MCQ 11 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 21 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 31 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 41 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 $DEF$ and $ABC$ is:
- ✓
$1 : 4$
- B
$1 : 2$
- C
$2 : 3$
- D
$4 : 5$
AnswerCorrect option: A. $1 : 4$
$1 : 4$
Given: In $\triangle ABC , D , E$ and F are the midpoints of $BC , CA$, and AB respectively.
To find: Ratio of the areas of $\triangle DEF$ and $\triangle ABC$
Since it is given that $D$ and, $E$ are the midpoints of $B C$, and $A C$ respectively.
Therefore $D E \| A B, D E| | F A$....(1)
Again it is given that $D$ and, $F$ are the midpoints of $B C$, and, $A B$ respectively.
Therefore, $D F \| C A, D F| | A E ~ . . .(2)$
From (1) and (2) we get AFDE is a parallelogram.
Similarly we can prove that BDEF is a parallelogram.
Now, in $\triangle ADE$ and $\triangle ABC$
$\angle FDE =\angle A$ (Opposite angles of $\|^{\text {gm }} AFDE$ )
$\angle DEF =\angle B$ (Opposite angles of $\|^{ gm } BDEF$ )
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 51 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 $\text{DE : BC} = 3 : 5,$ 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. $DE : BC = 3 : 5.$
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)
$\therefore\triangle\text{ADE}\sim\triangle\text{ABC} (AA$ 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 61 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 $\triangle\text{ADC},$
$AC^2 = AD^2 + DC^2 ($Pythagoras Theorem$)$
$AD^2 = AC^2 - DC^2$
$= BC^2 - DC^2 ( \because AC = BC = AB)$
$= (2DC)^2 - DC^2 ( \because D$ is mid point of $BC)$
$= 4DC^2 - DC^2 = 3DC^2$
$= 3CD^2$ View full question & answer→MCQ 71 Mark
If $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are similar such that $\text{2AB = DE}$ and $BC = 8\ cm,$ then $EF =$
- ✓
$16\ cm.$
- B
$12\ cm.$
- C
$8\ cm.$
- D
$4\ cm.$
AnswerCorrect option: A. $16\ cm.$
Given, $\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}$
$\Rightarrow\frac{\text{AB}}{2\text{AB}}=\frac{8}{\text{EF}}$
$\Rightarrow\text{EF}=8\times2$
$\Rightarrow\text{EF}=16\ cm$
View full question & answer→MCQ 81 Mark
A man goes $24\ m$ due west and then $7\ m$ due north. How far is he from the starting point?
- A
$31\ m.$
- B
$17\ m.$
- ✓
$25\ m.$
- D
$26\ m.$
AnswerCorrect option: C. $25\ m.$
Het a man be at $O$ and goes to $24\ m$ due west and then $7\ m$ due north.
Distance of man from starting point be $OB$
So,
In right $\triangle\text{ABO},$
$OB^2 = AB^2 + AO^2$
$\Rightarrow OB^2 = (7)^2 + (24)^2$
$\Rightarrow OB^2 = 49 + 576$
$\Rightarrow OB^2 = 625$
$\Rightarrow OB = 25\ m$
Thus, the distance of man from starting point is $25\ m.$ View full question & answer→MCQ 91 Mark
If $ABC$ is an isosceles triangle and $D$ is a point on $BC$ such that $\text{AD}\perp\text{BC},$ then:
- A
$AB^2 - AD^2 = BD \times DC$
- B
$AB^2 - AD^2 = BD^2 - DC^2$
- ✓
$AB^2 + AD^2 = BD \times DC$
- D
$AB^2 + AD^2 = BD^2 - DC^2$
AnswerCorrect option: C. $AB^2 + AD^2 = BD \times DC$
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 $\triangle\text{ABD},$
$AB^2 = AD^2 + BD^2$
$AB^2 - AD^2 = BD^2 = BD \times BD = BD \times DC (BD = DC).$
View full question & answer→MCQ 101 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 111 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 121 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 131 Mark
Two poles of height $6\ m$ and $11\ m$ stand vertically upright on a plane ground. If the distance between their foot is $12\ m$, the distance between their tops is:
- A
$12\ m.$
- B
$14\ m.$
- ✓
$13\ m.$
- D
$11\ m.$
AnswerCorrect option: C. $13\ m.$
Het $AB$ and $CD$ be two poles and distance blw their root is $12\ m$.
$ED = CD - CE = 11 - 6 = 5\ cm$
In right $\triangle\text{ADE},$
$AD^2 = AE^2 + DE^2$
$\Rightarrow AD^2 = (12)^2 + (5)^2$
$\Rightarrow AD^2 = 144 + 25$
$\Rightarrow AD^2 = 169$
$\Rightarrow AD = 13\ m$
Thus distance blw their tops is $13\ m$. View full question & answer→MCQ 141 Mark
In triangles $\text{ABC}$ and $\text{DEF}, \angle\text{A}=\angle\text{E}=40^\circ, \text{AB : ED = AC : EF}$ 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 $\triangle\text{DEF},$
$\angle\text{A}=\angle\text{E}=40^\circ$
$\text{AB : ED = AC : EF,} \angle\text{F}=65^\circ$
$\Rightarrow\frac{\text{AB}}{\text{ED}}=\frac{\text{AC}}{\text{EF}}$
$\because$ 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 151 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 161 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 oT their corresponding sides
Ratio in their areas $= (4)^2 : (9)^2 = 16 : 81$
View full question & answer→MCQ 171 Mark
The areas of two similar triangles are in respectively $9 \ cm^2$ and $16 \ cm^2$. 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 $= ($Ratio 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 181 Mark
In $\triangle\text{ABC}, D$ and $E$ are points on side $\text{AB}$ and $\text{AC}$ respectively such that $\ce{DE \| BC}$ and $\text{AD : DB} = 3 : 1$. If $\text{EA} = 3.3\ cm,$ then $\text{AC} =$
- A
$1.1\ cm.$
- B
$4\ cm.$
- ✓
$4.4\ cm.$
- D
$5.5\ cm.$
AnswerCorrect option: C. $4.4\ cm.$
Given: In $\triangle\text{ABC, D}$ and $E$ are points on the side $\text{AB}$ and $\text{AC}$ respectively such that $\ce{DE \| BC}$ and $\text{AD : DB} = 3 : 1$.
Also, $\text{EA} = 3.3\ cm.$
To find: $\text{AC}$
In $\triangle\text{ABC}, \ce{DE \| BC.}$
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→MCQ 191 Mark
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are two equilateral triangles such that $D$ is the mid$-$point of $\text{BC.}$ The ratio of the areas of triangle $\text{ABC}$ 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 equilateral triangles and $D$ is the mid$-$point of $PC.$
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are both equilateral triangles
$\therefore$ They are similar also
$\therefore\frac{\text{area of }\triangle\text{ABC}}{\text{area of }\triangle\text{BDE}}$
$=\frac{\text{BC}^2}{\text{BD}^2}=\frac{\text{BC}^2}{\big(\frac{1}{2}\text{BC}^2\big)} \{D$ is mid point of $BC\}$
$=\frac{\text{BC}^2}{\frac{1}{4}\text{BC}^2}=\frac{\text{4BC}^2}{\text{BC}^2}=\frac{4}{1}$
$\therefore$ Ratio is $4 : 1$ View full question & answer→MCQ 201 Mark
In an isosceles triangle $ABC$, if $AB = AC = 25 \ cm$ and $BC = 14\ cm$, then the measure of altitude from $A$ on $BC$ is:
- A
$20\ cm.$
- B
$22\ cm.$
- C
$18\ cm.$
- ✓
$24\ cm.$
AnswerCorrect option: D. $24\ cm.$
$\triangle\text{ABC}$ is an isosceles triangle in which $AB = AC = 25\ cm$, $BC = 14\ cm$
From A, draw $\text{AD}\perp\text{BC}$
$D$ is mid-point of $BC$
$\text{BD}=\frac{1}{2}\text{BC}=\frac{1}{2}\times14=7\text{cm}$
Now in right $\triangle\text{ABD}$
$AD^2 = AB^2 - BD^2$
$= (25)^2 - (7)^2 = 625 - 49 = 576 = (24)^2$
$AD = 24\ cm.$
View full question & answer→MCQ 211 Mark
In an equilateral triangle $ABC$ if $\text{AD}\perp\text{BC},$ then:
- A
$5AB^2 = 4AD^2$
- ✓
$3AB^2 = 4AD^2$
- C
$4AB^2 = 3AD^2$
- D
$2AB^2 = 3AD^2$
AnswerCorrect option: B. $3AB^2 = 4AD^2$
$\triangle\text{ABC}$ is an equilateral triangle and $\text{AD}\perp\text{BC}.$
In $\triangle\text{ABD},$ applying Pythagoras theorem, we get,
$\text{AB}^2=\text{AD}^2+\text{BD}^2$
$\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\Big(\because\text{BD}=\frac{1}{2}\text{BC}\Big)$
$\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{AB}\Big)^2\Big(\because\text{AB}=\text{BC}\Big)$
$\text{AB}^2=\text{AD}^2+\frac{1}{4}\text{AB}^2$
$3\text{AB}^2=4\text{AD}^2$
We got the result as $B$. View full question & answer→MCQ 221 Mark
In the figure, $\ce{RS \| DB \| PQ.}$ If $\text{CP = PD} = 11\ cm$ and $\text{DR = RA} = 3\ cm.$ Then the values of $x$ and $y$ are respectively:
- A
$12, 10.$
- B
$14, 6.$
- C
$10, 7.$
- ✓
$16, 8.$
AnswerCorrect option: D. $16, 8.$
the figure $\ce{RS \| DB \| PQ}$
$\text{CP = PD} = 11\ cm\ \text{DR = RA} = 3\ cm$
In $\triangle\text{ABD}$
$\ce{RS \| BD}$ and $\text{AR = RD}$
$\text{RS}=\frac{1}{2}\text{BD}$
$\text{y}=\frac{1}{2}\text{x}$ or ${x}=2\text{y}$
Only $16, 8$ is possible.
View full question & answer→MCQ 231 Mark
In a $\triangle\text{ABC},$ perpendicular $AD$ from $A$ on $BC$ meets $BC$ at $D$. If $BD = 8\ cm$, $DC = 2\ cm$ and $AD = 4\ cm$, then:
- A
$\triangle\text{ABC}$ is isosceles.
- B
$\triangle\text{ABC}$ is equilateral.
- C
$\text{AC} = 2\text{AB.}$
- ✓
$\triangle\text{ABC}$ is right$-$angled at $A.$
AnswerCorrect option: D. $\triangle\text{ABC}$ is right$-$angled at $A.$
in $\triangle\text{ABC},\ \text{AD}\perp\text{BC}$
$BD = 8 \ cm, DC = 2 \ cm, AD = 4 \ cm$
In right $\triangle\text{ACD},$
$AC^2 = AD^2 + CD^2 ($Pythagoras Theorem$)$
$= (4)^2 + (2)^2 = 16 + 4 = 20$
and in right $\triangle\text{ABD},$
$AB^2 = AD^2 + DB^2$
$= (4)^2 + (8)^2 = 16 + 64 = 80$
and $BC^2 = (BD + DC)^2$
$= (8 + 2 )^2 = (10)^2 = 100$
$AB^2 + AC^2 = 80 + 20 = 100 = BC^2$
$\triangle\text{ABC}$ is a right triangle whose $\angle\text{A}=90^\circ$ View full question & answer→MCQ 241 Mark
The areas of two similar triangles are $121\ cm^2$ and $64\ cm^2$ respectively. If the median of the first triangle is $12.1\ cm$, then the corresponding median of the other triangle is:
- A
$11\ cm.$
- ✓
$8.8\ cm.$
- C
$11.1\ cm.$
- D
$8.1\ cm.$
AnswerCorrect option: B. $8.8\ cm.$
Given: The area of two similar triangles is $121\ cm^2$ and $64\ cm^2$ respectively. The median of the first triangle is $12.1\ cm.$
To find: Corresponding medians of the other triangle.
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians.
$\frac{\text{ar}(\text{triangle 1})}{\text{ar}(\text{triangle 2})}=\Big(\frac{\text{median 1}}{\text{median 2}}\Big)^2$
$\frac{121}{64}=\Big(\frac{12.1}{\text{median 2}}\Big)^2$
Taking square root on both side, we get,
$\frac{11}{8}=\frac{12.1\text{cm}}{\text{median 2}}$
$\Rightarrow median2 = 8.8\ cm$
Hence the correct answer is $B$.
View full question & answer→MCQ 251 Mark
A chord of a circle of radius $10\ cm$ subtends a right angle at the centre. The length of the chord $($in $cm)$ is:
- A
$5\sqrt{2}$
- ✓
$10\sqrt{2}$
- C
$\frac{5}{\sqrt{2}}$
- D
$10\sqrt{3}$
AnswerCorrect option: B. $10\sqrt{2}$
In right $\triangle\text{OAB},$
$AB^2 = OA^2 + OB^2 ($Pythagoras Theorem$)$
$\Rightarrow AB^2 = (10)^2 + (10)^2 (OA = OB = 10cm)$
$\Rightarrow AB^2 = 100 + 100 = 200$
$\Rightarrow\text{AB}=\sqrt{200}=10\sqrt{2}\text{cm}$
Thus, the length of the chord is $10\sqrt{2}\text{cm}.$
Hence, the correct answer is option $B.$ View full question & answer→MCQ 261 Mark
The areas of two similar triangles $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are $144cm^2$ and $81cm^2$ respectively. If the longest side of larger $\triangle\text{ABC}$ be $36\ cm$, then the longest side of the smaller triangle $\triangle\text{DEF}$ is:
- A
$20\ cm.$
- B
$26\ cm.$
- ✓
$27\ cm.$
- D
$30\ cm.$
AnswerCorrect option: C. $27\ cm.$
Given: Areas of two similar triangles $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are $114\ cm^2$ and $81cm^2.$
If the longest side of larger $\triangle\text{ABC}$ is 36cm
To find: the longest side of the smaller triangle $\triangle\text{DEF}$
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\Big(\frac{\text{longest side of larger }\triangle\text{ABC}}{\text{longest side of smaller }\triangle\text{DEF}}\Big)^2$
$\frac{144}{81}=\Big(\frac{36}{\text{longest side of smaller}\ \triangle\text{DEF}}\Big)^2$
Taking aquare root on both sides, we get
$\frac{12}{9}=\frac{36}{\text{longest side of smaller}\ \triangle\text{DEF}}$
longest side of smaller $\triangle\text{DEF}=\frac{36\times9}{12}=27\text{cm}$
Hence the correct answer is $C.$
View full question & answer→MCQ 271 Mark
A vertical stick $20m$ long casts a shadow $10m$ long on the ground. At the same time, a tower casts a shadow $50m$ long on the ground. The height of the tower is:
- ✓
$100m.$
- B
$120m.$
- C
$25m.$
- D
$200m.$
AnswerCorrect option: A. $100m.$
Height of a stick $= 20m$
and length of its shadow $= 10m$
At the same time
Let height of tower $= x m$
and its shadow $= 50m$
$20 : x = 10 : 50$
$x \times 10 = 20 \times 50$
$\Rightarrow\text{x}=\frac{20\times50}{10}=100$
Height of tower $= 100m.$
View full question & answer→MCQ 281 Mark
In a $\triangle\text{ABC, AD}$ is the bisector of $\angle\text{BAC}.$ If $\text{AB} = 8\ cm, \text{BD} = 6\ cm$ and $\text{DC} = 3\ cm.$ Find $AC:$
- ✓
$4\ cm.$
- B
$6\ cm.$
- C
$3\ cm.$
- D
$8\ cm.$
AnswerCorrect option: A. $4\ cm.$
Given: In a $\triangle\text{ABC, AD}$ is the bisector of angle $\text{BAC. AB} = 8\ cm,$ and $\text{DC} = 3\ cm$ and $\text{BD} = 6\ cm.$
To find: $\text{AC}$
We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
Hence,
$\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\frac{8}{\text{AC}}=\frac{6}{3}$
$\text{AC}=\frac{8\times3}{6}$
$\text{AC}=4\ \text{cm}$
Hence we got the result $A.$
View full question & answer→MCQ 291 Mark
$\triangle\text{ABC}\sim\triangle\text{DEF}.$ If $BC = 3\ cm, EF = 4\ cm$ and $\text{ar}(\triangle\text{ABC})=54\text{cm}^2,$ then $\text{ar}(\triangle\text{DEF}):$
- A
$108\ cm^2$
- ✓
$96\ cm^2$
- C
$48\ cm^2$
- D
$100\ cm^2$
AnswerCorrect option: B. $96\ cm^2$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\text{BC}=3\text{cm},\ \text{EF}=4\text{cm}$
$\text{ar}(\triangle\text{ABC})=54\text{cm}^2$
$\because\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\Rightarrow\frac{54}{\text{ar}(\triangle\text{DEF})}=\frac{3^2}{4^2}=\frac{9}{16}$
$\therefore\text{ar}(\triangle\text{DEF})=\frac{16\times54}{9}=96\text{cm}^2$
View full question & answer→MCQ 301 Mark
The lenght of the hypotenuse of an isosceles right triangle whose one side is $4\sqrt{2}\text{cm}$ is:
- A
$12\text{cm}.$
- ✓
$8\text{cm}.$
- C
$8\sqrt{2}\text{ cm}.$
- D
$12\sqrt{2}\text{ cm}.$
AnswerCorrect option: B. $8\text{cm}.$
Het $\text{ABC}$ be an isosceles right triangle.
We have,
$\text{AB}=\text{BC}=4\sqrt{2}\ \text{cm}$
$\Rightarrow\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(4\sqrt{2})^2+(4\sqrt{2})^2$
$\Rightarrow\text{AC}^2=32+32$
$\Rightarrow\text{AC}^2=64$
$\Rightarrow\text{AC}=8\ \text{cm}$
Thus, the length of hypotenuse is $8\ cm.$ View full question & answer→MCQ 311 Mark
In the figure, if $\ce{PB \| CF}$ and $\ce{DP \| EF,}$ then $\frac{\text{AD}}{\text{DE}}=$ 
- A
$\frac{3}{4}.$
- ✓
$\frac{1}{3}.$
- C
$\frac{1}{4}.$
- D
$\frac{2}{3}.$
AnswerCorrect option: B. $\frac{1}{3}.$
In the figure, $\ce{PB \| CF, DP \| EF}$
$\text{AB} = 2\ cm, \text{AC} = 8\ cm$
$\text{BC = AC - AB}$
$= 8 - 2$
$= 6\ cm$
In $\triangle\text{ACF},\ \text{BP} || \text{CF}$
$\therefore\frac{\text{AB}}{\text{BC}}=\frac{\text{AP}}{\text{PF}}=\frac{2}{6}=\frac{1}{3}\ ....(1)$
In $\triangle\text{AEF},\ \text{DP}||\text{EF}$
$\therefore\frac{\text{AD}}{\text{DE}}=\frac{\text{AP}}{\text{PF}}=\frac{1}{3}\ \ [\text{From}\ (2)]$
$\frac{\text{AD}}{\text{DE}}=\frac{1}{3}.$
View full question & answer→MCQ 321 Mark
In the figure, the value of $x$ for which $\ce{DE \| AB}$ is:
AnswerIn $\triangle\text{ABC},\ \text{DE}||\text{BC}$
$\therefore\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{x}+3}{3\text{x}+19}=\frac{\text{x}}{3\text{x}+4}$
$\Rightarrow(\text{x}+3)(3\text{x}+4)=\text{x}(3\text{x}+19)$
$\Rightarrow3\text{x}^2+4\text{x}+9\text{x}+12=3\text{x}^2+19\text{x}$
$\Rightarrow3\text{x}^2+13\text{x}+12=3\text{x}^2+19\text{x}$
$\Rightarrow12=3\text{x}^2+19\text{x}-3\text{x}^2-13\text{x}$
$\Rightarrow12=6\text{x}$
$\Rightarrow\text{x}=\frac{12}{6}=2$
$\therefore\text{x}=2$
View full question & answer→MCQ 331 Mark
If in two triangles $\text{ABC}$ and $\text{DEF,} \frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{FE}}=\frac{\text{CA}}{\text{FD}},$ then:
- ✓
$\triangle\text{FDE}\sim\triangle\text{CAB}$
- B
$\triangle\text{FDE}\sim\triangle\text{ABC}$
- C
$\triangle\text{CBA}\sim\triangle\text{FDE}$
- D
$\triangle\text{BCA}\sim\triangle\text{FDE}$
AnswerCorrect option: A. $\triangle\text{FDE}\sim\triangle\text{CAB}$
In $\triangle\text{ABC}$ and $\triangle\text{DEF,}$
$\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{FE}}=\frac{\text{CA}}{\text{FD}} ($By $\text{SSS}$ axiom$)$
$\therefore\triangle\text{FDE}=\triangle\text{CAB}$ View full question & answer→MCQ 341 Mark
If $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are two triangles such that $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{CA}}{\text{FD}}=\frac{2}{5},$ then area $\text{Area }(\triangle\text{ABC}):\text{Area}(\triangle\text{DEF})=$
- A
$2 : 5$
- ✓
$4 : 25$
- C
$4 : 15$
- D
$8 : 125$
AnswerCorrect option: B. $4 : 25$
Given: $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are two triangles such that $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{CA}}{\text{FD}}=\frac{2}{5}$
To find: $\text{Ar}(\triangle\text{ABC}):\text{Ar}(\triangle\text{DEF})$
We know that if the sides of two triangles are proportional, then the two triangles are similar.
Since $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{CA}}{\text{FD}}=\frac{2}{5},$ therefore, $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are similar.
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}$
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{DEF})}=\frac{2^2}{5^2}$
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{DEF})}=\frac{4}{25}$
Hence the correct answer is $b.$
View full question & answer→MCQ 351 Mark
If $E$ is a point on side $CA$ of an equilateral triangle $\text{ABC}$ such that $\text{BE}\perp\text{CA},$ then $AB^2 + BC^2 + CA^2 =$
- A
$2BE^2$
- B
$3BE^2$
- ✓
$4BE^2$
- D
$6BE^2$
AnswerCorrect option: C. $4BE^2$
In triangle $ABC, E$ is a point on $AC$ such that $\text{BE}\perp\text{AC}.$
We need to find $AB^2 + BC^2 + AC^2$
Since $\text{BE}\perp\text{AC},$ $\text{CE} = \text{AE} = \frac{\text{AC}}{2} ($In a equilateral triangle, the perpendicular from the vertex bisects the base.$)$
In triangle ABE, we have,
$AB^2 = BE^2 + AE^2$
Since $AB = BC = AC$
Therefore, $AB^2 = BC^2 = AC^2 = BE^2 + AE^2$
$\Rightarrow AB^2 + BC^2 + AC^2 = 3BE^2 + 3AE^2$
Since in triangle $BE$ is an altitude, so $\text{BE}=\frac{\sqrt{3}}{2}\text{AB}$
$\text{BE}=\frac{\sqrt{3}}{2}\text{AB}$
$=\frac{\sqrt{3}}{2}\times\text{AC}$
$=\frac{\sqrt{3}}{2}\times2\text{AE}=\sqrt{3}\text{AE}$
$\Rightarrow\text{AB}^2+\text{BC}^2+\text{AC}^2=3\text{BE}^2+3\Big(\frac{\text{BE}}{\sqrt{3}}\Big)^2$
$=3\text{BE}^2+\text{BE}^2=4\text{BE}^2$
Hence option $C$ is correct. View full question & answer→MCQ 361 Mark
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $\text{AB} = 9.1\ cm$ and $\text{DE} = 6.5\ cm.$ If the perimeter of $\triangle\text{DEF}$ is $25\ cm,$ then the perimeter of $\triangle\text{ABC}$ is:
- A
$36\ cm.$
- B
$30\ cm.$
- C
$34\ cm.$
- ✓
$35\ cm.$
AnswerCorrect option: D. $35\ cm.$
Given: $\triangle\text{ABC}$ is similar to $\triangle\text{DEF}$ such that $\text{AB}= 9.1\ cm, \text{DE} = 6.5\ cm.$ Perimeter of $\triangle\text{DEF}$ is $25\ cm.$
To find: Perimeter of $\triangle\text{ABC}.$
We know that the ratio of corresponding sides of similar triangles is equal to the ratio of their perimeters.
Hence,
$\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DE}}=\frac{\text{P1}}{\text{P2}}$
$\frac{\text{AB}}{\text{DE}}=\frac{\text{P}(\triangle\text{ABC})}{\text{P}(\triangle\text{DEF})}$
$\frac{9.1}{6.5}=\frac{\text{P}(\triangle\text{ABC})}{25}$
$\text{P}(\triangle\text{ABC})=\frac{9.1\times25}{6.5}$
$\text{P}(\triangle\text{ABC})=35\text{cm}$
Hence the correct answer is $D.$
View full question & answer→MCQ 371 Mark
$\triangle\text{ABC}\sim\triangle\text{PQR}$ such that $\text{ar}(\triangle\text{ABC})=4\ \text{ar}(\triangle\text{PQR}).$ If $\text{BC} = 12\ cm,$ then $\text{QR} =$
- A
$9\ cm.$
- B
$10\ cm.$
- ✓
$6\ cm.$
- D
$8\ cm.$
AnswerCorrect option: C. $6\ cm.$
Given: $\triangle\text{ABC}\sim\triangle\text{PQR}$ and $\text{BC} = 12\ cm$
$\text{ar}(\triangle\text{ABC})=4\ \text{ar}(\triangle\text{PQR})$
Now, $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\Big(\frac{\text{BC}}{\text{QR}}\Big)^2$
$\Rightarrow\frac{4\text{ar}(\triangle\text{PQR})}{\text{ar}(\triangle\text{PQR})}=\frac{(12)^2}{\text{QR}^2}$
$\Rightarrow\text{QR}^2=\frac{144}{4}$
$\Rightarrow\text{QR}=\frac{12}{2}$
$\Rightarrow\text{QR}=6\text{cm}.$
View full question & answer→MCQ 381 Mark
$\triangle\text{ABC}\sim\triangle\text{DEF},$
$\text{ar}(\triangle\text{ABC})=9\text{cm}^2,\ \text{ar}(\triangle\text{DEF})=16\text{cm}^2.$ If $\text{BC} = 2.1\ cm,$ then the measure of $\text{EF}$ is:
- ✓
$2.8\ cm.$
- B
$4.2\ cm.$
- C
$2.5\ cm.$
- D
$4.1\ cm.$
AnswerCorrect option: A. $2.8\ cm.$
Given: $\text{Ar}(\triangle\text{ABC})=9\text{cm}^2,$
$\text{Ar}(\triangle\text{DEF})=16\text{cm}^2,$ and $\text{BC}=2.1\text{cm}$
To find: measure of $\text{EF}$
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\frac{9}{16}=\frac{2.1^2}{\text{EF}^2}$
$\frac{3}{4}=\frac{2.1}{\text{EF}}$
$\text{EF}=2.8\text{cm}$
Hence the correct answer is $A.$
View full question & answer→MCQ 391 Mark
In an equilateral triangle ABC if $\text{AD}\perp\text{BC},$ then $AD^2 =$
- A
$CD^2$
- B
$2CD^2$
- ✓
$3CD^2$
- D
$4CD^2$
AnswerCorrect option: C. $3CD^2$
In an equilateral $\triangle\text{ABC},\ \text{AD}\perp\text{BC}$
In $\triangle\text{ADC},$ applying Pythagoras theorem, we get,
$AC^2 = AD^2 + DC^2$
$BC^2 = AD^2 + DC^2 (\because AC = BC)$
$(2DC)^2 = AD^2 + DC^2 (\because BC = 2DC)$
$4DC^2 = AD^2 + DC^2$
$3DC^2 = AD^2$
$3CD^2 = AD^2$
Hence, the correct option is $C$. View full question & answer→MCQ 401 Mark
In a right triangle $\text{ABC}$ right$-$angled at $\text{B,}$ if $\text{P}$ and $\text{Q}$ are points on the sides $\text{AB}$ and $\text{AC}$ respectively, then:
- A
$\text{AQ}^2+\text{CP}^2=2(\text{AC}^2+\text{PQ}^2)$
- B
$2(\text{AQ}^2+\text{CP}^2)=\text{AC}^2+\text{PQ}^2$
- ✓
$\text{AQ}^2+\text{CP}^2=\text{AC}^2+\text{PQ}^2$
- D
$\text{AQ}+\text{CP}=\frac{1}{2}(\text{AC}+\text{PQ})$
AnswerCorrect option: C. $\text{AQ}^2+\text{CP}^2=\text{AC}^2+\text{PQ}^2$
Disclaimer: There is mistake in the problem. The question should be "In a right triangle $\text{ABC}$ right$-$angle at $B$, if $P$ and $Q$ are points on the sides $\text{AB}$ and $\text{BC}$ respectively, then"
Given: In the right $\triangle\text{ABC},$ right angled at $\text{B. P}$ and $Q$ are points on the sides $\text{AB}$ and $\text{BC}$ respectivelt.
Applying Pythagoras theorem,
In $\triangle\text{AQB},$
$\text{AQ}^2=\text{AB}^2+\text{BQ}^2\ ....(1)$
In $\triangle\text{PBC}$
$\text{CP}^2=\text{PB}^2+\text{BC}^2\ ....(2)$
Adding $(1)$ and $(2),$ we get
$\text{AQ}^2+\text{CP}^2=\text{AB}^2+\text{BQ}^2+\text{PB}^2+\text{BC}^2\ \ ...(3)$
In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2\ ....(4)$
In $\triangle\text{PBQ},$
$\text{PQ}^2=\text{PB}^2+\text{BQ}^2\ ....(5)$
From $(3), (4)$ and $(5),$ we get
$\text{AQ}^2+\text{CP}^2=\text{AC}^2+\text{PQ}^2$
We got the result as $C.$ View full question & answer→MCQ 411 Mark
In the given figure, if $\angle\text{ADE}=\angle\text{ABC},$ then $\text{CE =}$
- A
$2$
- B
$5$
- ✓
$\frac{9}{2}$
- D
$3$
AnswerCorrect option: C. $\frac{9}{2}$
Given: $\angle\text{ADE}=\angle\text{ABC}$
To find: The value of $\text{CE}$
Since $\angle\text{ADE}=\angle\text{ABC}$
$\therefore\text{DE}\|\text{BC} ($Two lines are parallel if the corresponding angles formed are equal$)$
According to basic proportionality theorem if a line is parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
In $\triangle\text{ABC},\ \text{DE}\|\text{BC}$
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\frac{2}{3}=\frac{3}{\text{EC}}$
$\text{EC}=\frac{3\times3}{2 }$
$\text{EC}=\frac{9}{2}$
Hence we got the result $C.$
View full question & answer→MCQ 421 Mark
If in $\triangle\text{ABC}$ and $\triangle\text{DEF},\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{FD}},$ then $\triangle\text{ABC}\sim\triangle\text{DEF}$ when:
- A
$\angle\text{A}=\angle\text{F}$
- B
$\angle\text{A}=\angle\text{D}$
- ✓
$\angle\text{B}=\angle\text{D}$
- D
$\angle\text{B}=\angle\text{E}$
AnswerCorrect option: C. $\angle\text{B}=\angle\text{D}$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
$\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{FD}}$
Then $\angle\text{B}=\angle\text{D} ($included angle $\text{SAS}$ axiom$).$ View full question & answer→MCQ 431 Mark
In a $\triangle\text{ABC, AD}$ is the bisector of $\angle\text{BAC}.$ If $\text{AB} = 6\ cm, \text{AC} = 5\ cm$ and $\text{BD} = 3\ cm,$ then $\text{DC} =$
- A
$11.3\ cm.$
- ✓
$2.5\ cm.$
- C
$3.5\ cm.$
- D
AnswerCorrect option: B. $2.5\ cm.$
In $\triangle\text{ABC, AD}$ is the bisector of $\angle\text{BAC}$
$\text{AB} = 6\ cm, \text{AC} = 5\ cm, \text{BD} = 3\ cm$
Let $\text{DC} = x$
In $\triangle\text{ABC}$
$\because \text{AD}$ is the bisector of $\angle\text{A}$
$\therefore\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\Rightarrow\frac{6}{5}=\frac{3}{\text{x}}$
$\Rightarrow\text{x}=\frac{3\times5}{6}$
$=\frac{5}{2}$
$=2.5$
$\therefore \text{DC} = 2.5\ cm$. View full question & answer→MCQ 441 Mark
In an isosceles triangle $ABC$ if $AC = BC$ and $AB^2 = 2AC^2$, then $\angle\text{C}=$
- A
$30^\circ$
- B
$45^\circ$
- ✓
$90^\circ$
- D
$60^\circ$
AnswerCorrect option: C. $90^\circ$
In isosceles $\triangle\text{ABC},\ \text{AC}=\text{BC}$
and $AB^2 = AC^2 + AC^2 = 2AC^2$
$= AC^2 + BC^2 (AC = BC)$
By converse of Pythagoras Theorem,
$\angle\text{C}=90^\circ$ View full question & answer→MCQ 451 Mark
If $\text{ABC}$ and $\text{DEF}$ are similar triangles such that $\angle\text{A}=47^\circ$ and $\angle\text{E}=83^\circ,$ then $\angle\text{C}=$
- ✓
$50^\circ$
- B
$60^\circ$
- C
$70^\circ$
- D
$80^\circ$
AnswerCorrect option: A. $50^\circ$
We have,
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\angle\text{A}=\angle\text{D}=47^\circ,\ \angle\text{B}=\angle\text{E}=83^\circ$ and $\angle\text{C}=\angle\text{F}=?$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ ($angle sum property$)$
$\Rightarrow47^\circ+83^\circ+\angle\text{C}=180^\circ$
$\Rightarrow130^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-130^\circ$
$\Rightarrow\angle\text{C}=50^\circ$
View full question & answer→MCQ 461 Mark
$\text{ABCD}$ is a trapezium such that $\ce{BC \| AD}$ and $\text{AD} = 4\ cm$. If the diagonals $\text{AC}$ and $\text{BD}$ intersect at $O$ such that $\frac{\text{AO}}{\text{OC}}=\frac{\text{DO}}{\text{OB}}=\frac{1}{2},$ then $\text{BC} =$
- A
$7\ cm.$
- ✓
$8\ cm.$
- C
$9\ cm.$
- D
$6\ cm.$
AnswerCorrect option: B. $8\ cm.$
We have,
$\frac{\text{AO}}{\text{OC}}=\frac{\text{DO}}{\text{OB}}=\frac{1}{2}$
In $\triangle\text{AOB}$ and $\triangle\text{DOC}$
$\angle\text{AOB}=\angle\text{DOC} ($Vertically oposite angle$)$
$\angle\text{OAB}=\angle\text{OCD} ($Alternate angle$)$
$\therefore\triangle\text{AOB}\sim\triangle\text{DOC}$
$\frac{\text{AO}}{\text{OC}}=\frac{\text{AB}}{\text{DC}}$
$\Rightarrow\frac{1}{2}=\frac{4}{\text{DC}}$
$\Rightarrow\text{DC}=4\times2$
$\Rightarrow\text{DC}=8\text{cm}.$ View full question & answer→MCQ 471 Mark
$\text{XY}$ is drawn parallel to the base $\text{BC}$ of a $\triangle\text{ABC}$ cutting $\text{AB}$ at $X$ and $\text{AC}$ at $Y$. If $\text{AB = 4 BX}$ and $\text{YC} = 2\ cm,$ then $\text{AY} =$
- A
$2\ cm.$
- B
$4\ cm.$
- ✓
$6\ cm.$
- D
$8\ cm.$
AnswerCorrect option: C. $6\ cm.$
In $\triangle\text{ABC}, \ce{XY \| BC}$
$\text{AB = 4BX, YC}= 2\ cm$
$\therefore \text{AB = 4BX}$
$\Rightarrow \text{AX + BX = 4BX}$
$\Rightarrow \text{AX = 4BX - BX = 3BX}$
Let $\text{AY = x}$
$\because$ In $\triangle\text{ABC}, \ce{XY \| BC}$
$\frac{\text{AX}}{\text{BX}}=\frac{\text{AY}}{\text{CY}}$
$\Rightarrow\frac{\text{3BX}}{\text{BX}}=\frac{\text{x}}{2}$
$\Rightarrow\frac{3}{1}=\frac{\text{x}}{2}$
$\Rightarrow\text{x}=3\times2=6$
$\therefore \text{AY} = 6\ cm.$ View full question & answer→MCQ 481 Mark
In the given figure the measure of $\angle\text{D}$ and $\angle\text{F}$ are respectively: 
- A
$50^\circ , 40^\circ$
- ✓
$20^\circ , 30^\circ$
- C
$40^\circ , 50^\circ$
- D
$30^\circ , 20^\circ$
AnswerCorrect option: B. $20^\circ , 30^\circ$
$\triangle\text{ABC}$ and $\triangle\text{DEF,}$
$\frac{\text{AB}}{\text{AC}}=\frac{\text{EF}}{\text{ED}}$
$\angle\text{A}=\angle\text{E}=130^\circ$
$\triangle\text{ABC}\sim\triangle\text{EFD} (\text{SAS}$ Similarity$)$
$\therefore\angle\text{F}=\angle\text{B}=30^\circ$
$\angle\text{D}=\angle\text{C}=20^\circ$
Hence the correct answer is $B.$ View full question & answer→MCQ 491 Mark
If $ABC$ is a right triangle right-angled at $B$ and $M$, $N$ are the mid-points of $AB$ and $BC$ respectively, then $4(AN^2 + CM^2) =$
- A
$4\text{AC}^2$
- ✓
$5\text{AC}^2$
- C
$\frac{5}{4}\text{AC}^2$
- D
$6\text{AC}^2$
AnswerCorrect option: B. $5\text{AC}^2$
$M$ is the mid-point of $AB.$
$\therefore\text{BM}=\frac{\text{AB}}{2}$
$N$ is the mid-point of $BC$.
$\therefore\text{BN}=\frac{\text{BC}}{2}$
Now,
$\text{AN}^2+\text{CM}^2=\bigg(\text{AB}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\bigg)+\bigg(\Big(\frac{1}{2}\text{AB}\Big)^2+\text{BC}^2\bigg)$
$=\text{AB}^2+\frac{1}{4}\text{BC}^2+\frac{1}{4}\text{AB}^2+\text{BC}^2$
$=\frac{5}{4}\Big(\text{AB}^2+\text{BC}^2\Big)$
$\Rightarrow4\Big(\text{AN}^2+\text{CM}^2\Big)=5\text{AC}^2$
Hence option $B$ is correct. View full question & answer→MCQ 501 Mark
$\triangle\text{ABC}$ is an isosceles triangle in which $\angle\text{C}=90^\circ$ If $AC = 6\ cm$, then $AB =$
- A
$6\sqrt{2}\text{cm}.$
- B
$6\text{cm}.$
- C
$2\sqrt{6}\text{cm}.$
- ✓
$4\sqrt{2}\text{cm}.$
AnswerCorrect option: D. $4\sqrt{2}\text{cm}.$
$\triangle\text{ABC}$ is an isosceles with $\angle\text{C}=90^\circ$
$AC = BC$
$AC = 6cm$
$AB^2 = AC^2 + BC^2 ($Pythagoras Theorem$)$
$(6)^2 + (6)^2 = 36 + 36 = 72 (AC = BC)$
$\text{AB}=\sqrt{72}=\sqrt{(36\times2}=6\sqrt{2}\text{cm}.$ View full question & answer→