MCQ 11 Mark
How many sides does a regular polygon have if the measure of an exterior angle is $24^\circ ?$
- A$6$
- B$9$
- ✓$15$
- D$12$
Answer
View full question & answer→Correct option: C.
$15$
Number of sides $\frac{360^\circ}{24^\circ}=15$
Questions · Page 1 of 5
M.C.Q. [1 Marks Each]
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MCQ 21 Mark
$PQRS$ is a trapezium in which $PQ || SR$ and $\angle\text{P}=130^\circ, \angle\text{Q}= 110^\circ.$ Then $\angle\text{R}$ is equal to:
- ✓$70^\circ$
- B$50^\circ$
- C$65^\circ$
- D$55^\circ$
Answer
View full question & answer→Correct option: A.
$70^\circ$
Since, $PQRS$ is a trapezium and $PQ \ || \ SR$
$\therefore\angle\text{Q}+\angle\text{R}=180^\circ$
$\Rightarrow\angle\text{R}= 180^\circ − 110^\circ= 70^\circ$
$\therefore\angle\text{Q}+\angle\text{R}=180^\circ$
$\Rightarrow\angle\text{R}= 180^\circ − 110^\circ= 70^\circ$
MCQ 31 Mark
If $AB$ and $CD$ are two parallel sides of a parallelogram, then:
- ✓$AB = CD$
- B$AB > CD$
- C$AB < CD$
- DNone of the above
Answer
View full question & answer→Correct option: A.
$AB = CD$
$AB = CD$
MCQ 41 Mark
In the quadrilateral $ABCD,$ the diagonals $AC$ and $BD$ are equal and perpendicular to each other. What type of a quadrilateral is $ABCD?$
- ✓A square
- BA parallelogram
- CA rhombus
- DA trapezium
Answer
View full question & answer→Correct option: A.
A square
A square
MCQ 51 Mark
Tick the correct answer in the following$?$
The measure of each exterior angle of a regular polygon is $40^\circ .$ How many sides does it have$?$
The measure of each exterior angle of a regular polygon is $40^\circ .$ How many sides does it have$?$
- A$8$
- ✓$9$
- C$6$
- D$10$
Answer
View full question & answer→Correct option: B.
$9$
Each exterior angle of a regular n-sided polygon $=\frac{160}{\text{n}}=40$
$\Rightarrow\text{n}=\frac{360}{\text{40}}=9$
MCQ 61 Mark
The number of sides of a regular polygon whose each interior angle is of $135^\circ $ is:
- A$6$
- B$7$
- ✓$8$
- D$9$
Answer
View full question & answer→Correct option: C.
$8$
We know that, the measures of each exterior angle of a polygon having n sides is given by $\frac{360^\circ}{\text{n}}$
$\therefore$ The number of sides, $\text{n}=\frac{360^\circ}{\text{Exterior angle}}=\frac{360^\circ}{180^\circ}=\frac{360^\circ}{45^\circ}=8$
MCQ 71 Mark
A simple closed curve made up of only line segments is called a $..............$
- ✓Polygon
- BQuadrilateral
- CHexagon
- DNone of these
Answer
View full question & answer→Correct option: A.
Polygon
Polygon
MCQ 81 Mark
How many diagonals does a triangle have?
- ✓$0$
- B$1$
- C$2$
- D$4$
Answer
View full question & answer→Correct option: A.
$0$
$0$
MCQ 91 Mark
Which of the following statement is true?
- AAll the rectangles are squares.
- BAll the parallelograms are rhombuses.
- ✓All the squares are rhombuses.
- DEach parallelogram is a trapezium.
Answer
View full question & answer→Correct option: C.
All the squares are rhombuses.
All the squares are rhombuses.
MCQ 101 Mark
The four angles of a pentagon are $40^\circ , 75^\circ , 125^\circ and 135^\circ .$ The measure of the fifth angle is:
- ✓$165^\circ$
- B$170^\circ $
- C$160^\circ$
- D$175^\circ$
Answer
View full question & answer→Correct option: A.
$165^\circ$
$ n= 5, (n - 2) 180^\circ = (5 - 2) 180^\circ = 540^\circ $
Fifth angle
$= 540^\circ - (40^\circ + 75^\circ + 125^\circ + 135^\circ )$
$= 540^\circ - 375^\circ = 165^\circ $
Fifth angle
$= 540^\circ - (40^\circ + 75^\circ + 125^\circ + 135^\circ )$
$= 540^\circ - 375^\circ = 165^\circ $
MCQ 111 Mark
How many non-overlapping triangles can we make in a $n-$gon (polygon having n sides), by joining the vertices$?$
- A$n - 1$
- ✓$n - 2$
- C$n - 3$
- D$n - 4$
Answer
View full question & answer→Correct option: B.
$n - 2$
The number of non-overlapping triangles in a $n-$gon $= n - 2,$ i.e., $2$ less than the number of sides.
MCQ 121 Mark
$\text{ABCD}$ is a quadrilateral. If $\text{AC}$ and $\text{BD}$ bisect each other, what is $\text{ABCD}?$
- AA square
- BA parallelogram
- CA rectangle
- ✓All the above
Answer
View full question & answer→Correct option: D.
All the above
All the above
MCQ 131 Mark
The perimeter of a parallelogram is $180\ cm.$ One side exceeds another by $10\ cm.$ The sides of the parallelogram are:
- ✓$40\ cm, 50\ cm$
- B$50\ cm$ each
- C$45\ cm$ each
- DCannot be determined
Answer
View full question & answer→Correct option: A.
$40\ cm, 50\ cm$
$40\ cm, 50\ cm$
MCQ 141 Mark
Length of one of the diagonals of a rectangle whose sides are $10\ cm$ and $24\ cm$ is.
- A$25\ cm$
- B$20\ cm$
- ✓$26\ cm$
- D$3.5\ cm$
Answer
View full question & answer→Correct option: C.
$26\ cm$
In $\triangle\text{BDC}=90^\circ$
Using Pythagoras Theorem, We have,
$\text{BC}^2=\text{BD}^2+\text{CD}^2$
$\Rightarrow\text{BC}^2=10^2+24^2=100+576$
$\Rightarrow\text{BC}^2=676$
$\Rightarrow\text{BC}=\sqrt{676}$
$\Rightarrow\text{BC}=256\text{cm}$
MCQ 151 Mark
A quadrialateral whose opposite sides and all the angles are equal is a.
- ✓rectangle
- Bparallelogram
- Csquare
- Drhombus
Answer
View full question & answer→Correct option: A.
rectangle
We know that, in a rectangle, opposite sides and all the angles are equal.
MCQ 161 Mark
Out of the three equal angles of a quadrilateral, each measures $70^\circ .$ The measure of the fourth angle is:
- A$90^\circ$
- B$140^\circ$
- ✓$150^\circ$
- D$70^\circ$
Answer
View full question & answer→Correct option: C.
$150^\circ$
Fourth angle $= 360^\circ - (70^\circ + 70^\circ + 70^\circ )$
$= 150^\circ .$
MCQ 171 Mark
If a quadrilateral has two adjacent sides equal and the other two sides equal, it is called:
- ASquare
- BParallelogram
- ✓Kite
- DRectangle
Answer
View full question & answer→Correct option: C.
Kite
Kite
MCQ 181 Mark
Which of the following quadrilaterals is a regular quadrilateral?
- ARectangle
- ✓Square
- CRhombus
- DKite
Answer
View full question & answer→Correct option: B.
Square
A rhombus is a quadrilateral with all the sides equal and its diagonals bisect each other, but all of its angles are not equal.
We already know that only the parallel sides or the opposite sides of a parallelogram are equal.
In a trapezium no side is equal to another.
A rectangle has only its opposite sides as equal. Therefore, square is the only regular quadrilateral.
So, the correct answer is “square”.
Note: Do not confuse that rectangle is also a regular quadrilateral.
It is not because all the sides are not equal in a rectangle whereas all the angles are equal and diagonals bisect each other.
For a quadrilateral to be regular it should have all the sides equal.
So the rectangle is not.
When all the sides are equal in a rectangle, then it can be considered as a square.
We already know that only the parallel sides or the opposite sides of a parallelogram are equal.
In a trapezium no side is equal to another.
A rectangle has only its opposite sides as equal. Therefore, square is the only regular quadrilateral.
So, the correct answer is “square”.
Note: Do not confuse that rectangle is also a regular quadrilateral.
It is not because all the sides are not equal in a rectangle whereas all the angles are equal and diagonals bisect each other.
For a quadrilateral to be regular it should have all the sides equal.
So the rectangle is not.
When all the sides are equal in a rectangle, then it can be considered as a square.
MCQ 191 Mark
The sum of angles of a concave quadrilateral is:
- Amore than $360^\circ$
- Bless than $360^\circ$
- ✓equal to $360^\circ$
- Dtwice of $360^\circ$
Answer
View full question & answer→Correct option: C.
equal to $360^\circ$
C. equal to $360^\circ$
Solution:
We know that, the sum of interior angles of any polygon (convex or concave) having n sides is $(n − 2) \times 180^\circ$
$\therefore$ The sum of angles of a concave quadrilateral is $(4 – 2) \times 180^\circ$, i.e. $360^\circ$
Solution:
We know that, the sum of interior angles of any polygon (convex or concave) having n sides is $(n − 2) \times 180^\circ$
$\therefore$ The sum of angles of a concave quadrilateral is $(4 – 2) \times 180^\circ$, i.e. $360^\circ$
MCQ 201 Mark
The diagonals of a rectangle are $2x + 1$ and $3x - 1,$ respectively. Find the value of $x.$
- ✓$2$
- B$4$
- C$1$
- D$3$
Answer
View full question & answer→Correct option: A.
$2$
The diagonals of a rectangle are equal in length.
$2x + 1 = 3x - 1$
$1 + 1 = 3x - 2x$
$2 = x$
Thus, the value of $x$ is $2.$
MCQ 211 Mark
Tick the correct answer in the following$?$ How many diagonals are there in a polygon having $12$ sides$?$
- A$12$
- B$24$
- C$36$
- ✓$54$
Answer
View full question & answer→Correct option: D.
$54$
For an n-sided polygon:
Number of diagonals $=\frac{\text{n}(\text{n}-3)}{2}$
$\therefore\text{n}=12$
$\Rightarrow\frac{12(12-3)}{2}=54$
MCQ 221 Mark
$ABCD$ is a rectangle and $AC$ & $BD$ are its diagonals. If $AC = 10\ cm,$ then $BD$ is:
- A$20\ cm$
- ✓$10\ cm$
- C$5\ cm$
- D$15\ cm$
Answer
View full question & answer→Correct option: B.
$10\ cm$
The diagonals of a rectangle are always equal.
MCQ 231 Mark
The closed curve which is also a polygon is:
- ✓
- B
- C
- D
Answer
is polygon as no two line segments intersect each other.
View full question & answer→Correct option: A.
is polygon as no two line segments intersect each other.
MCQ 241 Mark
The measures of the three angles of a quadrilateral are $65^\circ , 75^\circ $ and $85^\circ $. The measure of the fourth angle is:
- A$65^\circ$
- B$75^\circ$
- C$85^\circ$
- ✓$135^\circ$
Answer
View full question & answer→Correct option: D.
$135^\circ$
$135^\circ$
MCQ 251 Mark
Two adjacent angles of a quadrilateral measure $130^\circ $ and $40^\circ .$ The sum of the remaining two angles is:
- ✓$190^\circ$
- B$180^\circ$
- C$360^\circ$
- D$90^\circ$
Answer
View full question & answer→Correct option: A.
$190^\circ$
$190^\circ$
MCQ 261 Mark
The angle sum of a convex polygon with number of sides $8$ is:
- A$720^\circ$
- B$900^\circ$
- ✓$1080^\circ $
- D$1440^\circ$
Answer
View full question & answer→Correct option: C.
$1080^\circ $
$n = 8$
$(n - 2) 180^\circ = 1080^\circ .$
MCQ 271 Mark
The sides of a pentagon are produced in order. Which of the following is the sum of its exterior angles?
- A$540^\circ$
- B$180^\circ$
- C$720^\circ$
- ✓$360^\circ $
Answer
View full question & answer→Correct option: D.
$360^\circ $
$\because ABCD$ is a Quadrilateral
To Proof $\angle\text{1}+ \angle\text{2}+\angle\text{3}+\angle\text{4}=360^\circ$
$\angle\text{A}+ \angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$\angle\text{A}+ \angle\text{1}=180^\circ$
$\angle\text{B}+ \angle\text{2}=180^\circ$
$\angle\text{C}+ \angle\text{3}=180^\circ$
$\angle\text{D}+ \angle\text{4}=180^\circ$
$\angle\text{A}+ \angle\text{B}+\angle\text{C}+\angle\text{D}+ \angle\text{1}+ \angle\text{2}+\angle\text{3}+\angle\text{4}$
$= 180^\circ+180^\circ+180^\circ+180^\circ$
$=720^\circ$
$360^\circ+\angle1+\angle2+\angle3+\angle4=720^\circ$
$\angle1+\angle2+\angle3+\angle4=360^\circ$
Each polygon have $360^\circ $ as it's exterior angel but it's Interior angle is depend on it's sides, so the answer is $D.$
MCQ 281 Mark
In the given figure, $ABCD$ and $BDCE$ are parallelograms with common base $DC.$ If $BC ⊥ BD,$ then $\angle\text{BEC}=$
- ✓$60^\circ$
- B$30^\circ$
- C$150^\circ$
- D$120^\circ$
Answer
View full question & answer→Correct option: A.
$60^\circ$
$\angle\text{BCD}=30^\circ$
$\therefore\angle\text{BCD}=30^\circ$,
in $\triangle\text{CBD}$ by angle sum property of a triangle, we have
$\Rightarrow\angle\text{DBC}+\angle\text{BCD}+\angle\text{CDB}=180^\circ$
$\Rightarrow90^\circ+30^\circ+\angle\text{CDB}=180^\circ$
$\Rightarrow\angle\text{CDB}=180^\circ-120^\circ=60^\circ$
$\Rightarrow\angle\text{BEC}=60^\circ$
$\therefore\angle\text{BCD}=30^\circ$,
in $\triangle\text{CBD}$ by angle sum property of a triangle, we have
$\Rightarrow\angle\text{DBC}+\angle\text{BCD}+\angle\text{CDB}=180^\circ$
$\Rightarrow90^\circ+30^\circ+\angle\text{CDB}=180^\circ$
$\Rightarrow\angle\text{CDB}=180^\circ-120^\circ=60^\circ$
$\Rightarrow\angle\text{BEC}=60^\circ$
MCQ 291 Mark
If two adjacent angles of a parallelogram are $(5x – 5)^\circ$ and $(10x + 35)^\circ$, then the ratio of these angles is:
- ✓$1 : 3$
- B$2 : 3$
- C$1 : 4$
- D$1 : 2$
Answer
View full question & answer→Correct option: A.
$1 : 3$
A. $1 : 3$
Solution:
We know that, adjacent angles of a parallelogram are supplementary, i.e., their sum equals $180^{\circ}$
$\therefore(5-5)+(10 x+35)=180^{\circ} $
$\Rightarrow 15 x+30^{\circ} .=180^{\circ} . \Rightarrow 15 x=180^{\circ}$
$\Rightarrow x=10^{\circ}$. Thus, the angles are $(5 \times 10-5)$ and $(10 \times 10+35)$ i.e., $45^{\circ}$ and $135^{\circ}$. Hence, the required ratio is $45^{\circ}: 135^{\circ}$ i.e., $1: 3$.
Solution:
We know that, adjacent angles of a parallelogram are supplementary, i.e., their sum equals $180^{\circ}$
$\therefore(5-5)+(10 x+35)=180^{\circ} $
$\Rightarrow 15 x+30^{\circ} .=180^{\circ} . \Rightarrow 15 x=180^{\circ}$
$\Rightarrow x=10^{\circ}$. Thus, the angles are $(5 \times 10-5)$ and $(10 \times 10+35)$ i.e., $45^{\circ}$ and $135^{\circ}$. Hence, the required ratio is $45^{\circ}: 135^{\circ}$ i.e., $1: 3$.
MCQ 301 Mark
The angle between the two altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is $30^\circ$. The measure of the obtuse angle is.
- A$100^\circ$
- ✓$150^\circ$
- C$105^\circ$
- D$120^\circ$
Answer
View full question & answer→Correct option: B.
$150^\circ$
B. $150^\circ$
Solution:
Let EC and FC be altitudes and $\angle\text{EBC}=30^\circ$
let $\angle\text{EDC}=\text{x}=\angle\text{FBC}$
so, $\angle\text{EDC}=90-\text{x}\ \text{and}\ \angle\text{BCF}=90-\text{x}$
So, by property of the parallelogram,
$\Rightarrow\angle\text{ADC}+\angle\text{DCB}+180^\circ$
$\Rightarrow\angle\text{ADC}+(\angle\text{ECD}+\angle\text{ECF})=180^\circ$
$\Rightarrow\text{x}+90^\circ-\text{x}+30^\circ+90^\circ-\text{x}=180^\circ$
$\Rightarrow-\text{x}=180^\circ-210^\circ=-30^\circ$
$\Rightarrow\text{x}=30^\circ$
Hence, $\angle\text{DCB}=30^\circ+60^\circ+60^\circ=150^\circ$
Solution:
Let EC and FC be altitudes and $\angle\text{EBC}=30^\circ$
let $\angle\text{EDC}=\text{x}=\angle\text{FBC}$
so, $\angle\text{EDC}=90-\text{x}\ \text{and}\ \angle\text{BCF}=90-\text{x}$
So, by property of the parallelogram,
$\Rightarrow\angle\text{ADC}+\angle\text{DCB}+180^\circ$
$\Rightarrow\angle\text{ADC}+(\angle\text{ECD}+\angle\text{ECF})=180^\circ$
$\Rightarrow\text{x}+90^\circ-\text{x}+30^\circ+90^\circ-\text{x}=180^\circ$
$\Rightarrow-\text{x}=180^\circ-210^\circ=-30^\circ$
$\Rightarrow\text{x}=30^\circ$
Hence, $\angle\text{DCB}=30^\circ+60^\circ+60^\circ=150^\circ$
MCQ 311 Mark
What is the maximum exterior angle possible for a regular polygon$?$
- A$60^\circ$
- B$80^\circ$
- ✓$120^\circ$
- D$160^\circ$
Answer
View full question & answer→Correct option: C.
$120^\circ$
Since of increasing size of regular polygon $\Rightarrow $ its exterior angle decreases.
$\therefore$ Mass exterior angle is for a regular triangle (equilateral triangle) and $= 180 -$ interior angle
$= 180^\circ - 60^\circ = 120^\circ $
MCQ 321 Mark
The length and breadth of a rectangle is $4\ cm$ and $2\ cm$ respectively. Find the perimeter of the rectangle.
- A$8\ cm$
- B$16\ cm$
- ✓$12\ cm$
- D$6\ cm$
Answer
View full question & answer→Correct option: C.
$12\ cm$
Given, length of rectangle is $4\ cm$
Breadth of rectangle $= 2\ cm$
By the formula of perimeter of rectangle, we know that;
Perimeter $= 2 ($Length $+$ Breadth$)$
$P = 2(4 + 2)$
$P = 2 × 6$
$P = 12\ cm.$
MCQ 331 Mark
In a parallelogram $PQRS,$ if $\angle\text{P}=60^\circ$, then other three angles are:
- A$45^\circ, 135^\circ, 120^\circ$
- ✓$60^\circ, 120^\circ, 120^\circ$
- C$60^\circ, 135^\circ, 135^\circ$
- D$45^\circ, 135^\circ, 135^\circ$
Answer
View full question & answer→Correct option: B.
$60^\circ, 120^\circ, 120^\circ$
B. $ 60^\circ, 120^\circ, 120^\circ$
Solution:
Given,$\angle\text{P}=60^\circ$ Since, in a parallelogram, adjacent angles are supplementary,
$\Rightarrow\angle\text{P}+\angle\text{Q}=180^\circ$
$\Rightarrow60^\circ+\angle\text{Q}=180^\circ$
$\Rightarrow\angle\text{Q}=120^\circ$
Also, opposite angles are equal in a parallelogram Therefore, $\angle\text{R}=\angle\text{P}=60^\circ,\angle\text{S}=\angle\text{Q}=120^\circ$
Hence, other three angles are $60^\circ, 120^\circ, 120^\circ.$
Solution:
Given,$\angle\text{P}=60^\circ$ Since, in a parallelogram, adjacent angles are supplementary,
$\Rightarrow\angle\text{P}+\angle\text{Q}=180^\circ$
$\Rightarrow60^\circ+\angle\text{Q}=180^\circ$
$\Rightarrow\angle\text{Q}=120^\circ$
Also, opposite angles are equal in a parallelogram Therefore, $\angle\text{R}=\angle\text{P}=60^\circ,\angle\text{S}=\angle\text{Q}=120^\circ$
Hence, other three angles are $60^\circ, 120^\circ, 120^\circ.$
MCQ 341 Mark
What is the number of vertices of a quadrilateral?
- A$1$
- B$2$
- C$3$
- ✓$4$
Answer
View full question & answer→Correct option: D.
$4$
$4$
MCQ 351 Mark
A diagonal of a rectangle is inclined to one side of the rectangle at $25^\circ .$ The acute angle between the diagonals is:
- A$25^\circ$
- ✓$50^\circ$
- C$40^\circ$
- D$55^\circ$
Answer
View full question & answer→Correct option: B.
$50^\circ$
$50^\circ$
MCQ 361 Mark
Tick the correct answer in the following? In a regular polygon, each interior angle is thrice the exterior angle. The number os sides of the polygon is:
- A$6$
- ✓$8$
- C$10$
- D$12$
Answer
View full question & answer→Correct option: B.
$8$
For a regular polygon with $n$ sides:
Each exterior angle $=\frac{ 360}{ \text{n}}$
Each interior angle $=180-\frac{360}{\text{n}}$
$\therefore180-\frac{360}{\text{n}}=3\Big(\frac{360}{\text{n}}\Big)$
$\Rightarrow180=4\Big(\frac{360}{\text{n}}\Big)$
$\Rightarrow\text{n}=\frac{4\times360}{180}=8$
MCQ 371 Mark
The sum of the measures of the three angles of a triangle is ________.
- A$360^\circ$
- B$210^\circ$
- ✓$180^\circ$
- DNone of these
Answer
View full question & answer→Correct option: C.
$180^\circ$
$180^\circ$
MCQ 381 Mark
Which of the following is not true for an exterior angle of a regular polygon with $n$ sides?
- AEach exterior angle$=\frac{360^\circ}{\text{n}}$
- BExterior angle $=180^{\circ}$ – interior angle
- C$n =\frac{360^\circ}{\text{exterior angle}}$
- ✓Each exterior angle $= \frac{(\text{n}-2)\times180^\circ}{\text{n}}$
Answer
View full question & answer→Correct option: D.
Each exterior angle $= \frac{(\text{n}-2)\times180^\circ}{\text{n}}$
Each exterior angle $= \frac{(\text{n}-2)\times180^\circ}{\text{n}}$
MCQ 391 Mark
If the two diagonals of a rhombus are $8\ cm\ \& \ 6\ cm,$ its area is?
- A$28\ cm^2$
- B$48\ cm^2$
- C$14\ cm^2$
- ✓$24\ cm^2$
Answer
View full question & answer→Correct option: D.
$24\ cm^2$
D. $24\ cm^2$
MCQ 401 Mark
A parallelogram which has equal diagonals is a:
- ASquare
- BRhombus
- ✓Rectangle
- DNone
Answer
View full question & answer→Correct option: C.
Rectangle
Rectangle
MCQ 411 Mark
The lengths of the diagonals of a rhombus are $16\ cm$ and $12\ cm$. The length of each side of the rhombus is:
- A$8\ cm$
- B$9\ cm$
- ✓$10\ cm$
- D$12\ cm$
Answer
View full question & answer→Correct option: C.
$10\ cm$
$\text{AO}=\frac{1}{2}\text{AC}=\Big(\frac{1}{2}\times16\Big)=8\text{cm}$
$\text{BO}=\frac{1}{2}\text{BD}=\Big(\frac{1}{2}\times12\Big)=6\text{cm}$
From the right $\triangle\text{AOB},$ we have,
$\therefore\text{AB}^2=\text{AO}^2+\text{BO}^2$
$\Rightarrow\text{AB}^2=\big\{(8)^2+(6)^2\big\}\text{cm}^2$
$\Rightarrow\text{AB}=\sqrt{100}=10\text{cm}$
$\text{BO}=\frac{1}{2}\text{BD}=\Big(\frac{1}{2}\times12\Big)=6\text{cm}$
From the right $\triangle\text{AOB},$ we have,
$\therefore\text{AB}^2=\text{AO}^2+\text{BO}^2$
$\Rightarrow\text{AB}^2=\big\{(8)^2+(6)^2\big\}\text{cm}^2$
$\Rightarrow\text{AB}=\sqrt{100}=10\text{cm}$
MCQ 421 Mark
The number of sides of a regular polygon, whose each exterior angle has a measure of $45^\circ $, is:
- A$4$
- B$6$
- ✓$8$
- D$10$
Answer
View full question & answer→Correct option: C.
$8$
$8$
MCQ 431 Mark
The two diagonals are not necessarily equal in a:
- ARectangle
- BSquare.
- ✓Rhombus.
- DIsosceles trapezium.
Answer
View full question & answer→Correct option: C.
Rhombus.
All sides of Rhombus are equal in length but in case of angle it is not necessary to be equal.
If all the angles are equal then it will become a square.
That’s why diagonals of rhombus are not necessary to be equal in length.
If all the angles are equal then it will become a square.
That’s why diagonals of rhombus are not necessary to be equal in length.
MCQ 441 Mark
If an angle of a parallelogram is two-thirds of its adjacent angle, the smallest angle of the parallelogram is:
- A$54^\circ$
- ✓$72^\circ $
- C$81^\circ $
- D$108^\circ $
Answer
View full question & answer→Correct option: B.
$72^\circ $
Let the measure of the angle be $x^\circ .$
$\therefore\text{x}+\Big(\frac{2}{3}\times\text{x}\Big)=180$
$\Rightarrow\frac{3\text{x}+2\text{x}}{3}=180$
$\Rightarrow5\text{x}=3\times180$
$\Rightarrow\text{x}=\frac{3\times180}{5}=180$
Hence the anlge is $180^\circ .$
Its adjacent $= (180 - 108^\circ ) = 72^\circ .$
Therefore, the smallest angle is $72^\circ .$
$\therefore\text{x}+\Big(\frac{2}{3}\times\text{x}\Big)=180$
$\Rightarrow\frac{3\text{x}+2\text{x}}{3}=180$
$\Rightarrow5\text{x}=3\times180$
$\Rightarrow\text{x}=\frac{3\times180}{5}=180$
Hence the anlge is $180^\circ .$
Its adjacent $= (180 - 108^\circ ) = 72^\circ .$
Therefore, the smallest angle is $72^\circ .$
MCQ 451 Mark
What is the name of a regular polygon of $6$ sides?
- ASquare
- BEquilateral triangle
- ✓Regular hexagon
- DRegular octagon
Answer
View full question & answer→Correct option: C.
Regular hexagon
Regular hexagon
MCQ 461 Mark
Two adjacent angles of a parallelogram are of equal measure. The measure of each angle of the parallelogram is:
- A$45^\circ $
- B$30^\circ $
- C$60^\circ $
- ✓$90^\circ $
Answer
View full question & answer→Correct option: D.
$90^\circ $
$x^\circ + x^\circ = 180^\circ$
$\Rightarrow x^\circ = 90^\circ .$
$\Rightarrow x^\circ = 90^\circ .$
MCQ 471 Mark
The quadrilateral whose diagonals are perpendicular to each other is:
- ATrapezium
- ✓Rhombus
- CParallelogram
- DRectangle
Answer
View full question & answer→Correct option: B.
Rhombus
Rhombus
MCQ 481 Mark
The length and breadth of a rectangle are in the ratio $4 : 3.$ If the diagonal measures $25\ cm$ then the perimeter of the rectangle is:
- A$56\ cm$
- B$60\ cm$
- ✓$70\ cm$
- D$80\ cm$
Answer
View full question & answer→Correct option: C.
$70\ cm$
C. $70\ cm$
Solution:
Let the length $A B$ be $4 x$ and Breadth $B C$ be $3 x$.
Each angle of a rectangle is a right angle. We have,
$\therefore \angle ABC =90^{\circ}$
From the right $\triangle ABC$ :
$A C 2=A B 2+B C 2$
$\Rightarrow(25)^2=(4 x)^2+(3 x)^2$
$\Rightarrow 16 x^2+9 x^2=625$
$\Rightarrow 25 x^2=625$
$\Rightarrow x^2=25$
$\Rightarrow x=5$
Therefore, lenght $=4 \times 5=2\ cm$ and breadth $=3 \times 5=15\ cm$.
Solution:
Let the length $A B$ be $4 x$ and Breadth $B C$ be $3 x$.
Each angle of a rectangle is a right angle. We have,
$\therefore \angle ABC =90^{\circ}$
From the right $\triangle ABC$ :
$A C 2=A B 2+B C 2$
$\Rightarrow(25)^2=(4 x)^2+(3 x)^2$
$\Rightarrow 16 x^2+9 x^2=625$
$\Rightarrow 25 x^2=625$
$\Rightarrow x^2=25$
$\Rightarrow x=5$
Therefore, lenght $=4 \times 5=2\ cm$ and breadth $=3 \times 5=15\ cm$.
MCQ 491 Mark
If the two adjacent angles of a parallelogram are equal, then each of its angle is?
- A$70^\circ $
- B$80^\circ$
- ✓$90^\circ$
- D$100^\circ$
Answer
View full question & answer→Correct option: C.
$90^\circ$
$90^\circ $
MCQ 501 Mark
What is the number of sides of a quadrilateral?
- A$1$
- B$2$
- C$3$
- ✓$4$
Answer
View full question & answer→Correct option: D.
$4$
$4$