MCQ 11 Mark
Which of the following is the set of measures of the sides of a triangle$?$
- A
$8\ cm, 4\ cm, 20\ cm$
- ✓
$9\ cm, 17\ cm, 25\ cm$
- C
$11\ cm, 16\ cm, 28\ cm$
- D
AnswerCorrect option: B. $9\ cm, 17\ cm, 25\ cm$
We knwno that Triangle Inequality Theorem states that the sum of two side lengths of a triangle is always greater than the third side.
Using this in $(a),$ we get
$8 + 4 \geq20$
$\Rightarrow 12 \geq20$
So, triangle is not possible
Using this in $(b),$ we get
$9 + 17 > 25$
$\Rightarrow 26 > 25$
and,
$9 + 25 > 7$
$\Rightarrow 34 > 7$
and,
$17 + 25 > 9$
$\Rightarrow 42 > 9$
So, triangle is possible.
Using this in $(c),$ we get
$11 + 16\geq 28$
$\Rightarrow 27 \geq 28$
So, triangle is not possible.
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 21 Mark
Which of the following is/ are not Pythagorean triplet $(s)?$
- A
$3, 4, 5$
- B
$8, 15, 17$
- C
$7, 24, 25$
- ✓
$13, 26, 29$
AnswerCorrect option: D. $13, 26, 29$
In $(a)$
$3^2+4^2=5^2$
$\Rightarrow 9+16=25$
$\Rightarrow 25=25$
Since, the sum of the square of two smallest number is equal to the square of largest number.
Hence, it is a Pythagorean triplet.
In $(b)$
$8^2+15^2=17^2$
$\Rightarrow 64+225=289$
$\Rightarrow 289=289$
Since, the sum of the square of two smallest number is equal to the square of largest number.
Hence, it is a Pythagorean triplet.
In $(c)$
$7^2+24^2=25^2$
$\Rightarrow 49+576=625$
$\Rightarrow 625=625$
Since, the sum of the square of two smallest number is equal to the square of largest number.
Hence, it is a Pythagorean triplet.
In $(d)$
$13^2+26^2 \neq 29^2$
$\Rightarrow 169+676 \neq 841$
$\Rightarrow 845 \neq 841$
Since, the sum of the square of two smallest number is not equal to the square of largest number.
Hence, it is not a Pythagorean triplet.
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 31 Mark
In a $\triangle \text{ABC},$ if $2\angle \text{A}=3\angle \text{B}=6\angle \text{C},$ then th s measure of the smallest angle is:
- A
$90^\circ$
- B
$60^\circ $
- C
$40^\circ$
- ✓
$30^\circ$
AnswerCorrect option: D. $30^\circ$
We have,
$2\angle \text{A}=3\angle \text{B}=6\angle \text{C}$
$\therefore 3\angle \text{B}=2\angle \text{A}$ and $6\angle \text{C}=2\angle \text{A}$
$\Rightarrow \angle \text{B}=\frac{2}{3}\angle \text{A}$ and $\angle \text{C}=\frac{2}{6}\angle \text{A}=\frac{1}{3}\angle \text{A}$
Now, $\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow \angle \text{A}+\frac{2}{3}\angle \text{A}+\frac{1}{3}\angle \text{A}=180^\circ$
$\Rightarrow 3\angle \text{A}+2\angle \text{A}+\angle \text{A}=180^\circ\times3$
$\Rightarrow 6\angle \text{A}=540^\circ$
$\Rightarrow \angle \text{A}=90^\circ$
$\therefore$ Smallest angle $=\angle \text{C}=\frac{1}{3}\angle \text{A}=\frac{1}{3}\times90^\circ$
$=30^\circ$
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 41 Mark
If the exterior angles of a triangle are $(2x + 10)^\circ , (3x - 5)^\circ $ and $(2x + 40)^\circ ,$ then $x =$
AnswerSum of the exterior angles of a triangle is $360^\circ $
$\therefore (2x + 10)^\circ + (3x - 5)^\circ + (2x + 40)^\circ = 360^\circ $
$\Rightarrow 2x + 10 + 3x - 5 + 2x + 40 = 360$
$\Rightarrow 7x + 45 = 360$
$\Rightarrow 7x = 315$
$\Rightarrow x = 45$
Hence, the correct answer is option $(c)$
View full question & answer→MCQ 51 Mark
In Fig. the value of $x$ is:
Answer$\angle \text{TRS}+\angle \text{TRQ}=180^\circ$ [Linear angles]
$\Rightarrow 5\text{x}^\circ+\angle \text{TRQ}=180^\circ$
$\Rightarrow \angle \text{TEQ}=180^\circ-5\text{x}^\circ$
Now, $\angle \text{QTR}+\angle \text{TRQ}=\angle \text{PQT}$ [Exterior angle property of triangle]
$\Rightarrow 3\text{x}^\circ+180^\circ-5\text{x}^\circ=120^\circ$
$\Rightarrow 2\text{x}^\circ=60^\circ$
$\Rightarrow \text{x}=30$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 61 Mark
In Fig. the values of $x$ and $y$ are:
- A
$x = 120, y = 150$
- B
$x = 110, y = 160$
- ✓
$x = 150, y = 120$
- D
$x = 110, y = 160$
AnswerCorrect option: C. $x = 150, y = 120$
In $\triangle \text{DEF}$
$\angle \text{DEF}+\angle \text{DFE}+\angle \text{EDF}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 110^\circ+40^\circ+\angle \text{EDF}=180^\circ$
$\Rightarrow \angle \text{EDF}=30^\circ$
Now, $\angle \text{EDF}+\angle \text{FDA}=180^\circ$ [Linear pair angles]
$\Rightarrow 30^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}=150$
Now, $\angle \text{EDF}=\angle \text{ADB}=30^\circ$ [Vertically opposite angles]
Now, In $\triangle \text{ABD},$
$\angle \text{ADB}+\angle \text{DAB}+\angle \text{ABD}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 30^\circ+90^\circ+\angle \text{ABD}=180^\circ$
$\Rightarrow \angle \text{ABD}=60^\circ$
Now, $\angle \text{ABD}+\angle \text{DBC}=180^\circ$ [Linear pair angles]
$\Rightarrow 60^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}=120$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 71 Mark
Fig. if $AB || CD,$ then the values of $x, y$ and $z$ are:
- A
$x = 56, y = 47, z = 77$
- B
$x = 47, y = 56, z = 77$
- C
$x = 77, y = 56, z = 47$
- ✓
$x = 56, y = 77, z = 47$
AnswerCorrect option: D. $x = 56, y = 77, z = 47$
$\angle \text{AFE}+\angle \text{EFG}=180^\circ$ [Linear pair angles]
$\Rightarrow 124^\circ+\angle\text{EFG}=180^\circ$
$\Rightarrow \angle\text{EFG}=56^\circ$
Since, $AB || CD$
$\therefore \angle \text{EFG}=\angle \text{FHK}=56^\circ$ [Corresponding angles]
$\Rightarrow \text{x}=56$
Now, $\angle \text{QKH}+\angle \text{GKH}=180^\circ$ [Linear pair angles]
$\Rightarrow 103^\circ+\angle \text{GKH}=180^\circ$
$\Rightarrow \angle \text{GKH}=77^\circ$
Since, $AB || CD$
$\therefore \angle \text{EGF}=\angle \text{GKH}=77^\circ$ [Corresponding angles]
$\Rightarrow \text{y}=77$
In $\triangle \text{EHK},$
$\angle \text{EHK}+\angle \text{EKH}+\angle \text{HEK}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 56^\circ+77^\circ+\text{z}^\circ=180^\circ$
$\Rightarrow \text{z}=47$
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 81 Mark
Two poles of heights $6m$ and $11m$ stand vertically on a plane ground. If the distance between their feet is $12m,$ the distance between their tops is:
- ✓
$13m$
- B
$14m$
- C
$15m$
- D
$12.8m$
AnswerSuppose $AB$ and $CD$ are two poles.The is distance between $AB$ and $CD$ is $12m.$
In right traingle $B D E$,
$\mathrm{BD}^2=\mathrm{DE}^2+\mathrm{BE}^2$
$\Rightarrow \mathrm{BD}^2=(5)^2+(12)^2$
$\Rightarrow \mathrm{BD}^2=25+144$
$\Rightarrow \mathrm{BD}^2=169$
$\Rightarrow \mathrm{BD}^2=(13)^2$
$\Rightarrow \mathrm{BD}=13 \mathrm{~m}$
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 91 Mark
$\triangle \text{ABC}$ is a right triangle right angled at $A$. If $AB = 24\ cm$ and $AC = 7\ cm,$ then $BC =$
- A
$31\ cm$
- B
$17\ cm$
- ✓
$25\ cm$
- D
$28\ cm$
AnswerCorrect option: C. $25\ cm$
In right traingle $ABC,$
$B C^2=A C^2+A B^2$
$\Rightarrow B C^2=(7)^2+(24)^2$
$\Rightarrow B C^2=49+576$
$\Rightarrow B C^2=625$
$\Rightarrow B C^2=(25)^2$
$\Rightarrow B C=25 \mathrm{~cm}$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 101 Mark
If the measures of the angles of a triangle are $(2x)^\circ , (3x - 5)^\circ $ and $(4x -13)^\circ .$ Then the value of $x$ is:
Answer$(2x)^\circ + (3x - 5)^\circ + (4x - 13)^\circ = 180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 2x + 3x - 5 + 4x - 13 = 180^\circ $
$\Rightarrow 9x - 18 = 180$
$\Rightarrow 9x = 198$
$\Rightarrow x = 22$
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 111 Mark
The angles of a triangle are in the ratio $2 : 3 : 7.$ The measure of the largest angle is:
- A
$84^\circ$
- B
$91^\circ$
- ✓
$105^\circ$
- D
$98^\circ$
AnswerCorrect option: C. $105^\circ$
Let the angles of the triangle be $2x, 3x$ and $7x.$
Now, $2x + 3x + 7x = 180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 12x = 180^\circ $
$\Rightarrow x = 15^\circ $
$\therefore$ Largest angle $= 7x = 7 \times 15^\circ $
$= 105^\circ $
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 121 Mark
In Fig. the value of $x$ is:
Answer$\angle \text{CAD}=\angle \text{ABC}+\angle \text{ACB}$ [Exterior angle property]
$\Rightarrow 123^\circ = 39^\circ + \text{x}^\circ$
$\Rightarrow 84^\circ = \text{x}^\circ$
$\Rightarrow \text{x} = 84$
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 131 Mark
In Fig. the values of $x$ and $y$ are:
- A
$x = 20, y = 130$
- B
$x = 20, y = 140$
- ✓
$x = 40, y = 140$
- D
$x = 15, y = 140$
AnswerCorrect option: C. $x = 40, y = 140$
$\angle \text{ACB}+\angle \text{ACD}=180^\circ$
$\Rightarrow 40^\circ + \text{y}^\circ = 180^\circ$
$\Rightarrow \text{y}^\circ = 140^\circ$
$\Rightarrow \text{y} = 140$
Now, $\angle \text{ACD}=\angle \text{ABC}+\angle \text{BAC}$ [Exterior angle property]
$\Rightarrow 3\text{x}^\circ + 4\text{x}^\circ = \text{y}^\circ$
$⇒ 7\text{x}^\circ = 140^\circ$
$\Rightarrow \text{x} = 20$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 141 Mark
The hypotenuse of a right triangle is $26\ cm$ long. If one of the remaining two sides is $10\ cm$ long, the length of the other side is:
- A
$25\ cm$
- B
$23\ cm$
- ✓
$24\ cm$
- D
$22\ cm$
AnswerCorrect option: C. $24\ cm$
In right traingle $BOC,$
$\mathrm{BC}^2=\mathrm{OC}^2+\mathrm{OB}^2$
$\Rightarrow(26)^2=(10)^2+\mathrm{OB}^2$
$\Rightarrow 676=100+\mathrm{OB}^2$
$\Rightarrow \mathrm{OB}^2=576$
$\Rightarrow \mathrm{OB}^2=(24)^2$
$\Rightarrow \mathrm{OB}=24 \mathrm{~cm}$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 151 Mark
In Fig. the value of $x$ is:
- A
$72^\circ $
- B
$50$
- ✓
$58$
- D
$48$
Answer$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ [Angle sum property of triangle]
$ \Rightarrow 50^\circ+72^\circ+\angle \text{C}=180^\circ$
$\Rightarrow \angle \text{C}+122^\circ=180^\circ$
$\Rightarrow \angle \text{C}=58^\circ$
Now, $\text{x}^\circ=\angle\text{C}$ [Vertically opposite angles]
$\Rightarrow \text{x}^\circ= 58^\circ$
$\Rightarrow \text{x}=58$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 161 Mark
In a $\triangle \text{ABC},$ if $\angle \text{A}-\angle \text{B}=33^\circ$ and $\angle \text{B}-\angle \text{C}=10^\circ,$ then $\angle \text{B} =$
- A
$35^\circ$
- B
$45^\circ$
- C
$55^\circ$
- ✓
$25^\circ$
AnswerCorrect option: D. $25^\circ$
$\angle \text{A}-\angle \text{B}=33^\circ$ and $\angle \text{B}-\angle \text{C}=18^\circ$
$\Rightarrow \angle \text{A}=\angle \text{B}+33^\circ$ and $\angle \text{C}=\angle \text{B}-18^\circ$
Now, $\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow \angle \text{B}+33^\circ+\angle \text{B}+\angle \text{B}-18^\circ=180^\circ$
$\Rightarrow 3\angle \text{B}+15^\circ=180^\circ$
$\Rightarrow 3\angle \text{B}=165^\circ$
$\Rightarrow \angle \text{B}=55^\circ$
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 171 Mark
In Fig. if $AB || CD,$ the values of $x$ and $y$ are:
- A
$x = 21, y = 28$
- ✓
$x = 21, y = 38$
- C
$x = 38, y = 21$
- D
$x = 22, y = 38$
AnswerCorrect option: B. $x = 21, y = 38$
$\angle \text{AEC}+\angle \text{AEB}=180^\circ$ [Linear angles]
$\Rightarrow 79^\circ+\angle \text{AEB}=180^\circ$
$\Rightarrow \angle \text{AEB}=101^\circ$
Since, $AB || CD$
$\angle \text{ABE}=\angle \text{ECD}=58^\circ$ [Alternate angles]
Now, In $\triangle \text{AEB}$
$\angle \text{AEB}+\angle \text{EAB}+\angle \text{ABE}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 101^\circ+58^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}^\circ=21^\circ$
$\Rightarrow \text{x}=21$
Now, In $\triangle \text{AEB}$
$\angle \text{AEC}+\angle \text{CAE}+\angle \text{CEA}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 79^\circ+\text{y}^\circ+3\text{x}^\circ=180^\circ$
$\Rightarrow 79^\circ+\text{y}^\circ+3(21)^\circ=180^\circ$
$\Rightarrow 79^\circ+\text{y}^\circ+63^\circ=180^\circ$
$\Rightarrow \text{y}^\circ=38^\circ$
$\Rightarrow \text{y}=38$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 181 Mark
In Fig. if $AB || CD$ and $AE || BD,$ then the value of $x$ is:
View full question & answer→MCQ 191 Mark
If the measures of the angles of a triangle are $(2\text{x}-5)^\circ,\Big(3\text{x}-\frac{1}{2}\Big)$ and $\Big(30-\frac{\text{x}}{2}\Big)^\circ,$ then $x =$
- ✓
$\frac{311}{9}$
- B
$\frac{309}{11}$
- C
$\frac{310}{9}$
- D
$\frac{301}{9}$
AnswerCorrect option: A. $\frac{311}{9}$
$(2\text{x}-5)^\circ,\Big(3\text{x}-\frac{1}{2}\Big)+\Big(30-\frac{\text{x}}{2}\Big)^\circ=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 2\text{x}-5+3\text{x}-\frac{1}{2}+30-\frac{\text{x}}{2}=180$
$\Rightarrow 2\text{x}+3\text{x}-\frac{\text{x}}{2}-5-\frac{1}{2}+30=180$
$\Rightarrow 4\text{x}+6\text{x}-\text{x}-10-1+60=180\times2$
$\Rightarrow 9\text{x}+49=360$
$\Rightarrow \text{x}=\frac{311}{9}$
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 201 Mark
In Fig. if $AB || CD$, then the values of $x$ and $y$ are:
- ✓
$x = 106, y = 307$
- B
$x = 307, y = 106$
- C
$x =107, y = 306$
- D
$x = 105, y = 308$
AnswerCorrect option: A. $x = 106, y = 307$
In $\triangle \text{CDE}$
$\angle \text{CDE}+\angle \text{CED}+\angle \text{ECD}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 53^\circ+53^\circ+\angle \text{ECD}=180^\circ$
$\Rightarrow \angle \text{ECD}=74^\circ$
Since, $AB || CD$
$\therefore \angle \text{ECD}=\angle \text{CGB}=74^\circ$ [Corresponding angles]
Now, $\angle \text{CGB}+\angle \text{BGF}=180^\circ$ [Linear pair angles]
$\Rightarrow 74^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}=106$
Now, In $\triangle \text{EGB},$
$\angle \text{EGB}+\angle \text{BEG}+\angle \text{EBG}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 74^\circ+53^\circ+\angle \text{EBG}=180^\circ$
$\Rightarrow \angle \text{EBG}=53^\circ$
Now, $\angle \text{EBG}+\text{Reflex }\angle \text{EBG}=360^\circ$ [Complete angle]
$\Rightarrow 53^\circ+\text{y}^\circ=360^\circ$
$\Rightarrow \text{y}=307$
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 211 Mark
In Fig. if $AB || CD,$ then the values of $x$ and $y$ are:
- A
$x = 24, y = 48$
- ✓
$x = 34, y = 68$
- C
$x = 24, y = 68$
- D
$x = 34, y = 48$
AnswerCorrect option: B. $x = 34, y = 68$
$\angle \text{AGE}+\angle \text{BGE}=180^\circ$ [Linear pair angles]
$\Rightarrow 121^\circ+\angle \text{BGE}=180^\circ$
$\Rightarrow \angle \text{BGE}=59^\circ$
Since, $AB || CD$
$\therefore \angle \text{BGE}= \angle \text{GHD}=59^\circ$ [Corresponding angles]
$\Rightarrow \text{x}^\circ+25^\circ=59^\circ$
$\Rightarrow \text{x}=34$
In $\triangle \text{GHI},$
$\angle \text{GHI}+\angle \text{GIH}+\angle \text{HGI}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 34^\circ+78^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}=68$
Hence, the correct answer is option $(b).$ View full question & answer→MCQ 221 Mark
A $15m$ long ladder is placed against a wall in such away that the foot of the ladder is 9m away from the wall. Up to what height does the ladder reach the wall$?$
Answer
In right traingle $B O C$,
$B C^2=O C^2+O B^2$
$\Rightarrow(15)^2=(9)^2+O B^2$
$\Rightarrow 225=81+O B^2$
$\Rightarrow O B^2=144$
$\Rightarrow O B^2=(12)^2$
$\Rightarrow O B=12 m$
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 231 Mark
In Fig. the values of $x$ and $y$ are:
- ✓
$x = 130, y = 120$
- B
$x = 120, y = 130$
- C
$x = 120, y = 120$
- D
$x = 130, y = 130$
AnswerCorrect option: A. $x = 130, y = 120$
In $\triangle \text{ABD}$
$\angle \text{ADB}+\angle \text{BAD}+\angle \text{ABD}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 61^\circ+59^\circ+\angle \text{ABD}=180^\circ$
$\Rightarrow \angle \text{ABD}=60^\circ$
$\angle \text{ABD}+\angle \text{DBC}=180^\circ$ [Linear pair angles]
$\Rightarrow 60^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}=120$
Now, $\angle \text{ADB}=\angle \text{GDE}=61^\circ$ [Vertically opposite angles]
Now, In $\triangle \text{GDE},$
$\angle \text{GDE}+\angle \text{DGE}+\angle \text{GED}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 61^\circ+69^\circ+\angle \text{GED}=180^\circ$
$\Rightarrow \angle \text{GED}=50^\circ$
Now, $\angle \text{GED}+\angle \text{GEF}=180^\circ$ [Linear pair angles]
$\Rightarrow 50^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}=130$
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 241 Mark
In Fig. if $AB || CO,$ $\angle \text{CAB}=49^\circ, \angle \text{CBD}=27^\circ$ and $\angle \text{BDC}=112^\circ,$ then the values of $x$ and $y$ are:
- ✓
$x = 41, y = 90$
- B
$x = 41, y = 63$
- C
$x = 63, y = 41$
- D
$x = 90, y = 41$
AnswerCorrect option: A. $x = 41, y = 90$
Since, $AB || CD$
$\angle \text{ABD}+\angle \text{CDB}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow \text{x}^\circ+27^\circ+112^\circ=180^\circ$
$\Rightarrow \text{x}^\circ=41^\circ$
$\Rightarrow \text{x}=41$
Now, In $\triangle \text{ABC},$
$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 49^\circ+41^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}^\circ=90^\circ$
$\Rightarrow \text{y}=90$
Hence, the correct answer is option $(a).$ View full question & answer→MCQ 251 Mark
In Fig. the value of $x$ is:
Answer$\angle \text{TRS}+\angle \text{TRQ}=180^\circ$ [Linear angles]
$\Rightarrow 5\text{x}^\circ+\angle \text{TRQ}=180^\circ$
$\Rightarrow \angle \text{TEQ}=180^\circ-5\text{x}^\circ$
Now, $\angle \text{QTR}+\angle \text{TRQ}=\angle \text{PQT}$ [Exterior angle property of triangle]
$\Rightarrow 3\text{x}^\circ+180^\circ-5\text{x}^\circ=120^\circ$
$\Rightarrow 2\text{x}^\circ=60^\circ$
$\Rightarrow \text{x}=30$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 261 Mark
A ladder is placed in such a way that its foot is $15m$ away from the wall and its top reaches a window $20m$ above the ground. The length of the ladder is:
- A
$35m$
- ✓
$25m$
- C
$18m$
- D
$17.5m$
Answer Suppose $BC$ is the ladder which is placed againts the wall $OA.$ The foot of the ladder $C$ is $15m$ away from the foot $O$ of the wall and its top reaches the window which is $20m$ above the ground.
In right traingle $BOC ,$
$\mathrm{BC}^2=\mathrm{OC}^2+\mathrm{OB}^2$
$\Rightarrow B C^2=(15)^2+(20)^2$
$\Rightarrow B C^2=225+400$
$\Rightarrow B C^2=625$
$\Rightarrow B C^2=(25)^2$
$\Rightarrow B C=25 m$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 271 Mark
In which of the following cases can a right triangle $ABC$ be constructed$?$
- A
$AB = 5\ cm, BC = 7\ cm, AC = 10\ cm$
- B
$AB = 7\ cm, BC = 8\ cm, AC = 12\ cm$
- ✓
$AB = 8\ cm, BC = 17\ cm, AC = 15\ cm$
- D
AnswerCorrect option: C. $AB = 8\ cm, BC = 17\ cm, AC = 15\ cm$
$\text { In (c) }$
$\mathrm{BC}^2=A C^2+A B^2$
$\Rightarrow(17)^2=(15)^2+(8)^2$
$\Rightarrow 289=225+64$
$\Rightarrow 289=289$
Since, the sum of the square of two smallest side is equal to the square of largest side.
Hence, $A B C$ is a right angle triangle at $A$.
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 281 Mark
In which of the following cases, a right triangle cannot be constructed?
AnswerCorrect option: A. $12\ cm, 5\ cm, 13\ cm$
ln $(a)$
$12^2+5^2=13^2$
$\Rightarrow 144+25=169$
$\Rightarrow 169$
Since, the sum of the square of two smallest side is equal to the square of largest side.
Hence, a right triangle can be constructed.
ln $(b)$
$8^2+6^2=10^2$
$\Rightarrow 44+36=100$
$\Rightarrow 100=100$
Since, the sum of the square of two smallest side is equal to the square of largest side.
Hence, a right triangle can be constructed.
ln $(c)$
$5^2+9^2 \neq 11^2$
$\Rightarrow 25+81 \neq 121$
$\Rightarrow 106 \neq 121$
Since, the sum of the square of two smallest side is not equal to the square of largest side.
Hence, a right triangle can not be constructed.
Hence, the correct answer is option $(c)$.
View full question & answer→MCQ 291 Mark
In a $\triangle \text{ABC},$ if $\angle \text{A}+\angle \text{B}=150^\circ$ and $\angle \text{B}+\angle \text{C}=75^\circ,$ then $\angle \text{B} =$
- A
$35^\circ $
- ✓
$45^\circ$
- C
$55^\circ$
- D
$25^\circ$
AnswerCorrect option: B. $45^\circ$
$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 150^\circ+\angle \text{C}=180^\circ$
$\Rightarrow \angle \text{C}=30^\circ$
Now, $\angle \text{B}+\angle \text{C}=75^\circ$
$\Rightarrow \angle \text{B}+30^\circ=75^\circ$
$\Rightarrow \angle \text{B}=45^\circ$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 301 Mark
In a right triangle, one of the acute angles is four times the other. Its measure is:
- A
$68^\circ$
- B
$84^\circ$
- C
$80^\circ$
- ✓
$72^\circ$
AnswerCorrect option: D. $72^\circ$
Let the smallest angle be $x,$ then the other angle be $4x.$
Now,
$x + 4x + 90^\circ = 180^\circ $
$\Rightarrow 5x = 90^\circ $
$\Rightarrow x = 18^\circ $
Thus, the measure of the angles are $18^\circ ,$ and $4(18)^\circ = 72^\circ $
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 311 Mark
In Fig. if $AB || DE,$ then the value of $x$ is:
Answer$\angle \text{ACD}+\angle \text{ACB}=180^\circ$ [Linear angles]
$\Rightarrow 91^\circ+\angle \text{ACB}=180^\circ$
$\Rightarrow \angle \text{ACB}=89^\circ$
Since, $AB || DE$
$\angle \text{DEC}=\angle \text{CAB}=46^\circ$ [Alternate angles]
Now,
$\angle \text{ACB}+\angle \text{CAB}+\angle \text{ABC}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 89^\circ + 46^\circ + \text{x}^\circ = 180^\circ$
$⇒ \text{x}^\circ = 45^\circ$
$\Rightarrow \text{x} = 45$
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 321 Mark
In Fig. if $AF || DE,$ then $x =$
Answer
$\angle \text{EDC}=\angle \text{ACB}=109^\circ$ [Corresponding angles]
Now, In $\triangle \text{ABC}$
$\angle \text{ACB}+\angle \text{CAB}+\angle \text{CBA}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 109^\circ+24^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}^\circ=47^\circ$
$\Rightarrow \text{x}=47$
Hence, the correct answer is option $(c).$ View full question & answer→MCQ 331 Mark
If $\triangle \text{ABC}$ is an isosceles right-triangle right angled at $C$ such that $AC = 5\ cm.$ Then, $AB =$
- A
$2.5\ cm$
- ✓
$5\sqrt{2}\text{cm}$
- C
$10\ cm$
- D
$5\ cm$
AnswerCorrect option: B. $5\sqrt{2}\text{cm}$
Suppose $BC$ is the ladder which is placed againts the wall $OA.$
The foot of the ladder $C$ is $15m$ away from the foot $O$ of the wall and its top reaches the window which is $20m$ above the ground.
In right traingle $A B C$
$\mathrm{AB}^2=\mathrm{BC}^2+\mathrm{AC}^2$
$\Rightarrow \mathrm{AB}^2=(5)^2+(5)^2$
$\Rightarrow \mathrm{AB}^2=25+25$
$\Rightarrow \mathrm{AB}^2=50$
$\Rightarrow \mathrm{AB}^2=(5 \sqrt{2})^2$
$\Rightarrow \mathrm{AB}=5 \sqrt{2} \mathrm{~cm}$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 341 Mark
In Fig. if $AB || CD,$ the value of $x$ is:
AnswerSince, $AB || DE$
$\angle \text{DCB}=\angle \text{CBA}=3\text{x}^\circ$ [Alternate angles]
Now,
$\angle \text{ACB}+\angle \text{CAB}+\angle \text{CBA}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 55^\circ + 2\text{x}^\circ + 3\text{x}^\circ= 180$
$\Rightarrow 5\text{x}^\circ= 125^\circ$
$\Rightarrow \text{x} = 25$
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 351 Mark
A 15 m long ladder is placed against a wall in such a way that the foot of the ladder is 9 m away from the wall. Up to what height does the ladder reach the wall?
MCQ 361 Mark
The hypotenuse of a right triangle is 26 cm long. If one of the remaining two sides is 10 cm long, the length of the other side is
MCQ 371 Mark
A ladder is placed in such a way that its foot is 15 m away from the wall and its top reaches a window 20 m above the ground. The length of the ladder is
MCQ 381 Mark
Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, the distance between their tops is
MCQ 391 Mark
If $M B C$ is an isosceles right-triangle right angled at $C$ such that $A C=5 cm$. Then, $A B=$
AnswerCorrect option: B. $5 \sqrt{2} cm$
View full question & answer→MCQ 401 Mark
$\triangle A B C$ is a rioht trianole right anotled at $A$. If $A R=24 cm$ and $A C=7 cm$, then $B C=$
View full question & answer→MCQ 411 Mark
In which of the following cases can a right triangle ABC be constructed?
- A
$A B=5 cm, B C=7 cm, A C=10 cm$
- B
$A B=7 cm, B C=8 cm, A C=12 cm$
- ✓
$A B=8 cm, B C=17 cm, A C=15 cm$
- D
AnswerCorrect option: C. $A B=8 cm, B C=17 cm, A C=15 cm$
View full question & answer→MCQ 421 Mark
In a right triangle, one of the acute angles is four times the other. Its measure is
- A
$68^{\circ}$
- B
$84^{\circ}$
- C
$80^{\circ}$
- ✓
$72^{\circ}$
AnswerCorrect option: D. $72^{\circ}$
View full question & answer→MCQ 431 Mark
Which of the following is/are not Pythagorean triplet (s)?
MCQ 441 Mark
In which of the following cases, a right triangle cannot be constructed?
MCQ 451 Mark
Which of the following is the set of measures of the sides of a triangle?
MCQ 461 Mark
In Fig. if $A B \| C D, \angle C A B=49^{\circ}, \angle C B D=27^{\circ}$ and $\angle B D C=112^{\circ}$, then the values of $x$ and $y$ are
View full question & answer→MCQ 471 Mark
In Fig. the value of x is
MCQ 481 Mark
If the exterior angles of a triangle are $(2 x+10)^{\prime \prime},(3 x-5)^{\circ}$ and $(2 x-40)^{\circ}$, then $x=$
View full question & answer→MCQ 491 Mark
In Fig. if AB || CD and AE || BD, then the value of x is
MCQ 501 Mark
In Fig. if AB || CD, then the values of x, y and z are
MCQ 511 Mark
In Fig. if ABCD, then the values of x and y are
MCQ 521 Mark
In Fig. if AB || CD, then the values of x and y are
MCQ 531 Mark
In Fig. the values of x and y are
MCQ 541 Mark
In Fig. the values of x and y are
MCQ 551 Mark
In Fig. if $A F \| D E$, then $x=$
View full question & answer→MCQ 561 Mark
In Fig. if $A B \| C E$, then the values of $x$ and $y$ are
View full question & answer→MCQ 571 Mark
In Fig. if $A B \| C D$, the values of $x$ and $y$ are
View full question & answer→MCQ 581 Mark
In Fig. if $A B \| C D$ the value of x is
View full question & answer→MCQ 591 Mark
In Fig. if AB || DE then the value of x is
MCQ 601 Mark
In Fig. the value of x is
MCQ 611 Mark
In Fig. the values of x and y are
MCQ 621 Mark
In Fig. the value of x is
MCQ 631 Mark
If the measures of the angles of a triangle are $(2 x-5)^{\circ} ;\left(3 x-\frac{1}{2}\right)^{\circ}$ and $\left(3 x-\frac{x}{2}\right)^{\circ}$, then $x=$
- ✓
$\frac{311}{9}$
- B
$\frac{309}{11}$
- C
$\frac{310}{9}$
- D
$\frac{301}{9}$
AnswerCorrect option: A. $\frac{311}{9}$
View full question & answer→MCQ 641 Mark
In a $\triangle A B C$if $\angle A-\angle B=33^{\circ}$and $\angle B-\angle C=18^{\circ}$ $\angle B=$
- A
$35^{\circ}$
- B
$45^{\circ}$
- C
$56^{\circ}$
- ✓
$55^{\circ}$
AnswerCorrect option: D. $55^{\circ}$
View full question & answer→MCQ 651 Mark
In a $\triangle A B C$, if $\angle A+\angle B=150^{\circ}$and $\angle B+\angle C=75^{\circ}$then $\angle B=$
- A
$35^{\circ}$
- ✓
$45^{\circ}$
- C
$55^{\circ}$
- D
$25^{\circ}$
AnswerCorrect option: B. $45^{\circ}$
View full question & answer→MCQ 661 Mark
In a$\triangle A B C$ if $2 \angle A=3 \angle B=6 \angle C$ then the measure of the smallest angle is
- A
$90^{\circ}$
- B
$60^{\circ}$
- C
$40^{\circ}$
- ✓
$30^{\circ}$
AnswerCorrect option: D. $30^{\circ}$
View full question & answer→MCQ 671 Mark
The angles of a triangle are in the ratio 2:3:7. The measure of the largest angle is
- A
$84^{\circ}$
- B
$91^{\circ}$
- ✓
$105^{\circ}$
- D
$98^{\circ}$
AnswerCorrect option: C. $105^{\circ}$
View full question & answer→MCQ 681 Mark
If the measures of the angles of a triangle are $(2 x)^{\circ}$, $(3 x-5)^{\circ}$ and ($(4 x-13)^{\circ}$ Then the value of x is
View full question & answer→