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MATHS M.C.Q

Quadrilaterals · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 56 questions.

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M.C.Q

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Question 11 Mark
Three statements are given below:
  1. In a rectangle ABCD, the diagonal AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
  2. In a square ABCD, the diagonal AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
  3. In a rhombus ABCD, the diagonal AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
Which is true?
  1. I only
  2. II and III
  3. I and III
  4. I and II
Answer
  1. II and III
Solution:
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 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 ABCD, the diagonals AC bisects $\angle\text{A}$ as well as $\angle\text{C}.$
Thus, III is true.
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Question 21 Mark
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a:
  1. Rhombus.
  2. Square.
  3. Rectangle.
  4. Parallelogram.
Answer
  1. Rectangle.
Solution:
The figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle.
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Question 31 Mark
In the given figure, ABCD is a parallelogram in which $\angle\text{BDC}=45^{\circ}$ and $\angle\text{BAD}=75^{\circ}.$ Then, $\angle\text{CBD}=?$
  1. 45°
  2. 55°
  3. 60°
  4. 75°
Answer
  1. 60°.
Solution:
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)
$\Rightarrow\angle\text{CBD}=60^{\circ}$
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Question 41 Mark
The length of each side of a rhombus is 10cm and one if its diagonals is of length 16cm. The length of the other diagonal is:
  1. 13cm
  2. 12cm
  3. $2\sqrt{39}\text{cm}$
  4. 6cm
Answer
  1. 12cm.
Solution:

Let ABCD be the rhombus with diagonals AB = 10cm and Ac = 16cm.
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 12cm.
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Question 51 Mark
The figure formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a square, only if:
  1. ABCD is a rhombus.
  2. Diagonals of ABCD are equal.
  3. Diagonals of ABCD are perpendicular.
  4. Diagonals of ABCD are equal and perpendicular.
Answer
  1. Diagonals of ABCD are equal and perpendicular.
Solution:

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.
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Question 61 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:
  1. Rectangle.
  2. Parallelogram.
  3. Rhombus.
  4. Quadrilateral whose opposite angles are supplementary.
Answer
  1. Quadrilateral whose opposite angles are supplementary.
Solution:

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.
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Question 71 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, $\angle\text{DBC}=?$
  1. 40°
  2. 35°
  3. 45°
  4. 50°
Answer
  1. 40°
Solution:
$\text{AD || BC},$
$\Rightarrow\angle\text{DAO}=\angle\text{BCO}=30^{\circ}$ ...(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)
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Question 81 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:
  1. A trapezium.
  2. A parallelogram.
  3. A rectangle.
  4. A rhombus.
Answer
  1. A parallelogram.
Solution:

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.
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MCQ 91 Mark
In the given figure, $\text{ABCD}$ is a rhombus. Then:
  • A
    $AC^2 + BD^2 = AB^2$​​​​​​​
  • B
    $AC^2 + BD^2 = 2AB^2$
  • ✓
    $AC^2 + BD^2 = 4AB^2$
  • D
    $2(AC^2 + BD^2) = 3AB^{2x}$
Answer
Correct option: C.
$AC^2 + BD^2 = 4AB^2$
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$)$
$\Rightarrow\text{AB}^2=\frac{1}{4}\text{AC}^2+\frac{1}{4}\text{BD}^2$
$\Rightarrow\text{AC}^2+\text{BD}^2=4\text{AB}^2$
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Question 101 Mark
A diagonal of a rectangle is inclined to one side of the rectangle at 35°. The acute angle between the diagonals is:
  1. 55°
  2. 70°
  3. 45°
  4. 50°
Answer
  1. 70°
Solution:

$\angle\text{DAO}+\angle\text{OAB}=\angle\text{DAB}$
$\Rightarrow\angle\text{DAO}+35^{\circ}=90^{\circ}$
$\Rightarrow\angle\text{DAO}=55^{\circ}$
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,
$\angle\text{ODA}+\angle\text{DAO}+\angle\text{AOD}=180^{\circ}$
$\Rightarrow55^{\circ}+\angle55^{\circ}+\angle\text{AOD}=180^{\circ}$
$\Rightarrow\angle\text{AOD}=70^{\circ}$
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Question 111 Mark
The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rectangle, if:
  1. ABCD is a Parallelogram
  2. ABCD is rectangle
  3. Diagonals of ABCD are equal
  4. Diagonals of ABCD are perpendicular to each other.
Answer
  1. Diagonals of ABCD are perpendicular to each other.
Solution:

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 $\text{AC}\perp\text{BD}.$ 
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Question 121 Mark
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a:
  1. Rhombus.
  2. Square.
  3. Rectangle.
  4. Parallelogram.
Answer
  1. Parallelogram.
Solution:
The figure formed by joining the mid points of the adjacent sides of a parallelogram is a parallelogram.
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Question 131 Mark
Is quadrilateral ABCD a rhombus?
  1. Quadrilateral ABCD is a || gm.
  2. Diagonals AC and BD are perpendicular to each other.
  1. If the question can be answered by one of the given statements alone and not by the other;
  2. If the question can be answered by either statement alone;
  3. If the question can be answered by both the statements together but not by any one of the two;
  4. If the question cannot be answered by using both the statements together.
Answer
  1. If the question can be answered by both the statements together but not by any one of the two;
Solution:
If the quad. 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. ABCD is a rhombus.
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Question 141 Mark
Short Answer Questions.
What special name can be given to a quadrilateral PQRS if $\angle\text{P}+\angle\text{S}=180^{\circ}?$
Answer
In 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.
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Question 151 Mark
Short Answer Questions.
In a quadrilateral PQRS, the diagonals PR and QS bisect each other. If $\angle\text{Q}=56^{\circ},$ determine $\angle\text{R}.$
Answer
Since the diagonals PR and QS of quadrilateral PQRS bisect each, PQRS is a parallelogram.
Now, adjacent angles of parallelogram are supplementary.
$\Rightarrow\angle\text{Q}+\angle\text{R}=180^{\circ}$
$\Rightarrow56^{\circ}+\angle\text{R}=180^{\circ}$
$\Rightarrow\angle\text{R}=124^{\circ}$
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Question 161 Mark
Short Answer Questions.
All the angles of a quadrilateral can be obtuse. Is this statement true? Give reasons for your answer.
Answer
The 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°.
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Question 171 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.
Answer
D 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 $\text{DE}=\frac{1}{2}\text{AC}$
$\Rightarrow\text{DE}=\frac{1}{2}\times3.6=1.8\text{cm}$
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Question 181 Mark
Short Answer Questions.
Can we form a quadrilateral whose angles are 70°, 115°, 60° and 120°? Give reasons for your answer.
Answer
We 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.
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Question 191 Mark
Short Answer Questions.
In a quadrilateral PQRS, opposite angles are equal. If SR = 2cm and PR = 5cm then determine PQ.
Answer
Since opposite angles of quadrilateral are equal, PQRS is a parallelogram.
⇒ PQ = SR (opposite sides of parallelogram are equal)
⇒ PQ = 2cm
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Question 201 Mark
Short Answer Questions.
All the angles of a quadrilateral can be acute. Is this statement true? Give reasons for your answer.
Answer
The 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°. 
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Question 211 Mark
Short Answer Questions. In the adjoining figure, BDEF and AFDE are parallelograms. Is AF = FB? Why or why not?  
Answer
AFDE is a parallelogram
⇒ AF = ED …(i)
BDEF is a parallelogram.
⇒ FB = ED …(ii)
From (i) and (ii),
AF = FB
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Question 221 Mark
Short Answer Questions.
All the angles of a quadrilateral can be right angles. Is this statement true? Give reasons for your answer.
Answer
The 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°
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Question 231 Mark
Short Answer Questions.
What special name can be given to a quadrilateral whose all angles are equal?
Answer
A quadrilateral whose all angles are equal is a rectangle.
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Question 241 Mark
Short Answer Questions.
Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reasons for your answer.
Answer
The given statement is false.
Diagonals of a parallelogram bisect each other.
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Question 251 Mark
Which of the following is not true for a parallelogram?
  1. Opposite sides are equal.
  2. Opposite angles are equal.
  3. Opposite angles are bisected by the diagonals.
  4. Diagonals bisect each other.
Answer
  1. Opposite angles are bisected by the diagonals.
Solution:
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.
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MCQ 261 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:
  • A
    $\text{AF}=\frac{3}{2}\text{AB}$
  • ✓
    $AF = 2AB$
  • C
    $AF = 3AB$
  • D
    $AF^2 = 2AB^2$
Answer
Correct 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$)$
$\therefore\text{CD = BF} ...(C.P.C.T.)$
Now,
$\text{AF = AB + BF}$
$\Rightarrow\text{AF + AB + CD} ...($from $(i))$
$\Rightarrow\text{AF = 2AB}$
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Question 271 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:
  1. Square.
  2. Rhombus.
  3. Rectangle.
  4. Kite.
Answer
  1. Rectangle.
Solution:
The bisectors of $\angle\text{APQ},\angle\text{BPQ},\angle\text{CQP}$ and $\angle\text{PQD}$ enclose a rectengle.
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Question 281 Mark
The figure formed by joining the mid-points of the adjacent sides of a square is a:
  1. Rhombus.
  2. Square.
  3. Rectangle.
  4. Parallelogram.
Answer
  1. Square.
Solution:
The figure formed by joining the mid points of the adjacent sides of a square is a square.
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MCQ 291 Mark
In a trapezium $\text{ABCD,}$ if $AB \| CD,$ then $(AC^2 + BD^2) =$ ?
  • A
    $BC^2 + AD^2 + 2BC ⋅ AD$
  • B
    $AB^2 + CD^2+ 2AB ⋅ CD$
     
  • C
    $AB^2 + CD^2 + 2AD ⋅ BC$
  • ✓
    $BC^2 + AD^2 + 2AB ⋅ CD$
Answer
Correct option: D.
$BC^2 + AD^2 + 2AB ⋅ CD$
Solution:

Construction: Draw perpendicular from $D$ and $C$ on $AB$ which meets $AB$ at $E$ and $F,$ respectively.
So, $\text{DEFC}$ 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.
$\Rightarrow\text{BD}^2=\text{AD}^2+\text{AB}^2-2\text{AB}\times\text{AF}$
$\Rightarrow\text{AC}^2+\text{BD}^2=\text{BC}^2+\text{AD}^2+2\text{AB}(\text{AB}-\text{BE}-\text{AF})$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{EF}$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{CD}$
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Question 301 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:
  1. Rhombus.
  2. Kite.
  3. Trapezium.
  4. Parallelogram.
Answer
  1. Trapezium.
Solution:
Let 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°, we have
3x + 7x + 6x + 4x = 360
⇒ 20x = 360
⇒ x = 18°
$\therefore$ The angles of the quadrilateral are
3x = 3(18) = 54°
7x = 7(18) = 126°
6x = 6(18) = 108° and
4x = 4(18) = 72°
Now, 54 + 126 = 180° and 108 + 72 = 180°
So, the angles are interior angles and hence we get one pair of parallel sides of ABCD.
Hence, ABCD is a trapezium.
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Question 311 Mark
Three statements are given below:
  1. In a || gm, the angle bisectors of two adjacent angles enclose a right angle.
  2. The angle bisectors of a || gm form a rectangle.
  3. The triangle formed by joining the mid-points of the sides of an isosceles triangle is not necessarily an isosceles triangle.
Which is true?
  1. I only
  2. II only
  3. I and II
  4. II and III
Answer
  1. I and II
Solution:
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.
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Question 321 Mark
In which of the following figures are the diagonals equal?
  1. Parallelogram.
  2. Rhombus.
  3. Trapezium.
  4. Rectangle.
Answer
  1. Rectangle.
Solution:
The diagonals are equal in a rectangle.
The diagonals in a parallelogram, rhombus or trapezium need not be equal.
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Question 331 Mark
Is quad. ABCD a parallelogram?
  1. Its opposite sides are equal.
  2. Its opposite angles are equal.
  1. If the question can be answered by one of the given statements alone and not by the other;
  2. If the question can be answered by either statement alone;
  3. If the question can be answered by both the statements together but not by any one of the two;
  4. If the question cannot be answered by using both the statements together.
Answer
  1. If the question can be answered by either statement alone;
Solution:
If the opposite sides of a quad. ABCD are equal, the quadrilateral is a parallelogram.
If the opposite angles are equal, then the quad. ABCD is a parallelogram.
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Question 341 Mark
Two parallelograms stand on equal bases and between the same parallels. The ratio of their areas is:
  1. 1 : 2
  2. 2 : 1
  3. 1 : 3
  4. 1 : 1
Answer
  1. 1 : 1
Solution:
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.
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Question 351 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:
  1. A > B
  2. A = B
  3. A < B
  4. A ≥ B
Answer
  1. A < B
Solution:
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.
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Question 361 Mark
If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a:
  1. Trapezium.
  2. Parallelogram.
  3. Rectangle.
  4. Rhombus.
Answer
  1. Rhombus.
Solution:
If 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.
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Question 371 Mark
If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is:
  1. 108°
  2. 54°
  3. 72°
  4. 81°
Answer
  1. 72°
Solution:
Let one of the angle of the || gm be x°.
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°.)
$\Rightarrow\frac{3\text{x}+2\text{x}}{3}=180^{\circ}$
$\Rightarrow\frac{5\text{x}}{3}=180^{\circ}$
$\Rightarrow5\text{x}=540^{\circ}$
$\Rightarrow\text{x}=108^{\circ}$
⇒ the adjacent angles $=\frac{2}{3}(108)=36\times2=72^{\circ}$
Hence, the smallest angle is 72°.
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Question 381 Mark
The bisectors of any two adjacent angles of a parallelogram intersect at:
  1. 40°
  2. 45°
  3. 60°
  4. 90°
Answer
  1. 90°
Solution:

Consider parallelogram ABCD,
We know that, the sum if the adjacent angles of a parallelogram is 180°.
$\Rightarrow\angle\text{DAB}+\angle\text{CBA}=180^{\circ}$
$\Rightarrow\frac{1}{2}\angle\text{DAB}+\frac{1}{2}\angle\text{CBA}=\frac{1}{2}(180^{\circ})$
$\Rightarrow\angle\text{OAB}+\angle\text{OBA}=90^{\circ} ...(\text{i})$
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°.
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Question 391 Mark
The bisectors of the angles of a parallelogram enclose a:
  1. Rhombus.
  2. Square.
  3. Rectangle.
  4. Parallelogram.
Answer
  1. Rectangle
Solution:
The bisectors of the angles of a parallelogram encloses a rectangle.
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Question 401 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:
  1. $\frac{1}{2}(\text{a}-\text{b})$
  2. $\frac{1}{2}(\text{a}+\text{b})$
  3. $\frac{2\text{ab}}{(\text{a + b})}$
  4. $\sqrt{\text{ab}}$
Answer
  1. $\frac{1}{2}(\text{a}+\text{b})$
Solution:

E and F are the given to be the mid-points of AD and BC respectively.
$\therefore\text{EF} = \frac{1}{2}(\text{AB + DC})$
$=\frac{1}{2}(\text{a + b})$
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Question 411 Mark
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a:
  1. Rhombus.
  2. Square.
  3. Rectangle.
  4. Parallelogram.
Answer
  1. Parallelogram.
Solution:
The figure formed by joining the mid points of the adjacent sides of a quadrilateral is a parallelogram.
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Question 421 Mark
The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rhombus, if:
  1. ABCD is a Parallelogram.
  2. ABCD is rhombus.
  3. Diagonals of ABCD are equal.
  4. Diagonals of ABCD are perpendicular to each other.
Answer
  1. Diagonals of ABCD are equal.
Solution:

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.
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Question 431 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 = ?
  1. $\frac{1}{2}\text{AC}$
  2. $\frac{1}{3}\text{AC}$
  3. $\frac{2}{3}\text{AC}$
  4. $\frac{3}{4}\text{AC}$
Answer
  1. $\frac{1}{3}\text{AC}$
Solution:

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.)
$\Rightarrow\text{AF}=\frac{1}{2}\text{AC}$
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Question 441 Mark
If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angle of the parallelogram is:
  1. 68°
  2. 102°
  3. 112°
  4. 136°
Answer
  1. 112°
Solution:
Let the smallest angle be x°
⇒ its adjacent angle = (2x - 24)
Since sum of the adjacent angles = 180°
⇒ x + 2x - 24 = 180
⇒ 3x = 204
⇒ x = 68°
So, its adjacent angle = 2(68) - 24 = 136 - 24 = 112°.
Hence, the largest angle of the parallelogram is 112°.
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Question 451 Mark
In the given figure, ABCD is a parallelogram in which $\angle\text{BAD}=75{^\circ}$ and $\angle\text{CBD}=60^{\circ}.$ Then, $\angle\text{BDC}=?$
  1. 60°
  2. 75°
  3. 45°
  4. 50°
Answer
  1. 45°
Solution:
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)
$\Rightarrow75^{\circ}+\angle\text{BDC}+60^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{BDC}=45^{\circ}$
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Question 461 Mark
Is || gm ABCD a square?
  1. Diagonals of || gm ABCD are equal.
  2. Diagonals of || gm ABCD intersect at right angles.
  1. If the question can be answered by one of the given statements alone and not by the other;
  2. If the question can be answered by either statement alone;
  3. If the question can be answered by both the statements together but not by any one of the two;
  4. If the question cannot be answered by using both the statements together.
Answer
  1. If the question can be answered by both the statements together but not by any one of the two;
Solution:
If the diagonals of a || gm ABCD are equal, then ‖gm ABCD could either be a rectangle or a square.
If the diagonals of the ‖gm ABCD intersect at right angles, then the ‖gm ABCD could be a square or a rhombus.
However, if both the statements are combined, then ‖gm ABCD will be a square.
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Question 471 Mark
The lengths of the diagonals of a rhombus are 16cm and 12cm. The length of each side of the rhombus is:
  1. 10cm
  2. 12cm
  3. 9cm
  4. 8cm
Answer
  1. 10cm.
Solution:

We konw that, the diagonals of a rhombus bisect each other at right angles.
So, Ac = 16cm and BD = 12cm
⇒ OA = 8cm and OB = 6cm
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 10cm.
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Question 481 Mark
If ABCD is a parallelogram with two adjacent angles $\angle\text{A}=\angle\text{B}$ then the parallelogram is a:
  1. Rhombus.
  2. Trapezium.
  3. Rectangle.
  4. None of these.
Answer
  1. Rectangle.
Solution:
Given 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.
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Question 491 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, $\angle\text{AOB}=?$
  1. 40°
  2. 50°
  3. 80°
  4. 100°
Answer
  1. 50°
Solution:

We know that, sum of the angles of a quadrilateral is 360°.
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}$
$\Rightarrow\angle\text{A}+\angle\text{B}+70^{\circ}+30^{\circ}=360^{\circ}$
$\Rightarrow\angle\text{A}+\angle\text{B}=260^{\circ}$
$\Rightarrow\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=\frac{1}{2}(260)^{\circ}$
$\Rightarrow\angle\text{BAO}+\angle\text{ABO}=130^{\circ}...(\text{i})$
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))
$\Rightarrow\angle\text{AOB}=50^{\circ}$
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Question 501 Mark
Is quadrilateral ABCD a || gm?
  1. Diagonals AC and BD bisect each other.
  2. Diagonals AC and BD are equal.
  1. If the question can be answered by one of the given statements alone and not by the other;
  2. If the question can be answered by either statement alone;
  3. If the question can be answered by both the statements together but not by any one of the two;
  4. If the question cannot be answered by using both the statements together.
Answer
  1. If the question can be answered by one of the given statements alone and not by the other;
Solution:
If the diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a parallelogram.
So, I gives the answer.
If the diagonals are equal, then the quad. ABCD is a parallelogram.
So, II gives the answer.
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