MCQ 11 Mark
In an isosceles $\triangle\text{ABC},$ if $AC = BC$ and $AB^2= 2AC^2$ then $\angle\text{C}=?$
- A
$30^\circ $
- B
$45^\circ$
- C
$60^\circ$
- ✓
$90^\circ$
AnswerCorrect option: D. $90^\circ$
In an isosceles $\triangle\text{ABC},$ given $AC = BC$
$ A B^2=2 A C^2 $
$ \Rightarrow A B^2=B C^2+A C^2 \ldots .(\therefore A C=B C)$
by the Converse of Pythagoras theorem,
$\triangle\text{ABC}$ will be an isosceles right-angled triangle.
Since $AB$ will be the hypotenuse, the angle opposite $AB$ that is, $\angle\text{C}=90^\circ.$
View full question & answer→MCQ 21 Mark
In an equilateral triangle $ABC,$ if $\text{AD}\perp\text{BC}$ then which of the following is true?
- A
$ 2 A B^2 =3 A D^2 $
- B
$ 4 A B^2 =3 A D^2$
- ✓
$ 3 A B^2 =4 A D^2 $
- D
$ 3 A B^2 =2 A D^2 $
AnswerCorrect option: C. $ 3 A B^2 =4 A D^2 $
In an equilateral triangle, the perpendicular from the vertex to the base is bisects the base.
In right-angled $\triangle\text{ADC},$
$AB^2= AD^2+ BD^2$
$⇒ AB^2= AD^2+ BD^2$
$\Rightarrow\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{BC}\Big)^2$
$\Rightarrow\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{AB}\Big)^2\dots(\therefore\text{AB}=\text{BC} )$
$\Rightarrow\text{AB}^2=\text{AD}^2+\frac{1}{4}\text{AB}^2$
$\Rightarrow\text{AB}^2-\frac{1}{4}\text{AB}^2=\text{AD}^2$
$\Rightarrow\frac{3}{4}\text{AB}^2=\text{AD}^2$
$\Rightarrow3\text{AB}^2=4\text{AD}^2$
View full question & answer→MCQ 31 Mark
In $\triangle\text{ABC},\text{DE }||\text{ BC}$ such that $\frac{\text{AD}}{\text{DB}}=\frac{3}{5}.$ $AC = 5.6cm$ then $AE =?$
- A
$4.2cm$
- B
$3.1cm$
- C
$2.8cm$
- ✓
$2.1cm$
AnswerCorrect option: D. $2.1cm$
In $\triangle\text{ABC},\text{DE }||\text{ BC}$
By Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{AC}-\text{AE}}$
$\Rightarrow\frac{3}{5}=\frac{\text{AE}}{\text{5.6}-\text{AE}}$
$\Rightarrow3(5.6-\text{AE})=5\text{AE}$
$\Rightarrow16.8-3\text{AE})=5\text{AE}$
$\Rightarrow8\text{AE}=16.8$
$\Rightarrow\text{AE}=2.1\text{cm}$
View full question & answer→MCQ 41 Mark
In a $\triangle\text{ABC},$ it is given that $AD$ is the internal bisector of $\angle\text{A}.$ If $AB = 10cm, AC = 14cm$ and $BC = 6cm,$ then $CD = ?$
- A
$4.8cm$
- ✓
$3.5cm$
- C
$7cm$
- D
$10.5cm$
AnswerCorrect option: B. $3.5cm$
since $AD$ is the bisector of $\angle\text{A},$
by the angle bisector theorem,
$\frac{10}{14}=\frac{6-\text{x}}{\text{x}}$
$\Rightarrow10\text{x}=84-14\text{x}$
$\Rightarrow24\text{x}=84$
$\Rightarrow\text{x}=3.5$
So, $CD = 3.5cm$.
View full question & answer→MCQ 51 Mark
In a $\triangle\text{ABC},$ if $DE$ is drawn parallel to $BC,$ cutting $AB$ and $AC$ at $D$ and $E$ respectively such that $AB = 7.2cm, AC = 6.4cm$ and $AD = 4.5cm$. Then, $AE =?$
- A
$5.4cm$
- ✓
$4cm$
- C
$3.6cm$
- D
$3.2cm$
AnswerIn $\triangle\text{ABC},\text{DE }||\text{ BC}$
By Basic proportionality theorem,
$\frac{\text{AE}}{\text{AC}}=\frac{\text{AD}}{\text{AB}}$
$\Rightarrow\frac{\text{AE}}{6.4}=\frac{4.5}{7.2}$
$\Rightarrow\text{AE}=\frac{4.5\times6.4}{7.2}$
$\Rightarrow\text{AE}=4\text{cm}$
View full question & answer→MCQ 61 Mark
$\triangle\text{ABC}$ is an isosceles triangle with $AB = AC = 13cm$ and the length of altitude from $A$ on $BC$ is $5cm$. Then, $BC =?$
- A
$12cm$
- B
$16cm$
- C
$18cm$
- ✓
$24cm$
AnswerCorrect option: D. $24cm$
let $\triangle\text{ABC}$ be the isosceles triangle and AD be the altitude.
The height of an isosceles triangle is the same as its median.
So, BD = DC
$\triangle\text{ADB}$ is a right-angled triangle.
By Pythagoras theorem,
$ A B^2=A D^2+B D^2 $
$ \Rightarrow B D^2=A B^2-A D^2 $
$\Rightarrow B D^2=13^2-5^2 $
$Rightarrow B D^2=169-25$
$ \Rightarrow B D^2=144$
$⇒ BD = 12cm$
$⇒ DC = 2cm$
So, $BC = 12 + 12 = 24cm.$ View full question & answer→MCQ 71 Mark
In $\triangle\text{ABC},$ it is given that $AB = 9cm, BC = 6cm$ and $CA = 7.5cm$. Also, $\triangle\text{DEF}$ is given such that $EF = 8cm$ and $\triangle\text{DEF}\sim\triangle\text{ABC}.$ Then, perimeter of $\triangle\text{DEF}$ 1s:
- A
$22.5cm$
- B
$25cm$
- C
$27cm$
- ✓
$30cm$
AnswerCorrect option: D. $30cm$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{Perimeters of }\triangle\text{ABC}}=\frac{\text{EF}}{\text{BC}}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{AB+BC+AC}}=\frac{8}{6}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{9+6+7.5}=\frac{8}{6}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{22.5}=\frac{4}{3}$
$\Rightarrow\text{Perimeters of }\triangle\text{DEF}=\frac{4\times22.5}{3}$
$\Rightarrow\text{Perimeters of }\triangle\text{DEF}=30\text{cm}$
View full question & answer→MCQ 81 Mark
In a thombus of side 10cm, one of the diagonals is $12cm$ long. The length of the second diagonal is:
- A
$20cm$
- B
$18cm$
- ✓
$16cm$
- D
$22cm$
AnswerCorrect option: C. $16cm$
In an rhombus, the diagonals are perpendicular bisetoes of each other.
So, $\text{OD}=\frac{1}{2}\text{BD}=6\text{cm}$
In right-angled $\triangle\text{AOD},$
$ A O^2=A O^2+O D^2 $
$ \Rightarrow A O^2=A D^2-O D^2 $
$ \Rightarrow A O^2=10^2-6^2 $
$ \Rightarrow A O^2=100-36 $
$ \Rightarrow A O^2=64$
$\Rightarrow AO = 8cm$
So, $AC = 2AO = 2(8) = 16cm$
Thus, the length of the second diagonal is $16cm.$ View full question & answer→MCQ 91 Mark
The shadow of a $5$-m-long stick is 2m long. At the same time the langht of the shadow of a $12.5$-m-high (in m) is:
Answer
Let $AN$ be the long stick and $AW$ be its shadow.
Let $OB$ be the tree and $OW$ be its shadows.
$AW = 2cm$
$AN = 5m$
$OW = 12.5m$
Ratio of actual lengths = ratio of their shadows
$\Rightarrow\frac{\text{OB}}{\text{AN}}=\frac{\text{OW}}{\text{AW}}$
$\Rightarrow\frac{12.5}{5}=\frac{\text{OW}}{2}$
$\Rightarrow\text{OW}=\frac{12.5\times2}{5}$
$\Rightarrow\text{OW}=5.0\text{m}$
So, the height of the tower is $5.m$. View full question & answer→MCQ 101 Mark
A vertical pole 6m long casts a shadow of lenghth $3.6m$ on the ground. What is the height of a tower which casts a shadow of lenght $18m$ at the same time?
- A
$10.8m$
- B
$28.8m$
- C
$32.4m$
- ✓
$30m$
Answer
Let $AN$ be the vertical pole and $AW$ be its shadow.
Let $OB$ be the tower and $OW$ be its shadows.
$AW = 3.6cm$
$AN = 6m$
$OW = 18m$
Ratio of actual lengths = ratio of their shadows
$\Rightarrow\frac{\text{OB}}{\text{AN}}=\frac{\text{OW}}{\text{AW}}$
$\Rightarrow\frac{\text{h}}{6}=\frac{18}{3.6}$
$\Rightarrow\text{h}=\frac{6\times18}{3.6}$
$\Rightarrow\text{h}=30\text{m}$
So, the height of the tower is $30m$. View full question & answer→MCQ 111 Mark
In a triangle, the perpendicular from the vertex to the base bisect the base. The triangle is:
AnswerIn an isosceles triangle, the perpendicular from the vertex to the base bisects the base.
View full question & answer→MCQ 121 Mark
$ABC$ and $BDE$ are two equoilateral triangles such tha $D$ is the midpoint of $BC$. Ratio of the areas of triangles $ABC$ and $BDE$ is:
- A
$1 : 2$
- B
$2 : 1$
- C
$1 : 4$
- ✓
$4 : 1$
AnswerCorrect option: D. $4 : 1$
Given that $D$ is the mid-point of $BC,$
$\Rightarrow\text{BD}=\frac{1}{2}\text{BC}\dots(\text{i})$
Since $\triangle\text{ABC}$ and $\triangle\text{EBD}$ are equilateral triangles.
$\Rightarrow\triangle\text{ABC}\sim\triangle\text{EBD}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{EBD})}=\frac{\text{BC}^2}{\text{BD}^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{EBD})}=\frac{\text{BC}^2}{\Big(\frac{1}{ 2}\text{BC}\Big)^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{EBD})}=\frac{\text{BC}^2}{\frac{1}{4}\text{BC}^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{EBD})}=\frac{4}{1}$ View full question & answer→MCQ 131 Mark
In $\triangle\text{DEF}$ and $\triangle\text{PQR},$ it is given that $\angle\text{D}=\angle\text{Q}$ and $\angle\text{R}=\angle\text{E},$ then which of the following is not true?
- A
$\frac{\text{EF}}{\text{PR}}=\frac{\text{DF}}{\text{PQ}}$
- ✓
$\frac{\text{DE}}{\text{PQ}}=\frac{\text{EF}}{\text{RP}}$
- C
$\frac{\text{DE}}{\text{QR}}=\frac{\text{DF}}{\text{PQ}}$
- D
$\frac{\text{EF}}{\text{RP}}=\frac{\text{DE}}{\text{QR}}$
AnswerCorrect option: B. $\frac{\text{DE}}{\text{PQ}}=\frac{\text{EF}}{\text{RP}}$
In $\triangle\text{DEF}$ and $\triangle\text{PQR},$ it is given that $\angle\text{D}=\angle\text{Q}$ and $\angle\text{R}=\angle\text{E}$
So, $\text{D}\leftrightarrow\text{Q},\text{E}\leftrightarrow\text{R},\text{F}\leftrightarrow\text{P}$
$\Rightarrow\frac{\text{DE}}{\text{QR}}=\frac{\text{EF}}{\text{RP}}=\frac{\text{DF}}{\text{QP}}$
So, $\frac{\text{DE}}{\text{PQ}}=\frac{\text{EF}}{\text{RP}}$ is not true.
View full question & answer→MCQ 141 Mark
Two isosceles triangles have their corresponding angles equal and their areas are in the ratio $25 : 36$. The ratio of their corresponding heights is:
- A
$25 : 36$
- B
$36 : 25$
- ✓
$5 : 6$
- D
$6 : 5$
AnswerCorrect option: C. $5 : 6$
Since the triangles have correspondin angles equal, the triangles are similar.
Let the areas of the triangles be $A_1$ and $A_2$,
and let their corresponding heights be $h_1$ and $h_2$,
$\frac{\text{ar}(\text{A}_1)}{\text{ar}(\text{A}_2)}=\frac{\text{h}_1^2}{\text{h}_2^2}$
$\Rightarrow\frac{25}{36}=\frac{\text{h}_1^2}{\text{h}_2^2}$
$\Rightarrow\frac{\text{h}_1}{\text{h}_2}=\frac{5}{6}$
So, the ratio of their heights is $5 : 6$.
View full question & answer→MCQ 151 Mark
$\triangle\text{ABC}\sim\triangle\text{DEF}$ and their perimeters are $32cm$ and $24cm$ respectively. If $AB = 10cm$ then $DE =?$
- A
$8cm$
- ✓
$7.5cm$
- C
$15cm$
- D
$5\sqrt{3}\text{cm}$
AnswerCorrect option: B. $7.5cm$
$\therefore\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{Perimeter}(\triangle\text{ABC})}{\text{Perimeter}(\triangle\text{DEF})}=\frac{\text{AB}}{\text{DE}}$
$\Rightarrow\frac{32}{24}=\frac{10}{\text{DE}}$
$\Rightarrow\text{DE}=\frac{10\times24}{\text{32}}=7.5\text{cm}$
View full question & answer→MCQ 161 Mark
If the diagonals of a quadrilateral divide each other proportionally then it is a:
AnswerRecall that the diagonals of a trapezium divide each other proportionally.
Note that this happens even in a parallelogram, square and rectangle, but without additional information it is not possible to be sure.
View full question & answer→MCQ 171 Mark
In the given figure, $O$ is the point of intersection of two chords $AB$ and $CD$ such that $OB = OD$ and $\angle\text{AOC}=45^\circ.$ Then, $\triangle\text{OAC}$ and $\triangle\text{ODB}$ are:
- A
- B
Equilateral but not similar.
- ✓
- D
Isosceles but not similar.
AnswerIn $\triangle\text{AOC}$ and $\triangle\text{ODB}$
$\angle\text{AOC}=\angle\text{DOB}$ ....(Vertically opposite angles)
$\angle\text{OCA}=\angle\text{OBD}$ ....(angels in the same segment)
$\Rightarrow\triangle\text{OAC}\sim\triangle\text{ODB}$ ....(AA criterion for similarity)
The two triangles are surely not equil ateral,
Since the measure of every angle of an equilateral triangle is 60º.
So, the triangles are isosceles and similar.
View full question & answer→MCQ 181 Mark
In $\triangle\text{ABC},\text{DE }||\text{ BC}$ so that $AD = (7x - 4)cm, AE = (5x - 2)cm, DB = (3x + 4)cm$ and $EC = 3x cm$. Then, we have:
- A
$x = 3$
- B
$x = 5$
- ✓
$x = 4$
- D
$x = 2.5$
AnswerCorrect option: C. $x = 4$
In $\triangle\text{ABC},\text{DE }||\text{ BC}$
By Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{7\text{x}-4}{3\text{x}+4}=\frac{5\text{x}-2}{3\text{x}}$
$\Rightarrow21\text{x}^2-12\text{x}=15\text{x}^2+14\text{x}-8$
$\Rightarrow6\text{x}^2-26\text{x}+8=0$
$\Rightarrow3\text{x}^2-13\text{x}+4=0$
$\Rightarrow(\text{x}-4)(3\text{x}-1)=0$
$\Rightarrow\text{x}=4$ or $\text{x}=\frac{1}{3}$
If $\text{x}=\frac{1}{3},$ then $\text{AD}=7\text{x}-4=7\Big(\frac{1}{3}\Big)-4=\frac{-5}{3}<0$
This is not possible since length cannot be negative.
$\Rightarrow x = 4$
View full question & answer→MCQ 191 Mark
Which of the following is a true statement?
- A
Two similar triangles are always congruent.
- B
Two figures are similar if they have the same shape and size.
- ✓
Two triangles are similar if their corresponding sides are proportional.
- D
Two polygons are similar if their corresponding sides are proportional.
AnswerCorrect option: C. Two triangles are similar if their corresponding sides are proportional.
- Is incorrect. Since two similar triangles, may or may not be similar.
- Holds even if the size is not the same.
- Is surely true.
- Holds only if for the polygon, the corresponding sides are proportional and the corresponding angles are equal.
View full question & answer→MCQ 201 Mark
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$ it is given that $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{FD}}$ then:
- A
$\angle\text{B}=\angle\text{E}$
- B
$\angle\text{A}=\angle\text{D}$
- ✓
$\angle\text{B}=\angle\text{D}$
- D
$\angle\text{A}=\angle\text{F}$
AnswerCorrect option: C. $\angle\text{B}=\angle\text{D}$
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$ it is given that $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{FD}}$
So, $\text{A}\leftrightarrow\text{E},\text{B}\leftrightarrow\text{D},\text{C}\leftrightarrow\text{F}$
$\Rightarrow\angle\text{B}=\angle\text{D}$
View full question & answer→MCQ 211 Mark
Two poles of height $6m$ and $11m$ stand vertically upright upright on a plane ground. If distance between their fiit is $12m$ then the distance between their tops is:
- A
$12m$
- ✓
$13m$
- C
$14m$
- D
$15m$
Answer
Let the poles be $AB$ and $CD$.
It is given that:
$AB = 6m$ and $CD = 11m$
Let $AC$ be $12m$.
Draw a perpendicular from $B$ on $CD$ at $E$.
Then,
$BE = 12m$
We have to finf $BD$.
Applying Pythagoras theorem in right-angled triangle BED, we have:
$ B D^2=B E^2+E D$
$ =12^2+5^2(\therefore E D=C D-C E=11-6)$
$= 144 + 25 = 169$
$BD = 13m$ View full question & answer→MCQ 221 Mark
A ladder $25m$ long just reaches the top of a building $24m$ high from the ground. What is the distance of the foot of the ladder from the building?
- ✓
$7m$
- B
$14m$
- C
$21m$
- D
$24.5m$
Answer
Let $BW$ be the ladder and $OB$ be the building.
$\triangle\text{BOW}$ forms a righ-angled triangle.
By Pythagoras theorem,
${BW}^2=O W^2+O B^2 $
$ O W^2=B W^2-{OB}^2 $
${OW}^2=25^2-24^2 $
$ O W^2=(25-24)(25+24) \ldots\left(\text { Using }(a+b)^2=a^2+2 a b+b^2\right) $
$ O W^2=(1)(49) $
$ O W=7 {~m}$
So, the distance of the foot of the ladder from the bulling is $7m$.
View full question & answer→MCQ 231 Mark
In $\triangle\text{ABC},\text{DE }||\text{ BC}$ so that $AD = 2.4cm, AE = 3.2cm$ and $EC = 4.8cm$. Then, $AB =?$
- A
$3.6cm$
- ✓
$6cm$
- C
$6.4cm$
- D
$7.2cm$
AnswerIn $\triangle\text{ABC},\text{DE }||\text{ BC}$
By Basic Proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{2.4}{\text{DB}}=\frac{3.2}{4.8}$
$\Rightarrow\text{BD}=\frac{2.4\times4.8}{3.2}$
$\Rightarrow\text{BD}=3.6\text{cm}$
$\text{AB}=\text{AD}+\text{DB}=2.4+3.6=6\text{cm}$
View full question & answer→MCQ 241 Mark
In the given figure, $\angle\text{BAC}=90^\circ$ and $\text{AD}\perp\text{BC}.$ Then:
- A
$\text{BC}\cdot\text{CD}=\text{BC}^2$
- B
$\text{AB}\cdot\text{AC}=\text{BC}^2$
- ✓
$\text{BD}\cdot\text{CD}=\text{AD}^2$
- D
$\text{AB}\cdot\text{AC}=\text{AD}^2$
AnswerCorrect option: C. $\text{BD}\cdot\text{CD}=\text{AD}^2$
In $\triangle\text{ABC},$
$\angle\text{ABD}=90^\circ-\angle\text{C}$
Similarly, in $\triangle\text{ACD},$
$\angle\text{CAD}=90^\circ-\angle\text{C}$
In $\triangle\text{DBA}$ and $\triangle\text{DAC}$
$\angle\text{ADB}=\angle\text{CDA}=90^\circ$
$\angle\text{ABD}=\angle\text{CAD}=90^\circ-\angle\text{C}$
So, $\triangle\text{DBA}\sim\triangle\text{DAC}$ .....(AA criterion of similarity)
$\frac{\text{BD}}{\text{AD}}=\frac{\text{AD}}{\text{CD}}$
$\Rightarrow\text{BD}\cdot\text{CD}=\text{AD}^2$
View full question & answer→MCQ 251 Mark
A man goes $24m$ due west and then $10m$ due north. How far is he from the starting point?
- A
$34m$
- B
$17m$
- ✓
$26m$
- D
$28m$
Answer
Let $O$ be the starting point.
From $O$ the man goes west that is towards, $W$ till point $A$.
He then moves $10m$ due nirth, that is towards $N$ to point $B$.
$\triangle\text{OAB}$ forms a right - angled triangle.
By Pythagoras theorem,
$ \mathrm{OB}^2=\mathrm{OA}^2+\mathrm{AB}^2 $
$ \mathrm{OB}^2=24^2+10^2 $
$ \mathrm{OB}^2=576+100 $
$ \mathrm{OB}^2=676 $
$ \mathrm{OB}=26 \mathrm{~m}$
So, the man is $26m$ away from the the starting point. View full question & answer→MCQ 261 Mark
In the given figure, $O$ is a point inside a $\triangle\text{MNP}$ such that $\angle\text{MOP}=90^\circ,\text{OM}=16\text{cm}$ and $OP = 12cm$. If $MN = 21cm$ and $\angle\text{NMP}=90^\circ$ then $NP = ?$
- A
$25cm$
- ✓
$29cm$
- C
$33cm$
- D
$35cm$
AnswerCorrect option: B. $29cm$
$\triangle\text{MOP}$ is a right-angled triangle.
By Pythagoras theorem,
$ M P^2=M O^2+O P^2 $
$ M P^2=16^2+12^2 $
$ M P=20 \mathrm{~cm}$
$\triangle \mathrm{NMP}$ is a right-angled triangle.
By PYthagoras theorem,
$ N P^2=21^2+20^2 $
$ N P^2=441+400 $
$ N P=29 \mathrm{~cm}$
View full question & answer→MCQ 271 Mark
The line segments joining the midpoints of the adjacent side of a quadrilateral from:
AnswerThe line segment joining the midpoint of the adjacent sides of a quadrilateral form a parallelogram as shown below.
View full question & answer→MCQ 281 Mark
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$ we have $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}}=\frac{5}{7},$ then $\text{ar}(\triangle\text{ABC}):\text{ar}(\triangle\text{DEF})=?$
- A
$5 : 7$
- ✓
$25 : 49$
- C
$49 : 25$
- D
$125 : 343$
AnswerCorrect option: B. $25 : 49$
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
$\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}}=\frac{5}{7}$
$\Rightarrow\triangle\text{ABC}\sim\triangle\text{DEF}$ ....(SSS criterion for Similarity)
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}=\frac{5^2}{7^2}=\frac{25}{49}$
So, the ratio is $25 : 49$.
View full question & answer→MCQ 291 Mark
In the given figure, $ABCD$ is a trapezium whose diagonals $AC$ and $BD$ intersect at $O$ such that $OA = (3x - 1)cm, OB = (2x + 1)cm, OC = (5x - 3)cm$ and $OD (6x - 5)cm$. Then $x =?$
AnswerThe diagonals of a trapezium divide each other proportinally.
$\Rightarrow\frac{\text{OD}}{\text{OB}}=\frac{\text{OC}}{\text{OA}}$
$\Rightarrow\frac{6\text{x}-5}{2\text{x}+1}=\frac{5\text{x}-3}{3\text{x}-1}$
$\Rightarrow18\text{x}^2-21\text{x}+5=10\text{x}^2-\text{x}-3$
$\Rightarrow8\text{x}^2-20\text{x}+8=0$
$\Rightarrow2\text{x}^2-5\text{x}+2=0$
$\Rightarrow(\text{x}-2)(2\text{x}-1)=0$
$\Rightarrow\text{x}=2$ or $\text{x}=\frac{1}{2}$
if $\text{x}=\frac{1}{2},$ then $\text{OD}=6\text{x}-5=6\Big(\frac{1}{2}\Big)-5=-2$
RThis is not possible since length cannot be negative.
$\Rightarrow x = 2$
View full question & answer→MCQ 301 Mark
If the bisectore of an angle of a triangle bisects the opposite side then the triangle is:
Answer
Let $ABC$ be the triangle and $AD$ be the bisector of $\angle\text{A}.$
Also, $AD$ bisects the opposite side that is $BC$.
$\Rightarrow\text{ BD} =\text{ DC} \dots\text{(i)}$
$\Rightarrow\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\Rightarrow\frac{\text{AB}}{\text{AC}}=1\dots\text{(from(i))}$
$\Rightarrow\text{AB}=\text{AC}$
So, the triangle is an isosceles triangle. View full question & answer→MCQ 311 Mark
$\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $AB = 9.1cm$ and $DE = 6.5cm$. If the perimeter of $\triangle\text{DEF}$ is 25cm, what is the perimeter of $\triangle\text{ABC}?$
- ✓
$35cm$
- B
$28cm$
- C
$42cm$
- D
$40cm$
AnswerCorrect option: A. $35cm$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{ABC}}{\text{Perimeters of }\triangle\text{DEF}}=\frac{\text{AB}}{\text{DE}}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{ABC}}{25}=\frac{9.1}{6.5}$
$\Rightarrow\text{Perimeters of }\triangle\text{ABC}=\frac{9.1\times25}{6.5}$
$\Rightarrow\text{Perimeters of }\triangle\text{ABC}=35\text{cm}$
View full question & answer→MCQ 321 Mark
It is given that $\triangle\text{ABC}\sim\triangle\text{DFE}.$ If $\angle\text{A}=30^\circ,\angle\text{C}=50^\circ,\text{AB}=5\text{cm},\text{AC}=8\text{cm}$ and $\text{DF}=7.5\text{cm}$ then which of the following is true?
- A
$\text{DE}=12\text{cm},\angle\text{F}=50^\circ$
- ✓
$\text{DE}=12\text{cm},\angle\text{F}=100^\circ$
- C
$\text{EF}=12\text{cm},\angle\text{D}=100^\circ$
- D
$\text{EF}=12\text{cm},\angle\text{D}=30^\circ$
AnswerCorrect option: B. $\text{DE}=12\text{cm},\angle\text{F}=100^\circ$
Given that,
$\angle\text{A}=30^\circ,\angle\text{C}=50^\circ$
$\triangle\text{ABC}\sim\triangle\text{DFE}$
$\Rightarrow\angle\text{A}=\angle\text{D}=30^\circ$
$\angle\text{C}=\angle\text{E}=50^\circ$
Using angle sum property, we can find $\angle\text{B}=100^\circ$
So, $\angle\text{B}=\angle\text{F}=100^\circ$
Also, $AB = 5cm, AC = 8cm$ and $DF = 7.5cm$
$\frac{\text{AB}}{\text{DF}}=\frac{\text{BC}}{\text{FE}}=\frac{\text{AC}}{\text{DE}}$
$\Rightarrow\frac{5}{7.5}=\frac{\text{BC}}{\text{FE}}=\frac{8}{\text{DE}}$
$\Rightarrow\frac{5}{7.5}=\frac{8}{\text{DE}}\Rightarrow\frac{8\times7.5}{5}=12\text{cm}$
Hence, $DE = 12cm$ and $\angle\text{F}=100^\circ$
View full question & answer→MCQ 331 Mark
In $\triangle\text{ABC, }\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm}$ and $\text{BC}=6\text{cm}.$ Then, $\angle\text{B}$ is:
- A
$45^\circ $
- B
$60^\circ$
- ✓
$90^\circ$
- D
$120^\circ$
AnswerCorrect option: C. $90^\circ$
In $\triangle\text{ABC},$
$\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm}$ and $\text{BC}=6\text{cm}$
$\text{AB}^2+\text{BC}^2=3\sqrt{3}^2+6^2=108+36=144$
$\text{AC}^2=12^2=144$
$\Rightarrow\text{AB}^2+\text{BC}^2=\text{AC}^2$
So, by the Converse of Pythagoras theorem,
$\triangle\text{ABC}$ is a right angled triangle and since $AC$ is the hypotenuse,
$\angle\text{B}$ which is opposite $\text{AC} = 90^\circ.$
View full question & answer→MCQ 341 Mark
In a $\triangle\text{ABC}$ it is given that $AB = 6cm, AC = 8cm$ and $AD$ is the bisector of $\angle\text{A}.$ Then, $BD : DC =?$
- ✓
$3 : 4$
- B
$9 : 16$
- C
$4 : 3$
- D
$\sqrt{3}:2$
AnswerCorrect option: A. $3 : 4$
since $AD$ is the bisector of $\angle\text{A},$
by the angle bisector theorem,
$\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\Rightarrow\frac{6}{8}=\frac{\text{BD}}{\text{DC}}$
$\Rightarrow\frac{\text{BD}}{\text{DC}}=\frac{6}{8}=\frac{3}{4}$
So, $BD : DC = 3 : 4.$
View full question & answer→MCQ 351 Mark
In $\triangle\text{ABC}\sim\triangle\text{DEF}$ and the perimeters of $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are $30cm$ and $18cm$ respectively. If $BC = 9cm$ then $EF =?$
- A
$6.3cm$
- ✓
$5.4cm$
- C
$7.2cm$
- D
$4.5cm$
AnswerCorrect option: B. $5.4cm$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{ABC}}{\text{Perimeters of }\triangle\text{DEF}}=\frac{\text{BC}}{\text{EF}}$
$\Rightarrow\frac{30}{18}=\frac{9}{\text{EF}}$
$\Rightarrow\frac{9\times18}{\text{30}}$
$\Rightarrow\text{EF}=5.4\text{cm}$
View full question & answer→MCQ 361 Mark
Which of the following is a false statement?
- A
If the areas of two similar triangles are equal then the triangles are congruent.
- ✓
The ratio of the areas of two similar triangles is equal to the ratio of their correspondin sides.
- C
The ratio of the areas of two similar triangles is equal to the ratio of squares of their correspondin medians.
- D
The ratio of the areas of two similar triangles is equal to the ratio of squares of their correspondin altitudes.
AnswerCorrect option: B. The ratio of the areas of two similar triangles is equal to the ratio of their correspondin sides.
Is false since the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
View full question & answer→MCQ 371 Mark
It is given that $\triangle\text{ABC}\sim\triangle\text{PQR}$ and $\frac{\text{BC}}{\text{QR}}=\frac{2}{3}$ then $\frac{\text{ar}(\triangle\text{PQR})}{\text{ar}(\triangle\text{ABC})}=?$
- A
$\frac{2}{3}$
- B
$\frac{3}{2}$
- C
$\frac{4}{9}$
- ✓
$\frac{9}{4}$
AnswerCorrect option: D. $\frac{9}{4}$
$\triangle\text{ABC}\sim\triangle\text{PQR}$ and $\frac{\text{BC}}{\text{QR}}=\frac{2}{3}\Rightarrow\frac{\text{QR}}{\text{BC}}=\frac{3}{2}$
$\frac{\text{ar}(\triangle\text{PQR})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{QR}^2}{\text{BC}^2}=\frac{3^2}{2^2}=\frac{9}{4}$
So, the ratio is $9 : 4$.
View full question & answer→MCQ 381 Mark
Two poles of height $13\ m$ and $7m$ respectively stand vertically on a plane ground at a distance of 8m from each other. The distance between their tops is :
Answer
$OB$ and $AN$ are the two poles.
We have to find the distance between their tops
that is, $BN$
Construction: Draw $\text{NL}\perp\text{OB}$
$\text{OANL}$ is a rectangle $... ($Since all the angles are right anglrs$)$
$LN = OA = 8m$
$OL = AN = 7m$
$\Rightarrow BL = OB - OL = 13m - 7m = 6m$
$\triangle\text{BLN}$ forms a right $-$ angled triangle.
By Pythagoras theorem,
$ \mathrm{BN}^2=\mathrm{LN}^2+\mathrm{BL}^2 $
$ \mathrm{BN}^2=8^2+6^2 $
$ \mathrm{BN}^2=64+36 $
$ \mathrm{BN}^2=100 $
$ \mathrm{BN}=10 \mathrm{~m}$
So, the distance between their tops is $10m$. View full question & answer→MCQ 391 Mark
If $\triangle\text{ABC}\sim\triangle\text{EDF}$ and $\triangle\text{ABC}$ is not similar to $\triangle\text{DEF}$ then which of the following is not true?
- A
$\text{BC}\cdot\text{EF}=\text{AC}\cdot\text{FD}$
- B
$\text{AB}\cdot\text{EF}=\text{AC}\cdot\text{DE}$
- ✓
$\text{BC}\cdot\text{DE}=\text{AB}\cdot\text{EF}$
- D
$\text{BC}\cdot\text{DE}=\text{AB}\cdot\text{FD}$
AnswerCorrect option: C. $\text{BC}\cdot\text{DE}=\text{AB}\cdot\text{EF}$
In $\triangle\text{ABC}\sim\triangle\text{EDF},$ but $\triangle\text{ABC}$ is not similar $\triangle\text{DEF}.$
Since $\triangle\text{ABC}\sim\triangle\text{EDF},$
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{DF}}=\frac{\text{AC}}{\text{EF}}$
So, $\text{AB}\cdot\text{EF}=\text{AC}\cdot\text{DE}$
Hence, $\text{BC}\cdot\text{DE}\not=\text{AB}\cdot\text{EF}.$
View full question & answer→MCQ 401 Mark
$\Rightarrow\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $\text{ar}(\triangle\text{ABC})=36\text{cm}^2$ and $\text{ar}(\triangle\text{DEF})=49\text{cm}^2.$ Then, the ratio of their corresponding sides is:
- A
$36 : 49$
- ✓
$6 : 7$
- C
$7 : 6$
- D
$\sqrt{6}:\sqrt{7}$
AnswerCorrect option: B. $6 : 7$
$\Rightarrow\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}$
$\Rightarrow\frac{36}{49}=\frac{\text{AB}^2}{\text{DE}^2}$
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{6}{7}$
Since the triangle are similar,
$\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}}=\frac{6}{7}$
So, the ratio is $6 : 7$.
View full question & answer→MCQ 411 Mark
in the given figure, two line segments $AC$ and $BD$ intersect each other at the point $P$ such that $PA = 6cm, PB = 3cm, PC = 2.5cm, PD = 5cm,$ $\angle\text{APB}=50^\circ$ and $\angle\text{CDP}=30^\circ$ then $\angle\text{PBA}=?$
- A
$50^\circ$
- B
$30^\circ$
- C
$60^\circ$
- ✓
$100^\circ$
AnswerCorrect option: D. $100^\circ$
In $\triangle\text{PBA}\sim\triangle\text{PCD}$
$\frac{\text{AP}}{\text{PD}}=\frac{\text{PB}}{\text{PC}}\dots\Big(\therefore\frac{\text{AP}}{\text{PD}}=\frac{6}{5 }\text{ and }\frac{\text{BP}}{\text{ BC}}=\frac{3}{2.5}=\frac{6}{5}\Big)$
So, $\angle\text{APB}=\angle\text{DPC}$ ....(Vertically opposite angles)
$\Rightarrow\triangle\text{CAD}\sim\triangle\text{PQR}$ ....(AA criterion for similarity)
$\angle\text{PBA}=\angle\text{DCP}$
In $\triangle\text{PCD},$
$\angle\text{PCD}=180^\circ-\angle\text{DPC}-\angle\text{PDC}$
$\Rightarrow\angle\text{PCD}=180^\circ-50^\circ-30^\circ$
$\Rightarrow\angle\text{PCD}=100^\circ$
So, $\angle\text{PBA}=\angle\text{DCP}=100^\circ.$
View full question & answer→MCQ 421 Mark
The hypotenuse of a right triangle is $25\ cm$. The other two sides are such that one is $5\ cm$ longer than the other. The lengths of these sides are:
- A
$10\ cm, 15\ cm$
- ✓
$15\ cm, 20\ cm$
- C
$12\ cm, 17\ cm$
- D
$13\ cm, 18\ cm$
AnswerCorrect option: B. $15\ cm, 20\ cm$
The pythagoeas theorem states that, in a right-angled triangle, the hypotenuse square is equal to the sum of the squares of the opposite sides.
- $10^2+15^2=100+225=325$
hypotenuse$^2=25^2=625$
So, this is not possible by $(i).$
- $15^2+20^2=225+400=625$
hypotenuse$^2=25^2=626$
So, the lengths of the sides are 15cm and 20cm.
- $12^2+172=144+289=433$
hypotenuse$^2=25^2=625$
So, this is not possible by $(i)$
- $13^2+18^2=169+324=493$
hypotenuse$^2=25^2=626$
So, this is not possible by $(i).$ View full question & answer→MCQ 431 Mark
Corresponding sides of two similar triangles are in the ratio $4 : 9$. Areas of these triangles are in the ratio:
- A
$2 : 3$
- B
$4 : 9$
- C
$9 : 4$
- ✓
$16 : 81$
AnswerCorrect option: D. $16 : 81$
Let the areas of the triangle be $A_1$ and $A_2$.
$\frac{\text{ar}(\text{A}_1)}{\text{ar}(\text{A}_2)}=\frac{4^2}{9^2}=\frac{16}{81}$
So, the ratio is $16 : 81$.
View full question & answer→MCQ 441 Mark
A vertical stick $1.8m$ long casts a shadow $45cm$ long on the ground. At the same time, what is the lenght of the shadow of a pole 6m high?
- A
$2.4m$
- B
$1.35m$
- ✓
$1.5m$
- D
$13.5m$
AnswerCorrect option: C. $1.5m$
Let $AN$ be the vertical stick and $AW$ be its shadow.
Let $OB$ be the pole and $OW$ be its shadows.
$AW = 45cm = 0.45m$
$AN = 1.8m$
$OB = 6m$
Ratio of actual lengths = ratio of their shadows
$\Rightarrow\frac{\text{OB}}{\text{AN}}=\frac{\text{OW}}{\text{AW}}$
$\Rightarrow\frac{6}{1.8}=\frac{\text{OW}}{0.45}$
$\Rightarrow\text{OW}=\frac{6\times0.45}{1.8}$
$\Rightarrow\text{OW}=1.5\text{m}$ View full question & answer→MCQ 451 Mark
In $\triangle\text{ABC},$ if $AB = 16cm, BC = 12cm$ and $AC = 20cm$, then $\triangle\text{ABC}$ is:
AnswerNote that first check if the sum of any two sides is greater than the third side.
Since in this triangle, it holds, a triangle is possible.
In $\triangle\text{ABC},$
if $AB = 16cm, BC = 12cm$ and $AC = 20cm$
Consider,
$ A B^2+B C^2=16^2+12^2=400 $
$ A C^2=20^2=400$
By the Converse of Pythagoras theorem,
$\triangle\text{ABC}$ will be a right-angled triangle.
View full question & answer→MCQ 461 Mark
In $\triangle\text{ABC}$ and $\triangle\text{PQR},$ we have $\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}=\frac{\text{CA}}{\text{PQ}}$ then:
- ✓
$\triangle\text{PQR}\sim\triangle\text{CAB}$
- B
$\triangle\text{PQR}\sim\triangle\text{ABC}$
- C
$\triangle\text{CAB}\sim\triangle\text{PQR}$
- D
$\triangle\text{BCA}\sim\triangle\text{PQR}$
AnswerCorrect option: A. $\triangle\text{PQR}\sim\triangle\text{CAB}$
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
$\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}=\frac{\text{CA}}{\text{PQ}}$
So, $\angle\text{A}\leftrightarrow\angle\text{Q},\angle\text{B}\leftrightarrow\angle\text{R},\angle\text{C}\leftrightarrow\angle\text{P},$
$\Rightarrow\triangle\text{CAB}\sim\triangle\text{PQR}$
View full question & answer→MCQ 471 Mark
If $\triangle\text{ABC}\sim\triangle\text{QRP},\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{9}{4},\text{AB}=18\text{cm}$ and $\text{BC}=15\text{cm}$ then $\text{PR}=?$
- A
$8cm$
- ✓
$10cm$
- C
$12cm$
- D
$\frac{20}{3}\text{cm}$
AnswerCorrect option: B. $10cm$
$\triangle\text{ABC}\sim\triangle\text{QRP}$
$\Rightarrow\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}=\frac{\text{AC}}{\text{PQ}}$
Also, $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{AB}^2}{\text{BC}^2}$
$\Rightarrow\frac{9}{4}=\frac{\text{AB}^2}{\text{BC}^2}$
$\Rightarrow\frac{\text{AB}}{\text{QR}}=\frac{3}{2}$
So, $\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}\Rightarrow\frac{\text{BC}}{\text{PR}}=\frac{3}{2}$
$\Rightarrow\frac{15}{\text{PR}}=\frac{3}{2}$
$\Rightarrow\text{PR}=10\text{cm}.$
View full question & answer→MCQ 481 Mark
the lengths of the diagonals of a rhombus are $24cm$ and $10cm$. The length of each side of the rhombus is:
- A
$12cm$
- ✓
$13cm$
- C
$14cm$
- D
$17cm$
AnswerCorrect option: B. $13cm$
In an rhombus, the diagonals are perpendicular bisetoes of each other, and sides are equal to eachother.
So, $\text{AO}=\frac{1}{2}\text{AC}=12\text{cm}$
$\text{OD}=\frac{1}{2}\text{BD}=5\text{cm}$
In right-angled $\triangle\text{AOD},$
$ A D^2=A O^2+O D^2 $
$ \Rightarrow A D^2=12^2+5^2 $
$ \Rightarrow A D^2=144+25 $
$ \Rightarrow A D^2=169 $
$ \Rightarrow A D=13 \mathrm{~cm}$
So, the length of the each side of the rhombus is $13cm$. View full question & answer→MCQ 491 Mark
The height of an equilateral triangle having each side $12cm$, is:
- A
$6\sqrt{2}\text{ cm}$
- ✓
$6\sqrt{3}\text{ cm}$
- C
$3\sqrt{6}\text{ cm}$
- D
$6\sqrt{6}\text{ cm}$
AnswerCorrect option: B. $6\sqrt{3}\text{ cm}$
Let $\triangle\text{ABC}$ be the equilateral triangle and $AD$ be the height.
The height of an equilateral triangle is the same as its median.
So, $AD = 6m$
$\triangle\text{ABC}$ is a right-angled triangle.
By Pythagoras theorem,
$ \mathrm{AC}^2=\mathrm{AC}^2+\mathrm{AD}^2 $
$ \Rightarrow \mathrm{DC}^2=\mathrm{AC}^2-\mathrm{AD}^2 $
$ \Rightarrow \mathrm{DC}^2=12^2-6^2 $
$ \Rightarrow \mathrm{DC}^2=144-36 $
$ \Rightarrow \mathrm{DC}^2=108$
$\Rightarrow\text{DC}=\sqrt{3\times4\times9}$
$\Rightarrow\text{DC}=6\sqrt{3}\text{ cm}$
So, the height is $6\sqrt3\text{cm}.$ View full question & answer→MCQ 501 Mark
The areas of two similar triangles are $25cm^2$ and $36cm^2$ respectively. If the altitude of the first triangle is $3.5cm$ then the corresponding altitude of the other triangle is:
- A
$5.6cm$
- B
$6.3cm$
- ✓
$4.2cm$
- D
$7cm$
AnswerCorrect option: C. $4.2cm$
We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let h be the altitude of the other triangle.
Therefore,
$\frac{25}{36}=\frac{(3.5)^2}{\text{h}^2}$
$\Rightarrow\text{h}^2=\frac{(3.5)^2\times36}{25}$
$\Rightarrow\text{h}^2=17.64$
$\Rightarrow\text{h}=4.2\text{cm}$
View full question & answer→