MCQ 11 Mark
Three statements are given below:
$i.$ In a rectangle $\text{ABCD},$ the diagonal $AC$ bisects $\angle\text{A}$ as well as $\angle\text{C}.$
$ii.$ In a square $\text{ABCD}$, the diagonal $AC$ bisects $\angle\text{A}$ as well as $\angle\text{C}.$
$iii.$ In a rhombus $\text{ABCD},$ the diagonal $AC$ bisects $\angle\text{A}$ as well as $\angle\text{C}.$
Which is true$?$
- A
$I$ only
- ✓
$II$ and $III$
- C
$I$ and $III$
- D
$I$ and $II$
AnswerCorrect option: B. $II$ and $III$
Consider $I.$
We know that, in a rectangle the diagonals are not bisectors of each other, since the adjacent side.
Thus, $I$ is false.
Consider $II.$
We know that, in a square the diagonals bisect the opposite angles.
So, in a square $\text{ABCD},$ the diagonals $AC$ bisects $\angle\text{A}$ as well as $\angle\text{C}.$
Thus, $II$ is true.
Consider $III.$
We know that, in a rhombus the diagonals bisect the opposite angles.
So, in a rhombus $\text{ABCD},$ the diagonals $AC$ bisects $\angle\text{A}$ as well as $\angle\text{C}.$
Thus, $III$ is true.
View full question & answer→MCQ 21 Mark
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a:
AnswerThe figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle.
View full question & answer→MCQ 31 Mark
In the given figure, $ABCD$ is a parallelogram in which $\angle\text{BDC}=45^{\circ}$ and $\angle\text{BAD}=75^{\circ}.$ Then, $\angle\text{CBD}=?$ 
- A
$45^\circ$
- B
$55^\circ$
- ✓
$60^\circ$
- D
$75^\circ$
AnswerCorrect option: C. $60^\circ$
Since $ABCD$ is a parallelogram, $AB || CD$ since opposite angles of a parallelogram are equal.
$\Rightarrow\angle\text{ABD}=\angle\text{BCD}=45^{\circ}$ ...(Alternate angles)
In $\triangle\text{ADB},$
$\angle\text{ABD}+\angle\text{BDA}+\angle\text{DAB}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow45+\angle\text{BDA}+75=180$
$\Rightarrow\angle\text{BDA}+120=180$
$\Rightarrow\angle\text{BDA}=60^{\circ}$
$\Rightarrow\angle\text{CBD}=\angle\text{BDA}=60^{\circ}$ ...(Alternate angles)
$\Rightarrow\angle\text{CBD}=60^{\circ}$
View full question & answer→MCQ 41 Mark
Three statements are given below:
$i.$ In a $\| gm$, the angle bisectors of two adjacent angles enclose a right angle.
$ii.$ The angle bisectors of a $\| gm$ form a rectangle.
$iii.$ The triangle formed by joining the mid$-$points of the sides of an isosceles triangle is not necessarily an isosceles triangle.
Which is true?
- A
$I$ only
- B
$II$ only
- ✓
$I$ and $II$
- D
$II$ and $III$
AnswerCorrect option: C. $I$ and $II$
However, the triangle formed by joining the mid$-$point of the sides of an isosceles triangle is surely an isosceles triangle.
So, $III$ is false
Thus, $I$ and $II$ are true.
View full question & answer→MCQ 51 Mark
If area of a $||\ gm$ with sides a and $b$ is $A$ and that of a rectangle with sides a and $b$ is $B,$ then:
- A
$A > B$
- B
$A = B$
- ✓
$A < B$
- D
$A ≥ B$
AnswerCorrect option: C. $A < B$
Let $h$ be the heigth of the parallelogram.
Then, $h < b.$
We konw that, area of a parallelogram $=$ base $×$ height
If a is the base of the parallelogram, then area of a parallelogram $= a × h$
$⇒ A = a × h$
We know that, area of a rectangle $=$ length $×$ breadth
$⇒ A = a × b$
So, $a × h < a × b$
Hence, $A < B.$
View full question & answer→MCQ 61 Mark
In a trapezium $ABCD,$ if $E$ and $F$ be the mid-point of the diagonals $AC$ and $BD$ respectively. Then, $EF = ?$
AnswerCorrect option: D. $\frac{1}{2}(\text{AB}-\text{CD})$
Construction: join $CF$ and extend it to meet $AB$ at $G$.
In $\triangle\text{CDF}$ and $\triangle\text{GFB},$
$\angle\text{CDF}=\angle\text{GFB}$ ...(Vertically opposite angles)
$\text{AB || CD},$
So, $\angle\text{DCF}=\angle\text{GFB,}$ ...(alternate angles)
$\text{DF = FB} ...(F$ is the mid-point of $BD)$
$\Rightarrow\triangle\text{CDF}\cong\triangle\text{GFB} ...(ASA $ congruence criterion$)$
So, $\text{CD = GB}$ and $\text{CF = GF}...(C.P.C.T.)$
Since $E$ and $F$ are the mid-points of $CA$ and $CG$ respectively,
we have $\text{EF}=\frac{1}{2}\text{AG}$
$=\frac{1}{2}(\text{AB}-\text{GB})$
$=\frac{1}{2}(\text{AB}-\text{CD})$ View full question & answer→MCQ 71 Mark
The length of each side of a rhombus is $10\ cm$ and one if its diagonals is of length $16\ cm.$ The length of the other diagonal is:
- A
$13\ cm$
- ✓
$12\ cm$
- C
$2\sqrt{39}\text{cm}$
- D
$6\ cm$
AnswerCorrect option: B. $12\ cm$
Let $ABCD$ be the rhombus with diagonals $AB = 10\ cm$ and $Ac = 16\ cm.$
Since the diagonals of a rhombus are perpendicular bisectors of each other,
$\Rightarrow\text{OA}=8\text{cm},\text{BD = 2OB}$ and $\angle\text{AOB}=90^{\circ}$
In right $\triangle\text{AOB},$
By Pythagoras theorem,
$\text{AB}^2=\text{OA}^2+\text{OB}^2$
$\Rightarrow(10)^{2}=(8)^2+\text{OB}^2$
$\Rightarrow100=64+\text{OB}^2$
$\Rightarrow\text{OB}^2=36$
$\Rightarrow\text{OB}=\sqrt{36}$
$\Rightarrow\text{OB}=6\text{cm}$
$\Rightarrow\text{BD}=2\times\text{OB}$
$\Rightarrow\text{BD}=2\times6$
$\Rightarrow\text{BD}=12\text{cm}$
Hence, the lenght of the other diagonal is $12\ cm$ View full question & answer→MCQ 81 Mark
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a:
AnswerThe figure formed by joining the mid points of the adjacent sides of a parallelogram is a parallelogram.
View full question & answer→MCQ 91 Mark
If APB and CQD are two parallel lines, then the bisectors of $\angle\text{APQ},\angle\text{BPQ},\angle\text{CQP}$and $\angle\text{PQD}$ enclose a:
AnswerThe bisectors of $\angle\text{APQ},\angle\text{BPQ},\angle\text{CQP}$ and $\angle\text{PQD}$ enclose a rectengle.
View full question & answer→MCQ 101 Mark
In the given figure, $ABCD$ is a rhombus. Then:
AnswerCorrect option: C. $A C^2+B D^2=4 A B^2$
The diagonals of a rhombus bisect each other at right angles.
$\Rightarrow\text{OA}=\frac{1}{2}\text{AC}$ and $\text{OB}=\frac{1}{2}\text{BD}$
Also, $\angle\text{AOB}=90^{\circ}$
In right $\triangle\text{AOB},$
$\therefore\text{AB}^2=\text{OA}^2+\text{OB}^2$ ...(By Pythagoras theorem)
$\Rightarrow\text{AB}^2=\frac{1}{4}\text{AC}^2+\frac{1}{4}\text{BD}^2$
$\Rightarrow\text{AC}^2+\text{BD}^2=4\text{AB}^2$
View full question & answer→MCQ 111 Mark
The figure formed by joining the midpoints of the sides of a quadrilateral $ABCD$, taken in order, is a square, only if:
- A
$ABCD$ is a rhombus.
- B
Diagonals of $ABCD$ are equal.
- C
Diagonals of $ABCD$ are perpendicular.
- ✓
Diagonals of $ABCD$ are equal and perpendicular.
AnswerCorrect option: D. Diagonals of $ABCD$ are equal and perpendicular.
In $\triangle\text{ABC},P$ and $Q$ are the mid-points of sides $AB$ and $BC$ respectively.
$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC ...(i)}$
In $\triangle\text{BCD}, Q$ and $R$ are the mid-points of sides $BC$ and $CD$ respectively.
$\therefore\text{QR || BD}$ and $\text{QR}=\frac{1}{2}\text{BD ...(ii)}$
In $\triangle\text{ADC}, S$ and $R$ are the mid-points of sides $AD$ and $CD$ respectively.
$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC ...(iii)}$
In $\triangle\text{ABD}, P$ and $S$ are the mid-points of sides $AB$ and $AD$ respectively.
$\therefore\text{SP || BD}$ and $\text{SP}=\frac{1}{2}\text{BD ...(iv)}$
$\Rightarrow\text{PQ || RS}$ and $\text{QR || SP } [$From $(i), (ii), (iii)$ and $(iv)]$
Thus, $PQRS$ is a parallelogram.
Now, $\text{AC = BD}$ (given)
$\Rightarrow\frac{1}{2}\text{AC}=\frac{1}{2}\text{BD}$
$\Rightarrow\text{PQ = QR = RS = SP} [$From $(i), (ii), (iii)$ and $(iv)]$
Let the diagonals $AC$ and $BD$ intersect at $O.$
Now,
$\text{PS || BD}$
$\Rightarrow\text{PN || MO}$
Also, from (i), $\text{PQ || AC}$
$\Rightarrow\text{PM || NO}$
Thus, in quadrilateral $PMON, \text{PM || NO}$ and $\text{PN || MO}$
$\Rightarrow\text{PMON}$ is a parallelogram.
$\Rightarrow\angle\text{MPN}=\angle\text{MON}$ (opposite angles of a parallelogram are equal)
$\Rightarrow\angle\text{MPN}=\angle\text{BOA}$ $(\text{Since }\angle\text{BOA}=\angle\text{MON})$
$\Rightarrow\angle\text{MPN}=90^{\circ}$ $(\text{Since }\text{AC}\perp\text{BD},\angle\text{BOA}=90^{\circ})$
$\Rightarrow\angle\text{QPS}=90^{\circ}$
Thus, $PQRS$ is a parallelogram such that $PQ = QR = RS = SP$ and $\angle\text{QPS}=90^{\circ}.$
Hence, $PQRS$ is a square if diagonals of $ABCD$ are equal and perpendicular. View full question & answer→MCQ 121 Mark
If bisectors of $\angle\text{A}$ and $\angle\text{B}$ of a quadrilateral $ABCD$ intersect each other at $P,$ of $\angle\text{B}$ and $\angle\text{C}$ at $Q,$ of $\angle\text{C}$ and $\angle\text{D}$ at $R$ and of $\angle\text{D}$ and $\angle\text{A}$ at $S$ then $PQRS$ is a:
- A
- B
- C
- ✓
Quadrilateral whose opposite angles are supplementary.
AnswerCorrect option: D. Quadrilateral whose opposite angles are supplementary.
In $\triangle\text{APB},$ by angle sum property,
$\angle\text{APB}+\angle\text{PAB}+\angle\text{PBA}=180^{\circ}$
$\Rightarrow\angle\text{APB}+\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=180^{\circ}$
$\Rightarrow\angle\text{APB}=180^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}\Big)$
In $\triangle\text{CRD},$ by angle sum property,
$\angle\text{CRD}+\angle\text{RDC}+\angle\text{RCD}=180^{\circ}$
$\Rightarrow\angle\text{CRD}+\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{CRD}=180^{\circ}-\Big(\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}\Big)$
Now, $\angle\text{SPQ}+\angle\text{SRQ}=\angle\text{APB}+\angle\text{CRD}$
$=360^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\frac{1}{2}\angle\text{D}\Big)$
$=360^{\circ}=\frac{1}{2}(\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D})$
$=360^{\circ}-\frac{1}{2}\times360^{\circ}$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
Now, $\angle\text{PSR}+\angle\text{PQR}=360^{\circ}-(\angle\text{SPQ}+\angle\text{SRQ})$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
Hence, $PQRS$ is a quadrilateral whose opposite angles are supplementary. View full question & answer→MCQ 131 Mark
The diagonals $AC$ and $BD$ of a parallelogram $ABCD$ intersect each other at the point Osuch that $\angle\text{DAC}=30^{\circ}$ and $\angle\text{AOB}=70^{\circ}.$ Then, $\angle\text{DBC}=?$
- ✓
$40^\circ$
- B
$35^\circ$
- C
$45^\circ$
- D
$50^\circ$
AnswerCorrect option: A. $40^\circ$
$\text{AD || BC},$
$\Rightarrow\angle\text{DAO}=\angle\text{BCO}=30^{\circ} ...($Alternate angles$)$
$\Rightarrow\angle\text{BCO}=30^{\circ}$
$\angle\text{AOB}+\angle\text{BOC}=180^{\circ} ...($Linear pair of angles$)$
$\Rightarrow70+\angle\text{BOC}=180$
$\Rightarrow\angle\text{BOC}=110^{\circ}$
In $\triangle\text{CBO},$
$\angle\text{BOC}+\angle\text{BCO}+\angle\text{OBC}=180^{\circ} ...($Angle sum Property$)$
$\Rightarrow110+30+\angle\text{OBC}=180$
$\Rightarrow\angle\text{OBC}=40^{\circ}$
$\Rightarrow\angle\text{DBC}=40^{\circ} ...(D - O - B)$
View full question & answer→MCQ 141 Mark
$P$ is any point on the side $BC$ of a $\triangle\text{ABC}. P $ is joined to $A$. If $D$ and $E$ are the midpoints of the sides $AB$ and $AC$ respectively and $M$ and $N$ are the midpoints of $BP$ and $CP$ respectively then quadrilateral $DENM$ is:
Answer
In $\triangle\text{ABC}, D$ and $E$ are the mid-points of sides $AB$ and $AC$ respectively.
$\Rightarrow\text{DE || BC}$ and $\text{DE}=\frac{1}{2}\text{BC }...(\text{i})$
Now, $\text{MN = MP + PN}=\frac{1}{2}\text{BP}+\frac{1}{2}\text{CP}=\frac{1}{2}(\text{BP + CP})$
$\therefore\text{MN}=\frac{1}{2}\text{BC ...(ii)}$
From $(i)$ and $(ii),$
$\text{DE || MN}$ and $\text{DE = MN}$
Hence, $DENM$ is a parallelogram. View full question & answer→MCQ 151 Mark
A diagonal of a rectangle is inclined to one side of the rectangle at $35^\circ .$ The acute angle between the diagonals is:
- A
$55^\circ $
- ✓
$70^\circ $
- C
$45^\circ $
- D
$50^\circ $
AnswerCorrect option: B. $70^\circ $
$\angle\text{DAO}+\angle\text{OAB}=\angle\text{DAB}$
$\Rightarrow\angle\text{DAO}+35^{\circ}=90^{\circ}$
$\Rightarrow\angle\text{DAO}=55^{\circ}$
$ABCD$ is a rectangle and diagonals of a rectangle are equal and bisect each other.
$\text{OA = OD}$
$\Rightarrow\angle\text{ODA}=\angle\text{DAO}$ (angles opposte to equal sides are equal)
$\Rightarrow\angle\text{ODA}=55^{\circ}$
In $DODA,$ by angle sum property,
$\angle\text{ODA}+\angle\text{DAO}+\angle\text{AOD}=180^{\circ}$
$\Rightarrow55^{\circ}+\angle55^{\circ}+\angle\text{AOD}=180^{\circ}$
$\Rightarrow\angle\text{AOD}=70^{\circ}$ View full question & answer→MCQ 161 Mark
The bisectors of the angles of a parallelogram enclose a:
AnswerThe bisectors of the angles of a parallelogram encloses a rectangle.
View full question & answer→MCQ 171 Mark
If one angle of a parallelogram is $24^\circ$ less than twice the smallest angle, then the largest angle of the parallelogram is:
- A
$68^\circ $
- B
$102^\circ$
- ✓
$112^\circ$
- D
$136^\circ$
AnswerCorrect option: C. $112^\circ$
Let the smallest angle be $x^\circ $
$\Rightarrow $ its adjacent angle $= (2x - 24)$
Since sum of the adjacent angles $= 180^\circ $
$\Rightarrow x + 2x - 24 = 180$
$\Rightarrow 3x = 204$
$\Rightarrow x = 68^\circ $
So, its adjacent angle $= 2(68) - 24 = 136 - 24 = 112^\circ .$
Hence, the largest angle of the parallelogram is $112^\circ .$
View full question & answer→MCQ 181 Mark
The quadrilateral formed by joining the midpoints of the sides of a quadrilateral $ABCD,$ taken in order, is a rectangle, if:
- A
$ABCD$ is a Parallelogram
- B
$ABCD$ is rectangle
- C
Diagonals of $ABCD$ are equal
- ✓
Diagonals of $ABCD$ are perpendicular to each other.
AnswerCorrect option: D. Diagonals of $ABCD$ are perpendicular to each other.
In $\triangle\text{ABC}, P$ and $Q$ are the mid-point of sides $AB$ and $BC$ respectively.
$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC }...(\text{i})$
In $\triangle\text{ADC}, R$ and $S$ are the mid-point of sides $CD$ and $AD$ respectively.
$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC }...(\text{ii})$
From $(i)$ and $(ii),$
$\text{PQ ∥ RS}$ and $\text{PQ = RS}$
Thus, in quadrilateral $PQRS,$ a pair of opposite sides are equal are parallel.
So, $PQRS$ is a parallelogram.
Let the diagonals $AC$ and $BD$ intersect at $O$.
Now, in $\triangle\text{ABD}, P$ and $S$ are the mid-points of sides $AB$ and $AD$ respectively.
$\therefore\text{PS || BD}$
$\Rightarrow\text{PN || MO}$
Also, from $(i),$ $\text{PQ || AC}$
$\Rightarrow\text{PM || NO}$
Thus, in quadrilateral $PMON,$ $\text{PM || NO}$ and $\text{PN || MO},$
$⇒ PMON$ is a parallelogram.
$\Rightarrow\angle\text{MPN}=\angle\text{MON}$ (opposite angles of a parallelogram are equal)
$\Rightarrow\angle\text{MPN}=\angle\text{BOA}$ $(\text{Since }\angle\text{BOA}=\angle\text{MON})$
$\Rightarrow\angle\text{MPN}=90^{\circ}$ $(\text{Since }\text{AC}\perp\text{BD},\angle\text{BOA}=90^{\circ})$
$\Rightarrow\angle\text{QPS}=90^{\circ}$
Thus, $PQRS$ is a parallelogram whose one angle, i.e. $\angle\text{QPS}=90^{\circ}.$
Hence, $PQRS$ is a rectangle if $\text{AC}\perp\text{BD}.$ View full question & answer→MCQ 191 Mark
Is quadrilateral $\text{ABCD}$ a rhombus$?$
$i.$ Quadrilateral $\text{ABCD}$ is a $\|\ gm.$
$ii.$ Diagonals $AC$ and $BD$ are perpendicular to each other.
- A
If the question can be answered by one of the given statements alone and not by the other;
- B
If the question can be answered by either statement alone;
- ✓
If the question can be answered by both the statements together but not by any one of the two;
- D
If the question cannot be answered by using both the statements together.
AnswerCorrect option: C. If the question can be answered by both the statements together but not by any one of the two;
If the quad. $\text{ABCD}$ is a $\|gm,$ it could be a rectangle or square or rhombus.
So, statement $I$ is not sufficient to answer the question.
If the diagonals $AC$ and $BD$ are perpendicular to each other, then the $\|gm$ could be a square or rhombus.
So, statement $II$ is not sufficient to answer the question.
However, if the statements are combined, then the quad. $\text{ABCD}$ is a rhombus.
View full question & answer→MCQ 201 Mark
In the given figure, $ABCD$ is a parallelogram, $M$ is the mid-point of $BD$ and $BD$ bisects $\angle\text{B}$ as well as $\angle\text{D}.$ Then, $\angle\text{AMB}= ?$
- A
$45^\circ $
- B
$60^\circ $
- ✓
$90^\circ $
- D
$30^\circ$
AnswerCorrect option: C. $90^\circ $
$\angle\text{ABC}=\angle\text{ADC}$ ...(Opposite angles of a parallelogram are equal)
$\Rightarrow\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ADC}$
$\Rightarrow\angle\text{ABD}=\angle\text{ADB}$
So, $\text{AD = AB}$ ...(Sides opposite equal angles are equal.)
$\therefore\triangle\text{ABD}$ is isosceles
Also, $M$ is the mid-point of $BD.$
$\therefore\text{AM}\perp\text{BD}$
$\therefore\angle\text{AMB}=90^{\circ}$
View full question & answer→MCQ 211 Mark
Which of the following is not true for a parallelogram?
- A
Opposite sides are equal.
- B
Opposite angles are equal.
- ✓
Opposite angles are bisected by the diagonals.
- D
Diagonals bisect each other.
AnswerCorrect option: C. Opposite angles are bisected by the diagonals.
We know that, in a || gm opposite sides are equal, opposite angles are equal and also the diagonals bisect each other.
So, opposite angles are bisected by the diagonals is not true.
View full question & answer→MCQ 221 Mark
In the given figure, $ABCD$ is a $||\ gm$ and $E$ is the mid-point of $BC$. Also, $DE$ and $AB$ when produced meet at $F.$ Then:
AnswerCorrect option: B. $A F=2 A B$
In $\triangle\text{CDE}$ and $\triangle\text{BFE,}$
$\angle\text{DEC}=\angle\text{FEB} ...($Vertically opposite angles$)$
$\angle\text{DCE}=\angle\text{FBE} ...($Alternate angles$)$
and $\text{CE = BE}...(E$ is the mid-point$.)$
$\therefore\triangle\text{CDE}\cong\triangle\text{BFE} ...($By $AA$ congruence criterion$)$
$\therefore\text{CD = BF} ...(C.P.C.T.)$
Now,
$\text{AF = AB + BF}$
$\Rightarrow\text{AF + AB + CD} ...($from $(i))$
$\Rightarrow\text{AF = 2AB}$
View full question & answer→MCQ 231 Mark
The figure formed by joining the mid-points of the adjacent sides of a square is a:
AnswerThe figure formed by joining the mid points of the adjacent sides of a square is a square.
View full question & answer→MCQ 241 Mark
In a trapezium $A B C D$, if $A B \| C D$, then $\left(A C^2+B D^2\right)=$ ?
- A
$B C^2+A D^2+2 B C \cdot A D$
- B
$A B^2+C D^2+2 A B \cdot C D$
- C
$A B^2+C D^2+2 A D \cdot B C$
- ✓
$B C^2+A D^2+2 A B \cdot C D$
AnswerCorrect option: D. $B C^2+A D^2+2 A B \cdot C D$
Construction: Draw perpendicular from $D$ and $C$ on $AB$ which meets $AB$ at $E$ and $F,$ respectively.
So, $DEFC$ is a parallelogram, since one pair of opposite sides are parallel and equal.
In $\triangle\text{ABC},$
$\angle\text{B}$ is an acute angle.
$\Rightarrow\text{AC}^2=\text{BC}^2+\text{AB}^2-2\text{AB}\times\text{AE}$
In $\triangle\text{ABD},$
$\angle\text{A}$ is an acute angle.
$\Rightarrow\text{BD}^2=\text{AD}^2+\text{AB}^2-2\text{AB}\times\text{AF}$
$\Rightarrow\text{AC}^2+\text{BD}^2=\text{BC}^2+\text{AD}^2+2\text{AB}(\text{AB}-\text{BE}-\text{AF})$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{EF}$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{CD}$
View full question & answer→MCQ 251 Mark
If $\angle\text{A},\angle\text{B},\angle\text{C}$ and $\angle\text{D}$ of a quadrilateral $ABCD$ taken in order, are in the ratio $3 : 7 : 6 : 4$ then $ABCD$ is a:
AnswerLet the common multiple be $x.$
$\therefore$ The angle measure $3x, 7x, 6x$ and $4x.$
Since the sum of the angles of a quadrilateral is $360^\circ ,$ we have
$3x + 7x + 6x + 4x = 360$
$\Rightarrow 20x = 360$
$\Rightarrow x = 18^\circ $
$\therefore$ The angles of the quadrilateral are
$3x = 3(18) = 54^\circ $
$7x = 7(18) = 126^\circ $
$6x = 6(18) = 108^\circ $ and
$4x = 4(18) = 72^\circ $
Now, $54 + 126 = 180^\circ $ and $108 + 72 = 180^\circ $
So, the angles are interior angles and hence we get one pair of parallel sides of $ABCD.$
Hence, $ABCD$ is a trapezium.
View full question & answer→MCQ 261 Mark
In which of the following figures are the diagonals equal?
AnswerThe diagonals are equal in a rectangle.
The diagonals in a parallelogram, rhombus or trapezium need not be equal.
View full question & answer→MCQ 271 Mark
Is quad. $\text{ABCD}$ a parallelogram$?$
$i.$ Its opposite sides are equal.
$ii.$ Its opposite angles are equal.
- A
If the question can be answered by one of the given statements alone and not by the other;
- ✓
If the question can be answered by either statement alone;
- C
If the question can be answered by both the statements together but not by any one of the two;
- D
If the question cannot be answered by using both the statements together.
AnswerCorrect option: B. If the question can be answered by either statement alone;
If the opposite sides of a quad. $\text{ABCD}$ are equal, the quadrilateral is a parallelogram.
If the opposite angles are equal, then the quad. $\text{ABCD}$ is a parallelogram.
View full question & answer→MCQ 281 Mark
Two parallelograms stand on equal bases and between the same parallels. The ratio of their areas is:
- A
$1 : 2$
- B
$2 : 1$
- C
$1 : 3$
- ✓
$1 : 1$
AnswerCorrect option: D. $1 : 1$
Area of a parallelogram $=$ base $⨯$ height
Since two parallelogram stand on equal bases and between the same parallel lines,
their heights are same.
$\therefore$ Areas are also same.
$\therefore$ The ratio of their area is $1 : 1.$
View full question & answer→MCQ 291 Mark
If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a:
AnswerIf the diagonals of a quadrilateral bisect each other at right angles,
then the figure is a rhombus.
This is because in a rhombus, the diagonals are perpendicular bisectors of each other.
View full question & answer→MCQ 301 Mark
If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is:
- A
$108^\circ$
- B
$54^\circ$
- ✓
$72^\circ$
- D
$81^\circ$
AnswerCorrect option: C. $72^\circ$
Let one of the angle of the $||\ gm$ be $x^\circ .$
According to the given condition,
$\therefore$ the adjacent angle $=\frac{2}{3}\text{x}^{\circ}$
Now,
$\text{x}+\frac{2}{3}\text{x}=180^{\circ} ...($Sum of the adjacent angles of $||\ gm$ is $180^\circ .)$
$\Rightarrow\frac{3\text{x}+2\text{x}}{3}=180^{\circ}$
$\Rightarrow\frac{5\text{x}}{3}=180^{\circ}$
$\Rightarrow5\text{x}=540^{\circ}$
$\Rightarrow\text{x}=108^{\circ}$
$\Rightarrow $ the adjacent angles $=\frac{2}{3}(108)=36\times2=72^{\circ}$
Hence, the smallest angle is $72^\circ .$
View full question & answer→MCQ 311 Mark
The bisectors of any two adjacent angles of a parallelogram intersect at:
- A
$40^\circ$
- B
$45^\circ$
- C
$60^\circ$
- ✓
$90^\circ$
AnswerCorrect option: D. $90^\circ$
Consider parallelogram $ABCD,$
We know that, the sum if the adjacent angles of a parallelogram is $180^\circ .$
$\Rightarrow\angle\text{DAB}+\angle\text{CBA}=180^{\circ}$
$\Rightarrow\frac{1}{2}\angle\text{DAB}+\frac{1}{2}\angle\text{CBA}=\frac{1}{2}(180^{\circ})$
$\Rightarrow\angle\text{OAB}+\angle\text{OBA}=90^{\circ} ...(\text{i})$
Now,
In $\triangle\text{OAB},$
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow90^{\circ}+\angle\text{AOB}=180^{\circ}$ ...(from (i))
$\Rightarrow\angle\text{AOB}=90^{\circ}$
The bisectore of any adjacent angles of a parallelogram intersects at $90^\circ .$ View full question & answer→MCQ 321 Mark
The angles of a quadrilateral are in the ratio $3 : 4 : 5 : 6.$ The smallest of these angles is:
- A
$45^\circ $
- ✓
$60^\circ$
- C
$36^\circ$
- D
$48^\circ$
AnswerCorrect option: B. $60^\circ$
Let tha common multipal be $x.$
$\therefore$ The angle measure $3x, 4x, 5x$ and $6x.$
Since the sum of the angles of a quadrilateral being $360^\circ ,$ we have
$3x + 4x + 5x + 6x = 360^\circ $
$\Rightarrow 18x = 360^\circ $
$\Rightarrow x = 20^\circ $
$\therefore$ The angles of the quadrilateral are
$3x = 3(20) = 60^\circ ,$
$4x = 4(20) = 80^\circ ,$
$5x = 5(20)= 100^\circ ,$
$6x = 6(20) = 120^\circ ,$
$\therefore$ The smallest angle is $60^\circ .$
View full question & answer→MCQ 331 Mark
Three angles of a quadrilateral are $80^{\circ}, 95^{\circ}$ and $112^{\circ}$. Its fourth angle is:
- A
$78^{\circ}$
- ✓
$73^{\circ}$
- C
$85^{\circ}$
- D
$100^{\circ}$
AnswerCorrect option: B. $73^{\circ}$
Let the measure of the fourth angle be $x^0$.
We know that, the sum of the angles of a quadrilateral is $360^{\circ}$.
$\text { So, } 80^{\circ}+95^{\circ}+112^{\circ}+\mathrm{x}=360^{\circ}$
$\Rightarrow 287^{\circ}+\mathrm{x}=360^{\circ}$
$\Rightarrow \mathrm{x}=73^{\circ}$
$\therefore$ Its fourth angle is $73^{\circ}$.
View full question & answer→MCQ 341 Mark
The parallel sides of a trapezium are a and b respectively. The line joining the mid-points of its non-parallel sides will be:
- A
$\frac{1}{2}(\text{a}-\text{b})$
- ✓
$\frac{1}{2}(\text{a}+\text{b})$
- C
$\frac{2\text{ab}}{(\text{a + b})}$
- D
$\sqrt{\text{ab}}$
AnswerCorrect option: B. $\frac{1}{2}(\text{a}+\text{b})$
a
$E$ and $F$ are the given to be the mid-points of $AD$ and $BC$ respectively.
$\therefore\text{EF} = \frac{1}{2}(\text{AB + DC})$
$=\frac{1}{2}(\text{a + b})$ View full question & answer→MCQ 351 Mark
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a:
AnswerThe figure formed by joining the mid points of the adjacent sides of a quadrilateral is a parallelogram.
View full question & answer→MCQ 361 Mark
The quadrilateral formed by joining the midpoints of the sides of a quadrilateral $ABCD,$ taken in order, is a rhombus, if:
- A
$ABCD$ is a Parallelogram.
- B
$ABCD$ is rhombus.
- ✓
Diagonals of $ABCD$ are equal.
- D
Diagonals of $ABCD$ are perpendicular to each other.
AnswerCorrect option: C. Diagonals of $ABCD$ are equal.
In $\triangle\text{ABC}, P$ and $Q$ are the mid-points of sides $AB$ and $BC$ respectively.
$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC ...(i)}$
In $\triangle\text{BCD}, Q$ and $R$ are the mid-points of sides $BC$ and $CD$ respectively.
$\therefore\text{QR || BD}$ and $\text{QR}=\frac{1}{2}\text{BD ...(ii)}$
In $\triangle\text{ADC}, S$ and $R$ are the mid-points of sides $AD$ and $CD$ respectively.
$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC ...(iii)}$
In $\triangle\text{ABD}, P$ and $S$ are the mid-points of sides $AB$ and $AD$ respectively.
$\therefore\text{SP || BD}$ and $\text{SP}=\frac{1}{2}\text{BD ... (iv)}$
$\Rightarrow\text{PQ ∥ RS}$ and $\text{QR ∥ SP} [$From $(i), (ii), (iii)$ and $(iv)]$
Thus, $PQRS$ is a parallelogram.
Now, $AC = BD ($given$)$
$\Rightarrow\frac{1}{2}\text{AC}=\frac{1}{2}\text{BD}$
$\Rightarrow\text{PQ = QR = RS = SP } [$From $(i), (ii), (iii)$ and $(iv)]$
Hence, $PQRS$ is a rhombus if diagonals of $ABCD$ are equal. View full question & answer→MCQ 371 Mark
In the given figure, $AD$ is a median of $\triangle\text{ABC}$ and $E$ is the mid-point of $AD$. If $BE$ is joined and produced to meet $AC$ in $F,$ then $AF = ?$
- A
$\frac{1}{2}\text{AC}$
- ✓
$\frac{1}{3}\text{AC}$
- C
$\frac{2}{3}\text{AC}$
- D
$\frac{3}{4}\text{AC}$
AnswerCorrect option: B. $\frac{1}{3}\text{AC}$
Construction: Join $DG$ and $G$ be the mid-point of $FC.$
Now,
In $\triangle\text{BCF, D}$ is the mid-point of $BC$ and $G$ is the mid-point of $FC$ and $F$ is the mid-point of $AG.$
$\Rightarrow\text{DG || BF}$
$\Rightarrow\text{DG || EF}...(B - E - F)$
$\Rightarrow\text{AF}=\text{FG}=\text{GC} ...($Since $G$ is the mid-point.$)$
$\Rightarrow\text{AF}=\frac{1}{2}\text{AC}$ View full question & answer→MCQ 381 Mark
In the given figure, $ABCD$ is a parallelogram in which $\angle\text{BAD}=75{^\circ}$ and $\angle\text{CBD}=60^{\circ}.$ Then, $\angle\text{BDC}=?$
- A
$60^\circ$
- B
$75^\circ$
- ✓
$45^\circ$
- D
$50^\circ$
AnswerCorrect option: C. $45^\circ$
We know that, the opposite angles of a parallelogram are equal.
$\therefore\angle\text{C}=\angle\text{A}=75^{\circ}$
In $\triangle\text{BCD},$
$\angle\text{BCD}+\angle\text{BDC}+\angle\text{CBD}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow75^{\circ}+\angle\text{BDC}+60^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{BDC}=45^{\circ}$
View full question & answer→MCQ 391 Mark
Is $\| gm \text{ABCD}$ a square$?$
$i.$ Diagonals of $\| gm \text{ABCD}$ are equal.
$ii.$ Diagonals of $\| gm \text{ABCD}$ intersect at right angles.
- A
If the question can be answered by one of the given statements alone and not by the other;
- B
If the question can be answered by either statement alone;
- ✓
If the question can be answered by both the statements together but not by any one of the two;
- D
If the question cannot be answered by using both the statements together.
AnswerCorrect option: C. If the question can be answered by both the statements together but not by any one of the two;
If the diagonals of a $\| gm \text{ABCD}$ are equal,
then$\| gm \text{ABCD}$ could either be a rectangle or a square.
If the diagonals of the $\| gm \text{ABCD}$ intersect at right angles,
then the $\|gm \text{ABCD}$ could be a square or a rhombus.
However, if both the statements are combined,
then $\| gm \text{ABCD}$ will be a square.
View full question & answer→MCQ 401 Mark
The lengths of the diagonals of a rhombus are $16\ cm$ and $12\ cm.$ The length of each side of the rhombus is:
- ✓
$10\ cm$
- B
$12\ cm$
- C
$9\ cm$
- D
$8\ cm$
AnswerCorrect option: A. $10\ cm$
We konw that, the diagonals of a rhombus bisect each other at right angles.
So, $Ac = 16\ cm$ and $BD = 12\ cm$
$⇒ OA = 8\ cm$ and $OB = 6\ cm$
Also, $\angle\text{OAB}=90^{\circ}$
In right $\triangle\text{OAB},$
By Pythagoras theorem,
$\text{AB}^2=\text{OA}^2+\text{OB}^2$
$\Rightarrow\text{AB}^2=(8)^2+(6)^2$
$\Rightarrow\text{AB}^2=64+36$
$\Rightarrow\text{AB}^2=100$
$\Rightarrow\text{AB}=\sqrt{100}$
$\Rightarrow\text{AB}=10\text{cm}$
Hence, the length of each side of the rhombus is $10\ cm$ View full question & answer→MCQ 411 Mark
If $ABCD$ is a parallelogram with two adjacent angles $\angle\text{A}=\angle\text{B}$ then the parallelogram is a:
AnswerGiven that $ABCD$ is a parallelogram.
We konw that, opposite sides of a parallelogram are parallel.
$\Rightarrow\angle\text{A}+\angle\text{B}=180^{\circ}$ ...(interior angles)
Also, $\angle\text{A}=\angle\text{B}=90^{\circ} ...($Given$)$
Since opposite angles of a parallelogram are equal,
$\angle\text{A}=\angle\text{C}$ and $\angle\text{B}=\angle\text{D}$
So, $\angle\text{A}=\angle\text{C}=\angle\text{B}=\angle\text{D}=90^{\circ}$
$\therefore ABCD$ is a rectangle.
View full question & answer→MCQ 421 Mark
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a:
AnswerThe figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus.
View full question & answer→MCQ 431 Mark
$ABCD$ is a rhombus such that $\angle\text{ACB}=50^{\circ}.$ Then, $\angle\text{ADB}=?$
- ✓
$40^\circ$
- B
$25^\circ$
- C
$65^\circ$
- D
$130^\circ$
AnswerCorrect option: A. $40^\circ$
$ABCD$ is a rhombus.
$\Rightarrow\text{AD || BC}$ and $\text{AC}$ is the transversal.
$\Rightarrow\angle\text{DAC}=\angle\text{ACB}$ (alternate angles)
$\Rightarrow\angle\text{DAC}=50^{\circ}$
In $\triangle\text{AOD},$ by angle sum property,
$\angle\text{AOD}+\angle\text{DAO}+\angle\text{ADO}=180^{\circ}$
$\Rightarrow90^{\circ}+\angle\text{50}^{\circ}+\angle\text{ADO}=180^{\circ}$
$\Rightarrow\angle\text{ADO}=40^{\circ}$
$\Rightarrow\angle\text{ADB}=40^{\circ}$ View full question & answer→MCQ 441 Mark
In a quadrilateral $ABCD,$ if $AO$ and $BO$ are the bisectors of $\angle\text{A}$ and $\angle\text{B}$ respectively, $\angle\text{C}=70^{\circ}$ and $\angle\text{D}=30^{\circ}.$ Then, $\angle\text{AOB}=?$
- A
$40^\circ$
- ✓
$50^\circ$
- C
$80^\circ$
- D
$100^\circ $
AnswerCorrect option: B. $50^\circ$
We know that, sum of the angles of a quadrilateral is $360^\circ .$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}$
$\Rightarrow\angle\text{A}+\angle\text{B}+70^{\circ}+30^{\circ}=360^{\circ}$
$\Rightarrow\angle\text{A}+\angle\text{B}=260^{\circ}$
$\Rightarrow\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=\frac{1}{2}(260)^{\circ}$
$\Rightarrow\angle\text{BAO}+\angle\text{ABO}=130^{\circ}...(\text{i})$
In $\triangle\text{AOB},$
$\angle\text{BAO}+\angle\text{ABO}+\angle\text{AOB}=180^{\circ}...($ Angle sum Property$)$
$\Rightarrow130^{\circ}+\angle\text{AOB}=180^{\circ} ...($from $(i))$
$\Rightarrow\angle\text{AOB}=50^{\circ}$ View full question & answer→MCQ 451 Mark
Is quadrilateral $\text{ABCD}$ a $\| gm$?
$i.$ Diagonals $AC$ and $BD$ bisect each other.
$ii.$ Diagonals $AC$ and $BD$ are equal.
- ✓
If the question can be answered by one of the given statements alone and not by the other;
- B
If the question can be answered by either statement alone;
- C
If the question can be answered by both the statements together but not by any one of the two;
- D
If the question cannot be answered by using both the statements together.
AnswerCorrect option: A. If the question can be answered by one of the given statements alone and not by the other;
If the diagonals of a quad. $\text{ABCD}$ bisect each other, then the quad. $\text{ABCD}$ is a parallelogram.
So, $I$ gives the answer.
If the diagonals are equal, then the quad. $\text{ABCD}$ is a parallelogram.
So, $II$ gives the answer.
View full question & answer→