MCQ 11 Mark
In Figure, $\triangle\text{ABC}$ is an isosceles triangle, right-angled at $C$. Therefore.
- ✓
$\mathrm{AB}^2=2 \mathrm{AC}^2$
- B
$\mathrm{BC}^2=2 \mathrm{AB}^2$
- C
$\mathrm{AC}^2=2 \mathrm{AB}^2$
- D
$\mathrm{AB} 2=4 \mathrm{AC} 2$
AnswerCorrect option: A. $\mathrm{AB}^2=2 \mathrm{AC}^2$
Since $\triangle\text{ABC}$ is a right $-$ angled triangle at $C$.
Using pythagoras theorem in $\triangle\text{ABC}$
$\mathrm{Hypotenuse^2= (Height)^2+ (Base)^2}$
$\therefore\text{AB}^2=\text{AC}^2+\text{BC}^2$
$\Rightarrow\text{AB}^2=\text{AC}^2+\text{AC}^2$
$[\therefore\text{BC = AC}]$
$\Rightarrow\text{AB}^2=2\text{AC}^2$
Hence, proved. View full question & answer→MCQ 21 Mark
In fig. the graph of the polynomial $p(x)$ is given. The number of zeroes of the polynomial is :
AnswerThe number of zeroes is $2$ as the graph intersects the $x-$ axis at $2$ points.
View full question & answer→MCQ 31 Mark
If $\triangle\text{ABC} \sim \triangle\text{DEF}$ such that $AB = 1.2\ cm$ and $DE = 1.4\ cm,$ the ratio of the areas of $\triangle\text{ABC}$ and $\triangle\text{DEF}$ is :
- A
$49 : 36$
- B
$6 : 7$
- C
$7 : 6$
- ✓
$36 : 49$
AnswerCorrect option: D. $36 : 49$
We have,
$\triangle\text{ABC} \sim \triangle\text{DEF}$
$AB = 1.2\ cm$ and $DF = 1.4\ cm$
By area of similar triangle theorem
$\frac{\text{Area}(\triangle\text{ABC})}{\text{Area}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}$
$=\frac{1.2^2}{1.4^2}$
$=\frac{1.44}{1.96}$
$=\frac{36}{49}$
View full question & answer→MCQ 41 Mark
Choose the correct answer from the given four options : In $\angle\text{BAC}=90^\circ$ and $\text{ AD}\perp\text{BC}.$ Then, Traingles
- A
$ \mathrm{BD} \times \mathrm{CD}=\mathrm{BC}^2 $
- B
$ \mathrm{AB} \times \mathrm{AC}=\mathrm{BC}^2 $
- ✓
$ \mathrm{BD} \times \mathrm{CD}=\mathrm{AD}^2 $
- D
$ \mathrm{AB} \times \mathrm{AC}=\mathrm{AD}^2 $
AnswerCorrect option: C. $ \mathrm{BD} \times \mathrm{CD}=\mathrm{AD}^2 $
$\angle\text{D}=\angle\text{D}=90^\circ$
$\angle\text{DBA}=\angle\text{DAC}\ \ \big[$each equal to $ 90^\circ-<\text{c}\big]$
$\therefore\triangle\text{ADB}\sim\triangle\text{ADC}\ \ \big[$by $\text{AAA}$ simillariy criterion $\big]$
$\therefore\frac{BD}{AD}=\frac{AD}{CD}$
$\Rightarrow\text{BD}\times\text{CD}=\text{AD}^2$ View full question & answer→MCQ 51 Mark
In the given figure, $\text{DE}\parallel\text{BC} , AB = 15\ cm, BD = 6\ cm, AC = 25\ cm,$ then $AE$ is equal to :
- A
$20\ cm.$
- B
$18\ cm.$
- C
$10\ cm.$
- ✓
$15\ cm$.
AnswerCorrect option: D. $15\ cm$.
Since $\text{DE} \parallel\text{BC},$ then using Thales theorem,
$\Rightarrow\frac{\text{AB}}{\text{DB}}=\frac{\text{AC}}{\text{EC}}$
$\Rightarrow\frac{15}{6}=\frac{25}{\text{EC}}$
$\Rightarrow\text{EC}=10 \text{ cm}$
Now $, AE = AC -EC $
$= 25-10 = 15\ cm$
View full question & answer→MCQ 61 Mark
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
$81 : 16$
- ✓
$16 : 81$
AnswerCorrect option: D. $16 : 81$
Let $\text{ABC}$ and $\text{DEF}$ are two similar triangles, such that,
$\triangle\text{ABC}\sim\triangle\text{DEF}$
And $\frac{\text{AB}}{\text{DE}}=\frac{\text{AC}}{\text{DF}}=\frac{\text{BC}}{\text{EF}}=\frac{4}{9}$
As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,
$\therefore\frac{\text{Area}(\triangle\text{ABC})}{(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}$
$\therefore\frac{\text{Area}(\triangle\text{ABC})}{(\triangle\text{DEF})}=(\frac{4}{9})^2$
$=\frac{16}{81}=16:81$
View full question & answer→MCQ 71 Mark
In an isosceles $\triangle\text{ABC},$ if $\mathrm{AC}=\mathrm{BC}$ and $\mathrm{AB}^2=2 \mathrm{AC}{ }^2$ 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 $\mathrm{AC}=\mathrm{BC}$ and $\mathrm{AB}^2=2 \mathrm{AC}{ }^2$
$\Rightarrow AB^2= BC^2+ AC^2....(\therefore\text{AC}=\text{BC})$
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 81 Mark
$\triangle\text{ABC}$ is a right triangle right $-$ angled at $A$ and $\text{AD}\perp\text{BC}.$ Then, $\frac{\text{BD}}{\text{DC}}=$
- ✓
$\Big(\frac{\text{AB}}{\text{AC}}\Big)^2$
- B
$\frac{\text{AB}}{\text{AC}}$
- C
$\Big(\frac{\text{AB}}{\text{AD}}\Big)^2$
- D
$\frac{\text{AB}}{\text{AD}}$
AnswerCorrect option: A. $\Big(\frac{\text{AB}}{\text{AC}}\Big)^2$
In right angled $\triangle\text{ABC},\ \angle\text{A}=90^\circ$
$\text{AD}\perp\text{BC}$
$\therefore\triangle\text{ABD}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AB}}{\text{BC}}=\frac{\text{BD}}{\text{AB}}$
$\Rightarrow\text{AB}^2=\text{BD}\times\text{BC}\ \ ...(\text{i})$
Similarly $\triangle\text{ACD}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AC}}{\text{BC}}=\frac{\text{DC}}{\text{AC}}$
$\Rightarrow\text{AC}^2=\text{DC}\times\text{BC}\ \ ...(\text{ii})$
Dividing $(ii)$ by $(i)$
$\frac{\text{BD}\times\text{BC}}{\text{DC}\times{\text{BC}}}=\frac{\text{AB}^2}{\text{AC}^2}$
$\Rightarrow\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}^2}{\text{AC}^2}$
Hence $\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}^2}{\text{AC}^2}$ View full question & answer→MCQ 91 Mark
In an equilateral triangle $\text{ABC},$ 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 $\triangle\text{ADC},$
$ A B^2=A D^2+B D^2$
$ \Rightarrow A B^2=A D^2+B D^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 101 Mark
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $DE = 3\ cm, EF = 2\ cm, DF = 2.5\ cm, BC = 4\ cm,$ then perimeter of $\triangle\text{ABC}$ is :
- A
$18\ cm.$
- B
$20\ cm.$
- C
$12\ cm.$
- ✓
$15\ cm.$
AnswerCorrect option: D. $15\ cm.$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$DE = 3\ cm, EF = 2\ cm, DF = 2.5\ cm, BC = 4\ cm$
$\because\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore$ Perimeter of $\triangle\text{DEF}$
$= DE + EF + DF$
$= 3 + 2 + 2.5 = 7.5\ cm$
Now $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}}$
$=\frac{\text{AB}+\text{BC}+\text{CA}}{\text{DE}+\text{EF}+\text{DF}}$
$=\frac{4}{2}=\frac{\text{AB}+\text{BC}+\text{CA}}{7.5}$
$\Rightarrow\text{AB}+\text{BC}+\text{CA}=\frac{4\times7.5}{2}=15$
$\therefore$ Perimeter of $\triangle\text{ABC}=15\text{ cm}.$ View full question & answer→MCQ 111 Mark
In $\triangle\text{ABC},\text{DE }\|\text{ BC}$ such that $\frac{\text{AD}}{\text{DB}}=\frac{3}{5}. AC = 5.6\ cm$ then $AE =?$
- A
$4.2\ cm$
- B
$3.1\ cm$
- C
$2.8\ cm$
- ✓
$2.1\ cm$
AnswerCorrect option: D. $2.1\ cm$
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 121 Mark
$\triangle\text{ABC}$ is such that $AB = 3\ cm, BC = 2\ cm$ and $CA = 2.5\ cm$. If $\triangle\text{DEF}\sim\triangle\text{ABC}$ and $EF = 4\ cm,$ then perimeter of $\triangle\text{DEF}$ is :
- A
$7.5\ cm.$
- ✓
$15\ cm.$
- C
$22.5\ cm.$
- D
$30\ cm.$
AnswerCorrect option: B. $15\ cm.$
$\triangle\text{DEF}\sim\triangle\text{ABC}$
$AB = 3\ cm, BC = 2\ cm, CA = 2.5\ cm, EF = 4\ cm.$
$\triangle\text{s}$ are similar
$\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}}$
$\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5}$
Now $\frac{\text{DE}}{3}=\frac{4}{2}$
$\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{ cm}$
and $\text{FD}=\frac{4}{2}$
$\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{ cm}$
$\therefore$ Perimeter of $\triangle\text{DEF}$
$=6+4+5=15\text{ cm}.$
View full question & answer→MCQ 131 Mark
In a $\triangle\text{ABC},$ it is given that $AD$ is the internal bisector of $\angle\text{A}.$ If $AB = 10\ cm, AC = 14\ cm$ and $BC = 6\ cm,$ then $CD = ?$
- A
$4.8\ cm$
- ✓
$3.5\ cm$
- C
$7\ cm$
- D
$10.5\ cm$
AnswerCorrect option: B. $3.5\ cm$
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 $, CD = 3.5\ cm.$
View full question & answer→MCQ 141 Mark
If $D, E, F$ are the mid $-$ points of sides $BC, CA$ and $AB$ respectively of $\triangle\text{ABC},$ then the ratio of the areas of triangles $\text{DEF}$ and $\text{ABC}$ is :
- ✓
$1 : 4$
- B
$1 : 2$
- C
$2 : 3$
- D
$4 : 5$
AnswerCorrect option: A. $1 : 4$
Given : In $\triangle\text{ABC}, D, E$ and $F$ are the midpoints of $BC, CA,$ and $AB$ respectively.
To find : Ratio of the areas of $\triangle\text{DEF}$ and $\triangle\text{ABC}$
Since it is given that $D$ and, $E$ are the midpoints of $BC,$ and $AC$ respectively.
Therefore $DE \| AB, DE \| FA ….(1)$
Again it is given that $D$ and, $F$ are the midpoints of $BC,$ and, $AB$ respectively.
Therefore, $DF \| CA, DF \| AE …(2)$
From $(1)$ and $(2)$ we get $\text{AFDE}$ is a parallelogram.
Similarly we can prove that $\text{BDEF}$ is a parallelogram.
Now, in $\triangle\text{ADE}$ and $\triangle\text{ABC}$
$\angle\text{FDE}=\angle\text{A}\ ($Opposite angles of $\|^{g m} \ \text{AFDE})$
$\angle\text{DEF}=\angle\text{B} \ ($Opposite angles of $\|^{g m} \ \text{BDEF})$
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{\text{ar}(\triangle\text{DEF})}{\text{ar}(\triangle\text{ABC})}=\Big(\frac{\text{DE}}{\text{AB}}\Big)^2$
$\frac{\text{ar}(\triangle\text{DEF})}{\text{ar}(\triangle\text{ABC})}=\bigg(\frac{\frac{1}{2}\text{AB}}{\text{AB}}\bigg)^2\Big(\text{Since DE}=\frac{1}{2}\text{AB}\Big)$
$\frac{\text{ar}(\triangle\text{DEF})}{\text{ar}(\triangle\text{ABC})}=\Big(\frac{1}{4}\Big)$
Hence the correct option is $A.$ View full question & answer→MCQ 151 Mark
In a $\triangle\text{ABC},$ point $D$ is on side $AB$ and point $E$ is on side $AC,$ such that $\text{BCED}$ is a trapezium. If $DE : BC = 3 : 5,$ then $\text{Area}(\triangle\text{ADE}):\text{Area}(\Box\text{BCED})=$
- A
$3 : 4.$
- ✓
$9 : 16.$
- C
$3 : 5.$
- D
$9 : 25.$
AnswerCorrect option: B. $9 : 16.$
Given : In $\triangle\text{ABC}, D$ is on side $AB$ and point $E$ is on side $AC,$ such that $\text{BCED}$ is a trapezium. $E : BC = 3 : 5.$
To find : Calculate the ratio of the areas of $\triangle\text{ADE}$ and the trapezium $\text{BCED}$.
In $\triangle\text{ADE}$ and $\triangle\text{ABC},$
$\angle\text{ADE}=\angle\text{B} \ ($Corresponding angles$)$
$\angle\text{A}=\angle\text{A}\ ($Common$)$
$\therefore\triangle\text{ADE}\sim\triangle\text{ABC}\ (\text{AA}$ similarity$)$
We know that
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\triangle\text{ABC})}=\frac{\text{DE}^2}{\text{BC}^2}$
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\triangle\text{ABC})}=\frac{3^2}{5^2}$
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\triangle\text{ABC})}=\frac{9}{25}$
Let area of $\triangle\text{ADE}=9\text{x}\text{ sq}.$ units and area of $\triangle\text{ABC}=25\text{x}\text{ sq}.$ units
$\text{Ar[trap BCED]}=\text{Ar}(\triangle\text{ABC})-\text{Ar}(\triangle\text{ADE})$
$=25\text{x}-9\text{x}$
$=16\text{x sq. units}$
Now,
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\text{trap BCED})}=\frac{9\text{x}}{16\text{x}}$
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\text{trap BCED})}=\frac{9}{16}$
Hence the correct answer is $B.$ View full question & answer→MCQ 161 Mark
Tick the correct answer and justify : $\text{ABC}$ and $\text{BDE}$ are two equilateral triangles such that $D$ is the mid $-$ point of $BC$. Ratio of the areas of triangles $\text{ABC}$ and $\text{BDE}$ is :
- A
$2 : 1$
- B
$1 : 2$
- ✓
$4 : 1$
- D
$1 : 4$
AnswerCorrect option: C. $4 : 1$
$\triangle\text{ABC}$ and $ \triangle\text{BDE}$ are both equilateral triangles.
$\therefore\ \triangle\text{ABC}\sim\triangle\text{BDE}$
$[$Using $\text{AAA}$ similar condition$]$.
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
$\therefore\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{\text{AB}^2}{\text{BD}^2}$
$\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{\text{BC}}{\text{BD}}\Big)^2$
$[\therefore AB = BC = CA]$
$\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}= \Big(\frac{2\text{BD}}{\text{BD}}\Big)^2$
$\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{4}{1}$
View full question & answer→MCQ 171 Mark
In a $\triangle\text{ABC},$ if $DE$ is drawn parallel to $BC,$ cutting $AB$ and $AC$ at $D$ and $E$ respectively such that $AB = 7.2\ cm, AC = 6.4\ cm$ and $AD = 4.5\ cm$. Then $, AE =?$
- A
$5.4\ cm$
- ✓
$4\ cm$
- C
$3.6\ cm$
- D
$3.2\ cm$
AnswerCorrect option: B. $4\ cm$
In $\triangle\text{ABC},\text{DE }\|\text{ BC}$
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 181 Mark
$\triangle\text{ABC}$ is an isosceles triangle with $AB = AC = 13\ cm$ and the length of altitude from $A$ on $BC$ is $5\ cm$. Then $, BC =?$
- A
$12\ cm$
- B
$16\ cm$
- C
$18\ cm$
- ✓
$24\ cm$
AnswerCorrect option: D. $24\ cm$
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 $
$\Rightarrow BD = 12\ cm$
$\Rightarrow DC = 2\ cm$
So $, BC = 12 + 12 = 24\ cm.$ View full question & answer→MCQ 191 Mark
Choose the correct answer from the given four options : If in two traingles $\text{ABC}$ and $\text{PQR},$
$\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}=\frac{\text{CA}}{\text{PQ}},$ then :
- A
$\triangle\text{PQR}\sim\triangle\text{CAB}$
- B
$\triangle\text{PQR}\sim\triangle\text{ABC}$
- ✓
$\triangle\text{CBA}\sim\triangle\text{PQR}$
- D
$\triangle\text{BCA}\sim\triangle\text{PQR}$
AnswerCorrect option: C. $\triangle\text{CBA}\sim\triangle\text{PQR}$
Given, in two $\triangle\text{PQR}=\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}=\frac{\text{CA}}{\text{PQ}} $
Which shows that sidles of one triangle are proportional to the side of thew other triangle,
then their Corresponding angles are also equal, so by $\text{SSS}$ similarity, triangles are similar.
$\text{i.e.,}\ \triangle\text{CAB}\sim\triangle\text{PQR}$
View full question & answer→MCQ 201 Mark
If $\triangle\text{ABC}$ is an equilateral triangle such that $\text{AD}\perp\text{BC},$ then $AD^2 =$
- A
$\frac{3}{2}\text{DC}^2$
- B
$\text{2DC}^2$
- ✓
$3\text{CD}^2$
- D
$4\text{DC}^2$
AnswerCorrect option: C. $3\text{CD}^2$
In equilateral $\triangle\text{ABC},\ \text{AD}\perp\text{BC}$
$AD$ bisects $BC$ at $D$
$\therefore BD = DC$
Now in right $\triangle\text{ADC},$
$A C^2=A D^2+D C^2 \ ($Pythagoras Theorem$)$
$A D^2=A C^2-D C^2$
$=B C^2-D C^2\ (\because AC = BC = AB)$
$=(2 D C)^2-D C^2\ (\because D$ is mid point of $BC)$
$=4 D C^2-D C^2=3 D C^2$
$=3 C D^2$ View full question & answer→MCQ 211 Mark
If $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are similar such that $2AB = DE$ and $BC = 8\ cm,$ then $EF =$
- ✓
$16\ cm.$
- B
$12\ cm.$
- C
$8\ cm.$
- D
$4\ cm.$
AnswerCorrect option: A. $16\ cm.$
Given, $\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}$
$\Rightarrow\frac{\text{AB}}{2\text{AB}}=\frac{8}{\text{EF}}$
$\Rightarrow\text{EF}=8\times2$
$\Rightarrow\text{EF}=16\text{ cm}$
View full question & answer→MCQ 221 Mark
In $\triangle\text{ABC},$ it is given that $AB = 9\ cm, BC = 6\ cm$ and $CA = 7.5\ cm.$ Also, $\triangle\text{DEF}$ is given such that $EF = 8\ cm$ and $\triangle\text{DEF}\sim\triangle\text{ABC}.$ Then, perimeter of $\triangle\text{DEF} \ 1s $:
- A
$22.5\ cm$
- B
$25\ cm$
- C
$27\ cm$
- ✓
$30\ cm$
AnswerCorrect option: D. $30\ cm$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{Perimeters of }\triangle\text{ABC}}=\frac{\text{EF}}{\text{BC}}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{AB+BC+AC}}=\frac{8}{6}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{9+6+7.5}=\frac{8}{6}$
$\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{22.5}=\frac{4}{3}$
$\Rightarrow\text{Perimeters of }\triangle\text{DEF}=\frac{4\times22.5}{3}$
$\Rightarrow\text{Perimeters of }\triangle\text{DEF}=30\text{ cm}$
View full question & answer→MCQ 231 Mark
In a thombus of side $10\ cm,$ one of the diagonals is $12\ cm$ long. The length of the second diagonal is :
- A
$20\ cm$
- B
$18\ cm$
- ✓
$16\ cm$
- D
$22\ cm$
AnswerCorrect option: C. $16\ cm$
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^2 $
$ \Rightarrow A O^2=10^2-6^2 $
$ \Rightarrow A O^2=100-36 $
$ \Rightarrow A O^2=64 $
$\Rightarrow AO = 8\ cm$
So $, AC = 2AO = 2(8) = 16\ cm$
Thus, the length of the second diagonal is $16\ cm.$ View full question & answer→MCQ 241 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.
$AW = 2\ cm$
$AN = 5m$
$OW = 12.5m$
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 251 Mark
Choose the correct answer from the given four options : The lengths of the diagonals of a rhombus are $16\ cm$ and $12\ cm$. Then, the length of the side of the rhombus is :
- A
$9\ cm$
- ✓
$10\ cm$
- C
$8\ cm$
- D
$20\ cm$
AnswerCorrect option: B. $10\ cm$
We know that, the diagonals of a rhombus are perpendicular bisector of each other.
Given $, AC = 16\ cm$ and $BD = 12\ cm\ [$left$]$
$\therefore AO= 8\ cm, $
SO $= 6\ cm$
and $\angle\text{AOB}=90^\circ$
In right angled $\angle\text{AOB},$
$\mathrm{AB}^2=\mathrm{AO}^2+\mathrm{OB}^2\ [$by pythagoras theorem$]$
$\Rightarrow \mathrm{AB}^2=8^2+6^2$
$\Rightarrow \mathrm{AB}^2=64+36$
$\mathrm{AB}^2=100$
$\therefore AB = 10\ cm$ View full question & answer→MCQ 261 Mark
A man goes $24m$ due west and then $7m$ due north. How far is he from the starting point ?
- A
$31m.$
- B
$17m.$
- ✓
$25m.$
- D
$26m.$
AnswerCorrect option: C. $25m.$
Het a man be at $O$ and goes to $24m$ due west and then $7m$ due north.
Distance of man from starting point be $OB$
So,
In right $\triangle\text{ABO},$
$ O B^2=A B^2+A O^2 $
$ \Rightarrow O B^2=(7)^2+(24)^2 $
$ \Rightarrow O B^2=49+576 $
$\Rightarrow O B^2=625 $
$\Rightarrow OB = 25m$
Thus, the distance of man from starting point is $25m$. View full question & answer→MCQ 271 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.
$AW = 3.6\ cm$
$AN = 6m$
$OW = 18m$
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 281 Mark
In a triangle, the perpendicular from the vertex to the base bisect the base. The triangle is :
- A
Right $-$ angled
- ✓
- C
- D
Obtuse $-$ angled
AnswerIn an isosceles triangle, the perpendicular from the vertex to the base bisects the base.
View full question & answer→MCQ 291 Mark
Tick the correct answer and justify : 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
$81 : 16$
- ✓
$16 : 81$
AnswerCorrect option: D. $16 : 81$
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
Therefore,
Ratio of the areas of these triangles
$=4^2: 9^2=16: 81$
View full question & answer→MCQ 301 Mark
$\text{ABC}$ and $\text{BDE}$ are two equoilateral triangles such tha $D$ is the midpoint of $BC$. Ratio of the areas of triangles $\text{ABC}$ and $\text{BDE}$ 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 311 Mark
If $\text{ABC}$ is an isosceles triangle and $D$ is a point on $BC$ such that $\text{AD}\perp\text{BC},$ then :
- ✓
$ \mathrm{AB}^2-\mathrm{AD}^2=\mathrm{BD} \times \mathrm{DC} $
- B
$ \mathrm{AB}^2-\mathrm{AD}^2=\mathrm{BD}^2-\mathrm{DC}^2 $
- C
$ \mathrm{AB}^2+\mathrm{AD}^2=\mathrm{BD} \times \mathrm{DC} $
- D
$\mathrm{AB}^2+\mathrm{AD}^2=\mathrm{BD}^2-\mathrm{DC}^2 $
AnswerCorrect option: A. $ \mathrm{AB}^2-\mathrm{AD}^2=\mathrm{BD} \times \mathrm{DC} $
If $\triangle\text{ABC}, AB = AC$
$D$ is a point on $BC$ such that
$\text{AD}\perp\text{BC}\ , AD$ bisects $BC$ at $D$
In right $\triangle\text{ABD},$
$ A B^2=A D^2+B D^2 $
$ A B^2-A D^2=B D^2$
$=B D \times B D=B D \times D C(B D=D C) .$ View full question & answer→MCQ 321 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 331 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 $\mathrm{A}_1$ and $\mathrm{A}_2$,
and let their corresponding heights be $\mathrm{h}_1$ and $\mathrm{h}_2$,
$\frac{\text{ar}(\text{A}_1)}{\text{ar}(\text{A}_2)}=\frac{\text{h}_1^2}{\text{h}_2^2}$
$\Rightarrow\frac{25}{36}=\frac{\text{h}_1^2}{\text{h}_2^2}$
$\Rightarrow\frac{\text{h}_1}{\text{h}_2}=\frac{5}{6}$
So, the ratio of their heights is $5 : 6$.
View full question & answer→MCQ 341 Mark
If in two $\triangle\text{ABC}$ and $\text{PQR},\frac{\text{AB}}{\text{QR}} =\frac{\text{BC}}{\text{PR}}=\frac{\text{CA}}{\text{PQ}},$ then,
- ✓
$\triangle{\text{PQR}}\sim\triangle\text{CAB}$
- B
$\triangle{\text{PQR}}\sim\triangle\text{ABC}$
- C
$\triangle{\text{CBA}}\sim\triangle\text{PQR}$
- D
$\triangle{\text{BCA}}\sim\triangle\text{PQR}$
AnswerCorrect option: A. $\triangle{\text{PQR}}\sim\triangle\text{CAB}$
Given that, If in two $\triangle\text{ABC}$ and $\text{PQR},$
$\frac{\text{AB}}{\text{QR}} =\frac{\text{BC}}{\text{PR}}=\frac{\text{CA}}{\text{PQ}}$
If sides of one triangle are proportional to the side of the other triangle, and their corresponding angles are also equal, then both the triangles are similar by $\text{SSS}$ similarity.
$\therefore\triangle\text{PQR}\sim \text{CAB}$
View full question & answer→MCQ 351 Mark
Two isosceles triangles have equal angles and their areas are in the ratio $16 : 25$. The ratio of their corresponding heights is :
- ✓
$4 : 5.$
- B
$5 : 4.$
- C
$3 : 2.$
- D
$5 : 7.$
AnswerCorrect option: A. $4 : 5.$
Given : two isosceles triangles have equal vertical angles and their areas are in the ratio of $16 : 25.$
To find : Ratio of their corresponding heights.
Let $\triangle\text{ABC}$ and $\triangle\text{PQR}$ be two isosceles triangles such that $\angle\text{A}=\angle\text{P}.$
Suppose $\text{AD}\perp\text{BC}$ and $\text{PS}\perp\text{QR}.$
In $\triangle\text{ABC}$ and $\triangle\text{PQR},$
$\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}$
$\angle\text{A}=\angle\text{P}$
$\therefore\triangle\text{ABC}\sim\triangle\text{PQR}\ (\text{SAS}$ similarity)
We know that the ratio of areas of two similar triangles is equal to the squares of their corresponding altitudes.
Hence,
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{PQR})}=\Big(\frac{\text{AD}}{\text{PS}}\Big)^2$
$\Rightarrow\frac{16}{25}=\Big(\frac{\text{AD}}{\text{PS}}\Big)^2$
$\Rightarrow\frac{\text{AD}}{\text{PS}}=\frac{4}{5}$
Hence we got the result as $A.$ View full question & answer→MCQ 361 Mark
In a $\triangle\text{ABC},\ \angle\text{A}=90^\circ, AB = 5\ cm$ and $AC = 12\ cm$. If $\text{AD}\perp\text{BC},$ then $AD =$
- A
$\frac{13}{2}\text{ cm}.$
- ✓
$\frac{60}{13}\text{ cm}.$
- C
$\frac{13}{60}\text{ cm}.$
- D
$\frac{2\sqrt{15}}{13}\text{ cm}.$
AnswerCorrect option: B. $\frac{60}{13}\text{ cm}.$
In$\triangle\text{ABC},$
$\angle\text{A}=90^\circ, AB = 5\ cm, AC = 12\ cm$
$\text{AD}\perp\text{BC}$
$\text{BC}^2=\text{AB}^2+\text{AC}^2 \ ($Pythagoras Theorem$)$
$=(5)^2+(12)^2$
$=25+144=169=(13)^2$
$\therefore\text{BC}=13\text{ cm}$
Now area of $\triangle\text{ABC}=\frac{1}{2}\text{AB}\times\text{AC}$
$=\frac{1}{2}\times5\times12=30\text{ cm}^2$
and also area of $\triangle\text{ABC}=\frac{1}{2}\text{BC}\times\text{AD}$
$\Rightarrow30=\frac{1}{2}\times13\times\text{AD}$
$\Rightarrow\text{AD}=\frac{30\times2}{13}=\frac{60}{13}\text{ cm}.$ View full question & answer→MCQ 371 Mark
If in two triangle $\text{ABC}$ and $\text{DEF}, \angle\text{A}=\angle\text{E},\ \angle\text{B}=\angle\text{F},$ then which of the following is not true ?
- A
$\frac{\text{BC}}{\text{DF}}=\frac{\text{AC}}{\text{DE}}$
- ✓
$\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{DF}}$
- C
$\frac{\text{AB}}{\text{EF}}=\frac{\text{AC}}{\text{DE}}$
- D
$\frac{\text{BC}}{\text{DF}}=\frac{\text{AB}}{\text{EF}}$
AnswerCorrect option: B. $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{DF}}$
In $\triangle\text{ABC}$ and $\triangle\text{DEF}$
$\angle\text{A}=\angle\text{E}$
$\angle\text{B}=\angle\text{F}$
$\therefore\triangle\text{ABC}$ and $\triangle\text{DEF}$ are similar triangles
Hence $\frac{\text{AB}}{\text{EF}}=\frac{\text{BC}}{\text{FD}}=\frac{\text{CA}}{\text{DE}}$
Hence the correct answer is $B.$ View full question & answer→MCQ 381 Mark
Two poles of height $6m$ and $11m$ stand vertically upright on a plane ground. If the distance between their foot is $12m,$ the distance between their tops is :
- A
$12m.$
- B
$14m.$
- ✓
$13m.$
- D
$11m.$
AnswerCorrect option: C. $13m.$
Het $AB$ and $CD$ be two poles and distance blw their root is $12m.$
$ED = CD - CE = 11 - 6 = 5\ cm$
In right $\triangle\text{ADE},$
$ A D^2=A E^2+D E^2 $
$\Rightarrow A D^2=(12)^2+(5)^2 $
$ \Rightarrow A D^2=144+25 $
$ \Rightarrow A D^2=169 $
$ \Rightarrow A D=13m$
Thus distance blw their tops is $13m.$ View full question & answer→MCQ 391 Mark
$\triangle\text{ABC}\sim\triangle\text{DEF}$ and their perimeters are $32\ cm$ and $24\ cm$ respectively. If $AB = 10\ cm$ then $DE =?$
- A
$8\ cm$
- ✓
$7.5\ cm$
- C
$15\ cm$
- D
$5\sqrt{3}\text{ cm}$
AnswerCorrect option: B. $7.5\ cm$
$\therefore\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{Perimeter}(\triangle\text{ABC})}{\text{Perimeter}(\triangle\text{DEF})}=\frac{\text{AB}}{\text{DE}}$
$\Rightarrow\frac{32}{24}=\frac{10}{\text{DE}}$
$\Rightarrow\text{DE}=\frac{10\times24}{\text{32}}=7.5\text{ cm}$
View full question & answer→MCQ 401 Mark
It is given tha $\triangle\text{ABC}\sim\triangle\text{DFE},\angle\text{A} = 30^\circ,\angle\text{C}=40^\circ , \text{AB}= 5\text{ cm} , \text{AC}=8\text{ cm}$ and $DF = 7.5\ cm$. Then, which of the following is true :
- ✓
$\angle\text{F}=110^\circ , \text{DE}=12\text{ cm}$
- B
$\angle\text{D}=30^\circ , \text{EF}=12\text{ cm}$
- C
$\angle\text{F}=110^\circ , \text{EF}=12\text{ cm}$
- D
$\angle\text{F}=40^\circ , \text{DE}=12\text{ cm}$
AnswerCorrect option: A. $\angle\text{F}=110^\circ , \text{DE}=12\text{ cm}$
In $\triangle\text{ABC},\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow30^\circ+40^\circ\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=180^\circ$
$\text{since }\triangle\text{ABC}\sim\triangle\text{DFE}$
$\therefore\angle\text{B}=\angle\text{F}=110^\circ$
$\frac{\text{DF}}{\text{DE}}=\frac{\text{AB}}{\text{AC}} $
$\Rightarrow\frac{7.5}{\text{DE}}=\frac{5}{8}$
$\Rightarrow\text{DE}=12\text{ cm}$
View full question & answer→MCQ 411 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 421 Mark
In triangles $\text{ABC}$ and $\text{DEF} , \angle\text{A}=\angle\text{E}=40^\circ, AB : ED = AC : EF$ and $\angle\text{F}=65^\circ,$ then $\angle\text{B}=$
- A
$35^\circ$
- B
$65^\circ$
- ✓
$75^\circ$
- D
$85^\circ$
AnswerCorrect option: C. $75^\circ$
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
$\angle\text{A}=\angle\text{E}=40^\circ$
$AB : ED = AC : EF, \angle\text{F}=65^\circ$
$\Rightarrow\frac{\text{AB}}{\text{ED}}=\frac{\text{AC}}{\text{EF}}$
$\because$ In $\triangle\text{ABC}$ and $\triangle\text{EDF},$
$\angle\text{A}=\angle\text{E} \ ($each $= 40^\circ )$
$\frac{\text{AB}}{\text{ED}}=\frac{\text{AC}}{\text{EF}} \ ($given$)$
$\therefore\triangle\text{ABC}\sim\triangle\text{EDF} \ (\text{SAS}$ criterion$)$
$\therefore\angle\text{C}=\angle\text{F}=65^\circ$
and $\angle\text{B}=\angle\text{D}$
But $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ ($Sum of angles of a triangle$)$
$\Rightarrow40^\circ+65^\circ+\angle\text{C}=180^\circ$
$\Rightarrow105^\circ+\angle\text{C}=180^\circ$
$\therefore\angle\text{C}=180^\circ-105^\circ=75^\circ$ View full question & answer→MCQ 431 Mark
Choose the correct answer from the given four options: It $S$ is a point on side $PQ$ of a $\triangle\text{PQR}$ such that $PS = QS = RS,$ then :
- A
$ \mathrm{PR} \times \mathrm{QR}=\mathrm{RS}^2 $
- B
$ \mathrm{QS}+\mathrm{RS}=\mathrm{QR}^2 $
- ✓
$ \mathrm{PR}+\mathrm{QR}=\mathrm{PQ}^2 $
- D
$ \mathrm{PS}+\mathrm{RS}=\mathrm{PR}^2 $
AnswerCorrect option: C. $ \mathrm{PR}+\mathrm{QR}=\mathrm{PQ}^2 $
Given, in $\triangle\text{PQR}$
$\text{PS}=\text{QS}=\text{RS}\ .......(\text{i})$
$\text{In }\triangle\text{ PSR},\ \ \text{PS}=\text{RS}\ [$from Eq.$(i)]$
$\Rightarrow\angle1=\angle2\ ......(\text{ii})$
Similarly, in $\triangle\text{RSQ},$
$\Rightarrow\angle3=\angle4\ ......(\text{iii})$
$[$Corresponding angles of equal sides are equal$]$
Now, in $\triangle\text{PQR},$ sum of angles $=180^\circ$
$\Rightarrow\angle{\text{P}}+\angle{\text{Q}}+\angle{\text{R}}=180^\circ$
$\Rightarrow\angle2+\angle4+\angle1+\angle3=180^\circ$
$\Rightarrow\angle1+\angle3+\angle1+\angle3=180^\circ$
$\Rightarrow2(\angle1+\angle3)=180^\circ$
$\Rightarrow\angle1+\angle3=\frac{180^\circ}{2}=90^\circ$
$\therefore\angle\text{R}=90^\circ$
In $\triangle\text{PQR},$ by Pythagoras theorem,
$\text{PR}^2=\text{QR}^2=\text{PQ}^2$ View full question & answer→MCQ 441 Mark
If triangles $\text{ABC}$ and $\text{DEF}$ are similar and $AB = 4\ cm, DE = 6\ cm, EF = 9\ cm$ and $FD = 12\ cm,$ the perimeter of triangle is :
- A
$22\ cm$
- B
$20\ cm$
- C
$21\ cm$
- ✓
$18\ cm$
AnswerCorrect option: D. $18\ cm$
$\text{ABC}$ and $\text{DEF}$
$AB = 4\ cm, DE = 6\ cm, EF = 9\ cm$ and $FD = 12\ cm$
$\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}}$
$\frac{4}{6}=\frac{\text{BC}}{9}=\frac{\text{AC}}{12}$
$\text{BC}=(\frac{4.9}{6})=6\text{ cm}$
$\text{AC}=(\frac{12.4}{6})=8\text{ cm}$
Perimeter $= AB + BC + AC$
$= 4 + 6 + 8$
$= 18\ cm$
View full question & answer→MCQ 451 Mark
In a $\triangle\text{ABC},\ \angle\text{A}=90^\circ, AB = 5\ cm$ and $AC = 12\ cm$. If $\text{AD}\perp\text{BC},$ then $AD =$
- A
$\frac{13}{2}\text{ cm}.$
- ✓
$\frac{60}{13}\text{ cm}.$
- C
$\frac{13}{60}\text{ cm}.$
- D
$\frac{2\sqrt{15}}{13}\text{ cm}.$
AnswerCorrect option: B. $\frac{60}{13}\text{ cm}.$
In $\triangle\text{ABC}$ and $\triangle\text{BDA}$
$\angle\text{BAC}=\angle\text{ADC}=90^\circ$
$\angle\text{B}=\angle\text{B}\ ($Common$)$
$\triangle\text{ABC}\sim\triangle\text{BDA}$
$\frac{\text{AC}}{\text{AD}}=\frac{\text{BC}}{\text{AB}}\ ....(1)$
Using Pythagoras theorem in $\triangle\text{ABC}$ we get
$\text{BC}=\sqrt{(12)^2+(5)^2}$
$\Rightarrow\text{BC}=\sqrt{144+25}$
$\Rightarrow\text{BC}=\sqrt{169}$
$\Rightarrow\text{BC}={13}\text{ cm}$
From $(1)$
$\Rightarrow\frac{12}{\text{AD}}=\frac{13}{5}$
$\Rightarrow\text{AD}=\frac{12\times5}{13}$
$\Rightarrow\text{AD}=\frac{60}{13}\text{ cm}.$ View full question & answer→MCQ 461 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 $60^\circ .$
So, the triangles are isosceles and similar.
View full question & answer→MCQ 471 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 481 Mark
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
$81 : 16$
- ✓
$16 : 81$
AnswerCorrect option: D. $16 : 81$
Triangles are similar and the ratio of their sides is $4 : 9$
The ratio of the areas of two similar triangles are proportion to the square to their corresponding sides
Ratio in their areas $= (4)^2:(9)^2=16: 81$
View full question & answer→MCQ 491 Mark
The areas of two similar triangles are in respectively $9 \mathrm{ cm}^2$ and $16 \mathrm{ cm}^2$. The ratio of their corresponding sides is :
- ✓
$3 : 4$
- B
$4 : 3$
- C
$2 : 3$
- D
$4 : 5$
AnswerCorrect option: A. $3 : 4$
Given, two similar $\triangle\text{s}$
Ratio of areas $= (\text{Ratio of corresponding sides})^2$
$\Rightarrow\frac{9}{16}=(\text{Ratio of corresponding sides})^2$
$\Rightarrow\text{ Ratio corresponding sides}=\frac{3}{4}$
View full question & answer→MCQ 501 Mark
In $\triangle\text{ABC}, D$ and $E$ are points on side $AB$ and $AC$ respectively such that $DE \| BC$ and $AD : DB = 3 : 1.$ If $EA = 3.3\ cm,$ then $AC =$
- A
$1.1\ cm.$
- B
$4\ cm.$
- ✓
$4.4\ cm.$
- D
$5.5\ cm.$
AnswerCorrect option: C. $4.4\ cm.$
Give $n$ : In $\triangle\text{ABC}, D$ and $E$ are points on the side $AB$ and $AC$ respectively such that $DE \| BC$ and $AD : DB = 3 : 1$.
Also $, EA = 3.3\ cm.$
To find : $AC$
In $\triangle\text{ABC}, DE \| BC.$
Using corollory of basic proportionality theorem, we have,
$\frac{\text{AD}}{\text{AB}}=\frac{\text{EA}}{\text{AC}}$
$\frac{\text{AD}}{\text{AD}+\text{BD}}=\frac{3.3}{\text{AC}}$
$\frac{\text{AD}}{\text{AD}+\frac{1}{3}\text{AD}}=\frac{3.3}{\text{AC}}$
$\text{EC}=4.4\text{ cm}$
Hence the correct answer is $C$. View full question & answer→