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 $, \text{ABCD}$ 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 $\text{ABCD}$ 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→Question 41 Mark
Short Answer Questions.
What special name can be given to a quadrilateral whose all angles are equal?
AnswerA quadrilateral whose all angles are equal is a rectangle.
View full question & answer→MCQ 51 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 61 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 $\times$ height
If a is the base of the parallelogram, then area of a parallelogram $= a \times h$
$\Rightarrow A = a \times h$
We know that, area of a rectangle $=$ length $\times$ breadth
$\Rightarrow A = a \times b$
So $, a \times h < a \times b$
Hence $, A < B.$
View full question & answer→Question 71 Mark
Short Answer Questions.
What special name can be given to a quadrilateral PQRS if $\angle\text{P}+\angle\text{S}=180^{\circ}?$
AnswerIn quadrilateral PQRS, $\angle\text{P}$ and $\angle\text{S}$ are adjacent angles.
Since the sum of adjacent angles $\neq180^{\circ},$ PQRS is not a parallelogram.
Hence, PQRS is a trapezium.
View full question & answer→MCQ 81 Mark
In a trapezium $\text{ABCD}$, 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$)$
$ AB \| CD,$
So, $\angle\text{DCF}=\angle\text{GFB,} ...($alternate angles$)$
$\text{DF = FB} ... (F$ is the mid $-$ point of $BD)$
$\Rightarrow\triangle\text{CDF}\cong\triangle\text{GFB} ...(\text{ASA}$ congruence criterion$)$
So, $\text{CD = GB}$ and $\text{CF = GF} ...(\text{C.P.C.T}.)$
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→Question 91 Mark
Short Answer Questions.
In a quadrilateral PQRS, opposite angles are equal. If SR = 2cm and PR = 5cm then determine PQ.
AnswerSince opposite angles of quadrilateral are equal, PQRS is a parallelogram.
⇒ PQ = SR (opposite sides of parallelogram are equal)
⇒ PQ = 2cm
View full question & answer→MCQ 101 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 $\text{ABCD}$ 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→Question 111 Mark
Short Answer Questions.
In a quadrilateral PQRS, the diagonals PR and QS bisect each other. If $\angle\text{Q}=56^{\circ},$ determine $\angle\text{R}.$
AnswerSince the diagonals PR and QS of quadrilateral PQRS bisect each, PQRS is a parallelogram. Now, adjacent angles of parallelogram are supplementary.$\Rightarrow\angle\text{Q}+\angle\text{R}=180^{\circ}$
$\Rightarrow56^{\circ}+\angle\text{R}=180^{\circ}$
$\Rightarrow\angle\text{R}=124^{\circ}$
View full question & answer→MCQ 121 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 131 Mark
If $\text{APB}$ and $\text{CQD}$ 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 141 Mark
In the given figure, $\ce{ABCD}$ is a rhombus. Then:
- A
$\ce{AC^2 + BD^2 = AB^2}$
- B
$\ce{AC^2 + BD^2 = 2AB^2}$
- ✓
$\ce{AC^2 + BD^2 = 4AB^2}$
- D
$\ce{2\left(AC^2 + BD^2\right)=3AB^2}$
AnswerCorrect option: C. $\ce{AC^2 + BD^2 = 4AB^2}$
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 151 Mark
The figure formed by joining the midpoints of the sides of a quadrilateral $\text{ABCD},$ taken in order, is a square, only if:
- A
$\text{ABCD}$ is a rhombus.
- B
Diagonals of $\text{ABCD}$ are equal.
- C
Diagonals of $\text{ABCD}$ are perpendicular.
- ✓
Diagonals of $\text{ABCD}$ are equal and perpendicular.
AnswerCorrect option: D. Diagonals of $\text{ABCD}$ are equal and perpendicular.
In $\triangle\text{ABC}, P$ and $Q$ are the mid $-$ points of sides $AB$ and $BC$ respectively.
$\therefore \ PQ \| AC$ 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.
$\therefore\ QR \| BD$ 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.
$\therefore \ RS \| AC$ 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.
$\therefore \ SP \| BD$ and $\text{SP}=\frac{1}{2}\text{BD ...(iv)}$
$\Rightarrow \ PQ \| RS$ and $QR \| SP \ [$From $(i), (ii), (iii)$ and $(iv)]$
Thus, $\text{PQRS}$ 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,
$PS \| BD$
$\Rightarrow PN \| MO$
Also, from $(i), PQ \| AC$
$\Rightarrow PM \| NO$
Thus, in quadrilateral $\text{PMON}, PM \| NO$ and $PN \| MO$
$\Rightarrow \text{PMON}$ 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, $\text{PQRS}$ is a parallelogram such that $PQ = QR = RS = SP$ and $\angle\text{QPS}=90^{\circ}.$
Hence, $\text{PQRS}$ is a square if diagonals of $\text{ABCD}$ are equal and perpendicular. View full question & answer→MCQ 161 Mark
If bisectors of $\angle\text{A}$ and $\angle\text{B}$ of a quadrilateral $\text{ABCD}$ 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 $\text{PQRS}$ 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, $\text{PQRS}$ is a quadrilateral whose opposite angles are supplementary. View full question & answer→MCQ 171 Mark
The diagonals $AC$ and $BD$ of a parallelogram $\text{ABCD}$ 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. $40^\circ$
$AD \| BC,$
$\Rightarrow\angle\text{DAO}=\angle\text{BCO}=30^{\circ} ... ($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$)$
$\Rightarrow110+30+\angle\text{OBC}=180$
$\Rightarrow\angle\text{OBC}=40^{\circ}$
$\Rightarrow\angle\text{DBC}=40^{\circ} ... \text{(D - O - B)}$
View full question & answer→Question 181 Mark
Short Answer Questions.
All the angles of a quadrilateral can be obtuse. Is this statement true? Give reasons for your answer.
AnswerThe given statement is false.
We know that the sum of all the four angles of a quadrilateral is 360°.
If all the angles of a quadrilateral are obtuse, the sum will be more than 360°.
View full question & answer→MCQ 191 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 $\text{DENM}$ is:
Answer
In $\triangle\text{ABC}, D$ and $E$ are the mid $- $ points of sides $AB$ and $AC$ respectively.
$\Rightarrow\ DE\| BC$ 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 $(ii),$
$DE \| MN$ and $\text{DE = MN}$
Hence, $\text{DENM}$ is a parallelogram. View full question & answer→MCQ 201 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. $70^\circ$
$\angle\text{DAO}+\angle\text{OAB}=\angle\text{DAB}$
$\Rightarrow\angle\text{DAO}+35^{\circ}=90^{\circ}$
$\Rightarrow\angle\text{DAO}=55^{\circ}$
$\text{ABCD}$ 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 $\text{DODA},$ 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 211 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→Question 221 Mark
Short Answer Questions.
All the angles of a quadrilateral can be acute. Is this statement true? Give reasons for your answer.
AnswerThe given statement is false.
We know that the sum of all the four angles of a quadrilateral is 360°.
If all the angles of a quadrilateral are acute, the sum will be less than 360°.
View full question & answer→MCQ 231 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 241 Mark
The quadrilateral formed by joining the midpoints of the sides of a quadrilateral $\text{ABCD},$ taken in order, is a rectangle, if:
- A
$\text{ABCD}$ is a Parallelogram
- B
$\text{ABCD}$ is rectangle
- C
Diagonals of $\text{ABCD}$ are equal
- ✓
Diagonals of $\text{ABCD}$ are perpendicular to each other.
AnswerCorrect option: D. Diagonals of $\text{ABCD}$ are perpendicular to each other.
In $\triangle\text{ABC}, P$ and $Q$ are the mid $-$ point of sides $AB$ and $BC$ respectively.
$\therefore PQ \| AC$ 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.
$\therefore RS \| AC$ and $\text{RS}=\frac{1}{2}\text{AC }...(\text{ii})$
From $(i)$ and $(ii),$
$PQ \| RS$ and $\text{PQ = RS}$
Thus, in quadrilateral $\text{PQRS},$ a pair of opposite sides are equal are parallel.
So, $\text{PQRS}$ 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.
$\therefore PS \| BD$
$\Rightarrow PN \| MO$
Also, from $(i), PQ \| AC$
$\Rightarrow PM \| NO$
Thus, in quadrilateral $\text{PMON}, PM \| NO$ and $ 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, $\text{PQRS}$ is a parallelogram whose one angle, i.e. $\angle\text{QPS}=90^{\circ}.$
Hence, $\text{PQRS}$ is a rectangle if $\text{AC}\perp\text{BD}.$ View full question & answer→MCQ 251 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 261 Mark
In the given figure, $\text{ABCD}$ 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 271 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 281 Mark
In the given figure, $\text{ABCD}$ is $a \| gm$ and $E$ is the mid$-$point of $BC$. Also, $DE$ and $AB$ when produced meet at $F.$ Then:
AnswerCorrect option: B. $AF = 2AB$
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$)$
$\therefore \ce{CD = BF ...(C.P.C.T.)}$
Now,
$\text{AF = AB + BF}$
$\Rightarrow\text{AF + AB + CD} ...($from $(i))$
$\Rightarrow\text{AF = 2AB}$
View full question & answer→MCQ 291 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 301 Mark
In a trapezium $ABCD$, if $AB || CD$, then $(AC^2 + BD^2) = ?$
- A
$BC^2 + AD^2 + 2BC ⋅ AD$
- B
$AB^2 + CD^2^+ 2AB ⋅ CD$
- C
$AB^2 + CD^2 + 2AD ⋅ BC$
- ✓
$BC^2 + AD^2 + 2AB ⋅ CD$
AnswerCorrect option: D. $BC^2 + AD^2 + 2AB ⋅ CD$
$BC^2 + AD^2 + 2AB ⋅ CD$
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→Question 311 Mark
Short Answer Questions.
If D and E are respectively the midpoints of the sides AB and BC of $\triangle\text{ABC}$ in which AB = 7.2cm, BC = 9.8cm and AC = 3.6cm then determine the length of DE.
AnswerD and E are respectively the midpoints of the sides AB and BC of $\triangle\text{ABC}.$ Thus, by mid-point theorem, we have$\text{DE || AC}$ and $\text{DE}=\frac{1}{2}\text{AC}$
$\Rightarrow\text{DE}=\frac{1}{2}\times3.6=1.8\text{cm}$
View full question & answer→MCQ 321 Mark
If $\angle\text{A},\angle\text{B},\angle\text{C}$ and $\angle\text{D}$ of a quadrilateral $\text{ABCD}$ taken in order, are in the ratio $\{3 : 7 : 6 : 4\}$ then $\text{ABCD}$ 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 $\text{ABCD}$.
Hence $, \text{ABCD}$ is a trapezium.
View full question & answer→MCQ 331 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 341 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 351 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 $\times$ 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 361 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 371 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→Question 381 Mark
Short Answer Questions.
Can we form a quadrilateral whose angles are 70°, 115°, 60° and 120°? Give reasons for your answer.
AnswerWe know that the sum of all the four angles of a quadrilateral is 360°. Here,$70^{\circ} + 115^{\circ} + 60^{\circ} + 120^{\circ} = 365^{\circ} \neq 360^{\circ}$
Hence, we cannot form a quadrilateral with given angles.
View full question & answer→MCQ 391 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 $\text{ABCD},$
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 401 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→Question 411 Mark
Short Answer Questions. In the adjoining figure, BDEF and AFDE are parallelograms. Is AF = FB? Why or why not?
AnswerAFDE is a parallelogram
⇒ AF = ED …(i)
BDEF is a parallelogram.
⇒ FB = ED …(ii)
From (i) and (ii),
AF = FB
View full question & answer→MCQ 421 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 $x^{\circ}$.
We know that, the sum of the angles of a quadrilateral is $360^{\circ}$.
So, $80^{\circ}+95^{\circ}+112^{\circ}+x=360^{\circ}$
$\Rightarrow 287^{\circ}+x=360^{\circ}$
$\Rightarrow x=73^{\circ}$
$\therefore$ Its fourth angle is $73^{\circ}$.
View full question & answer→Question 431 Mark
Short Answer Questions.
All the angles of a quadrilateral can be right angles. Is this statement true? Give reasons for your answer.
AnswerThe given statement is true.
We know that the sum of all the four angles of a quadrilateral is 360°.
If all the angles of a quadrilateral are right angles,
Sum of all angles of a quadrilateral = 4 × 90° = 360°
View full question & answer→MCQ 441 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. $\frac{1}{2}(\text{a}+\text{b})$
$E$ 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 451 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 461 Mark
The quadrilateral formed by joining the midpoints of the sides of a quadrilateral $\text{ABCD}$, taken in order, is a rhombus, if:
- A
$\text{ABCD}$ is a Parallelogram.
- B
$\text{ABCD}$ is rhombus.
- ✓
Diagonals of $\text{ABCD}$ are equal.
- D
Diagonals of $\text{ABCD}$ are perpendicular to each other.
AnswerCorrect option: C. Diagonals of $\text{ABCD}$ are equal.
In $\triangle\text{ABC}, P$ and $Q$ are the mid $-$ points of sides $AB$ and $BC$ respectively.
$\therefore PQ \| AC$ 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.
$\therefore QR \| BD$ 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.
$\therefore RS \| AC$ 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.
$\therefore SP \| BD$ and $\text{SP}=\frac{1}{2}\text{BD ... (iv)}$
$\Rightarrow PQ \| RS$ and $QR \| SP \ [$From $(i), (ii), (iii)$ and $(iv)]$
Thus, $\text{PQRS}$ 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, $\text{PQRS}$ is a rhombus if diagonals of $\text{ABCD}$ are equal. View full question & answer→MCQ 471 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 $AG$.
$\Rightarrow DG \| BF$
$\Rightarrow DG \| EF ...\text{(B - E - F)}$
$\Rightarrow\text{AF}=\text{FG}=\text{GC} ...($Since $G$ is the mid $-$ point.$)$
$\Rightarrow\text{AF}=\frac{1}{2}\text{AC}$ View full question & answer→MCQ 481 Mark
In the given figure $, \text{ABCD}$ 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 491 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 501 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 $BD = 12\ cm$
$\Rightarrow OA = 8\ cm$ 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→