and the longest sides of the triangle are ________ .
- ABC, AB
- BBC, AC
- CAB, BC
- DAB, AC
Answer
View full question & answer→- BC, AC
Solution:
Smallest angle is A and greatest angle is B and hence sides opposite to these angles are BC and AC and they are shortest and longest respectively.
MCQ 331 Mark
- A30º
- B105º
- C150º
- D72º
Answer
View full question & answer→- 105º
Solution:
Join AC. We get two isosceles triangles. $\triangle\text{ABC}$ and $\triangle\text{ACD}.$
In $\triangle\text{ABC},\ \angle\text{ABC}= 108^\circ$
$\therefore\angle\text{BAC} = \angle\text{BCA} = \frac{(180^\circ -108^\circ)}{2} = \frac{72^\circ}{2} = 36^\circ$
In $\triangle\text{ACD},\ \angle\text{ADC}= 42^\circ$
$\therefore\angle\text{DAC} = \angle\text{DCA} = \frac{(180^\circ -42^\circ)}{2} = \frac{138^\circ}{2} = 69^\circ$
Now, $\angle\text{BCD} = \angle\text{BCA} + \angle\text{DCA} = 36^\circ + 69^\circ = 105^\circ$
MCQ 341 Mark
In $\angle\text{ABC}, \angle\text{A}=40^\circ$ and $\angle\text{B}=60^\circ$ Then, the longest side of $\triangle\text{ABC}$ is:
- AAC
- BCannot be determined
- CBC
- DAB
Answer
View full question & answer→- AB
Solution:
The third angle $\angle\text{C}=80^\circ$ [Angle sum property]
The side opposite to the longest angle is the longest side.
Therefore, AB is the longest side of the triangle.
MCQ 351 Mark
In $\triangle\text{ABC,}$ if $\angle\text{C}>\angle\text{B},$ then:
- ABC > AC
- BAB > AC
- CAB < AC
- DBC < AC
Answer
View full question & answer→- AB > AC
Solution:
We know that in a triangle, the greater angle has the longer side opposite to it.
In $\triangle\text{ABC},$
$\angle\text{C}>\angle\text{B}$
$\Rightarrow\text{AB > AC}$
MCQ 361 Mark
- ANone of these
- B8cm
- C4cm
- D32cm
Answer
View full question & answer→- 8cm
Solution:
Using relation
perimeter, $\triangle\text{DEF}=\frac{1}{2}$
perimeter, $\triangle\text{ABC}$
$=\frac{1}{2}×16=8\text{cm}$
MCQ 371 Mark
- ASSS
- BASA
- CRHS
- DSAS
Answer
View full question & answer→- SSS
Solution:
Given that two sides are equal and third side is common I.e. AD hence all three corresponding sides are equal.
MCQ 381 Mark
In a $\triangle\text{ABC},$ if $3\angle\text{A}=4\angle\text{B}=6\angle\text{C}$ then A : B : C = ?
- A4 : 3 : 2
- B6 : 4 : 3
- C2 : 3 : 4
- D3 : 4 : 6
Answer
View full question & answer→- 3 : 4 : 6
Solution:
In the given figure, $\angle\text{ACB}+\angle\text{ACD}=180^\circ$ (Linear pair of angles)
$∴$ 5yº + 7yº = 180º
⇒ 12yº = 180º
⇒ y = 15
In $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$∴$ 3yº + xº + 5yº = 180º
⇒ xº + 8yº = 180º
⇒ xº + 8 × 15º = 180º
⇒ xº + 120º = 180º
⇒ xº = 180º − 120º = 60º
Thus, the value of x is 60º.
Hence the correct answer is 3 : 4 : 6
MCQ 391 Mark
- A320º
- B240º
- C300º
- D360º
Answer
View full question & answer→- 360º
Solution:
We have:
$\angle1+\angle\text{BAE}=180^\circ\ ...\ \text{(i)}$
$\angle2+\angle\text{CBF}=180^\circ ...\ \text{(ii)}$
$\angle\text{3}+\angle\text{ACD}=180^\circ ...\ \text{(iii)}$
Adding (i),(ii) and (iii), we get:
$(\angle1+\angle2+\angle3)+(\angle\text{BAE}+\angle\text{CBF}+\angle\text{ACD})=540^\circ$
$⇒180^\circ+\angle\text{BAE}+\angle\text{CBF}+\angle\text{ACD}=540^\circ$ $ [∵ \angle1+\angle2+\angle3=180^\circ]$
$⇒\angle\text{BAE}+\angle\text{CBF}+\angle\text{ACD}=360^\circ.$
MCQ 401 Mark
- ASAS
- BASA
- CSSS
- DRHS
Answer
View full question & answer→- RHS
Solution:
In $\triangle\text{ABD}$ and $\triangle\text{ACD},$ we have
$\angle\text{ADB} = \angle\text{ADC}$ (Right angles)
AB = AC (Given and hypotenuses)
AD = AD (common in both)
Therefore, $\triangle\text{ABD}\cong\triangle\text{ACD}$ by RHS.
MCQ 411 Mark
If $\triangle\text{PQR}≅\triangle\text{EFD},$ then ED =
- APR
- BQR
- CNone of these
- DPQ
Answer
View full question & answer→- PR
Solution:
Since, by corresponding part of congruent triangle ED of $\triangle\text{EFD}$ is equal to the PR of $\triangle\text{PQR}.$
MCQ 421 Mark
- A40º
- B70º
- C110º
- D80º
Answer
View full question & answer→- 40º
Solution:
Since, It is given that PQ = QR, then $\angle\text{Q} = \angle\text{R}$ (Isosceles trangle property)
As $\angle\text{Q} = 70^\circ,$ therefore $\angle\text{R} = 70^\circ,$
Sum of all the three angles of triangle = 180º, therefore $\angle\text{P}+\angle\text{Q}+\angle\text{R}=180^\circ$
$\angle\text{P} = 180 - 70 - 70\ = \ 40^\circ$
MCQ 431 Mark
- A$\angle\text{BAD}$
- B$\angle\text{BCA}$
- C$\angle\text{BDA}$
- D$\angle\text{BCD}$
Answer
View full question & answer→- $\angle\text{BDA}$
Solution:
In triangle ABD and CBD
AB = BC and $\angle\text{ABD} = \angle\text{CBD},$ (Given)
BD (Common)
Therefore In triangle ABD and CBD are congruent by SAS criteria.
Therefore, $\angle\text{BDA}=30^\circ$ (by CPCT)
MCQ 441 Mark
Two sides of a triangle are of lengths 5cm and 1.5cm. The length of the third side of the triangle cannot be.
- A3.4cm
- B4cm
- C3.8cm
- D3.6cm
Answer
View full question & answer→- 3.4cm
Solution:
Given that: Two sides of triangle are 5cm and 1.5cm. We know that the sum of two sides of the triangle is always greater than the third side. Hence, 3.4cm cannot be the third side. If it is the third side the sum of 3.4cm and 1.5cm will be smaller than 5cm, so, the triangle will not be possible.
MCQ 451 Mark
In a $\triangle\text{ABC},$ if $\text{AB}=\text{AC}$ and BC is produced to D such that $\angle\text{ACD}=100^\circ$ then $\angle\text{A}=$
- A20°
- B40°
- C60°
- D80°
MCQ 461 Mark
- ASAS
- BASA
- CSSS
- DRHS
MCQ 471 Mark
- A70º
- B90º
- C100º
- D60º
MCQ 481 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 $\angle\text{E}=$
- A25º
- B50º
- C100º
- D75º
MCQ 491 Mark
- A70º
- B50º
- C57.5º
- D65º
Answer
View full question & 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 180°
Therefore, $\angle\text{A} + \angle\text{B} + \angle\text{ACB} = 180^\circ$
$\angle\text{A} = \angle\text{B} = 57.5$
MCQ 501 Mark
In a $\triangle\text{ABC},$ If $\angle\text{A}= 45^\circ$ and $\angle\text{B}= 70^\circ.$ Determine the shortest sides of the triangles.
- AAC
- BBC
- CCA
- DAll are equal.
Answer
View full question & answer→- BC
Solution:
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{A}=45^\circ$
$\angle\text{B}=70^\circ$
$\angle\text{C}+45^\circ+70^\circ=180^\circ$
$\angle\text{C}+115^\circ=180^\circ$
$\angle\text{C}=180^\circ-115^\circ$
$\angle\text{C}=65^\circ$
$\angle\text{A}$ is shortest angle and the side opp to shortest angle is shortest.
So, BC is the shotest side.