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Maths M.C.Q. [1 Marks Each]

Perimeter and Area · Jharkhand Board (JAC) STD 7 (Jharkhand - English Medium). 34 questions.

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M.C.Q. [1 Marks Each]

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34 questions · auto-graded multiple-choice test.

MCQ 11 Mark
Area of shaded portion in Fig. is:
  • A
    $25 \mathrm{~cm}^2$
  • B
    $15 \mathrm{~cm}^2$
  • C
    $14 \mathrm{~cm}^2$
  • ✓
    $10 \mathrm{~cm}^2$
Answer
Correct option: D.
$10 \mathrm{~cm}^2$

From the given,
Length of rectangle $(l) = 5\ cm$ and breadth of rectangle $(b) = 3 + 1 = 4\ cm$
$\therefore$ Area of shaded portion $=\frac{1}{2}\times\text{Area of rectangle}$
$=\frac{1}{2}\times(\text{l}\times\text{b})$
$=\frac{1}{2}\times5\times4=10\text{cm}^2$

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MCQ 21 Mark
Area of a rectangle and the area of a circle are equal. If the dimensions of the rectangle are $14\ cm \times 11\ cm$, then radius of the circle is:
  • A
    $21\ cm$
  • B
    $10.5\ cm$
  • C
    $14\ cm$
  • ✓
    $7\ cm$
Answer
Correct option: D.
$7\ cm$
Given, dimensions of rectangle, $l = 14\ cm$ and $b = 11\ cm$
According to the question,
Are of rectangle = Area of circle
$\Rightarrow\text{l}\times\text{b}=\pi\text{r}^2$
$\Rightarrow14\times11=\frac{22}{7}\times\text{r}^2$
$\Rightarrow\text{r}^2=\frac{14\ \times\ 11\ \times\ 7}{22}$ $\Big[\because\pi=\frac{22}{7}\Big]$
$\Rightarrow\text{r}^2=49$
$\Rightarrow\text{r}=\sqrt{49}$
$\Rightarrow\text{r}=7\text{cm}$
Hence, the redius of circle is $7\ cm.$
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MCQ 31 Mark
Observe the shapes $1, 2, 3$ and $4$ in the figures. Which of the following statements is not correct?
  • ✓
    Shapes $1, 3$ and $4$ have different areas and different perimeters.
  • B
    Shapes $1$ and $4$ have the same area as well as the same perimeter.
  • C
    Shapes $1, 2$ and $4$ have the same area.
  • D
    Shapes $1, 3$ and $4$ have the same perimeter.
Answer
Correct option: A.
Shapes $1, 3$ and $4$ have different areas and different perimeters.
Shape 1:

Perimeter $= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
Perimeter $= 22$ units
Area $= 18 \times 1$
Area $= 18$sq. units
Shape 2:

Perimeter $= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
Perimeter $= 18$ units
Area $= 18 \times 1$
Area $= 18$sq. units
Shape 3:

Perimeter $= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
Perimeter $= 22$ units
Area $= 16 \times 1$
Area $= 18$sq. units
Shape 4:

Perimeter $= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
Perimeter $= 22$ units
Area $= 18 \times 1$
Area $= 18$sq. units
So, only option $(a)$ is false.
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MCQ 41 Mark
If p squares of each side $1mm$ makes a square of side $1\ cm$, then $p$ is equal to:
  • A
    $10$
  • ✓
    $100$
  • C
    $1000$
  • D
    $10000$
Answer
Correct option: B.
$100$

$\because$ Area of 1 square of side $1 \mathrm{~mm}=1 \times 1 \mathrm{~mm}^2=1 \mathrm{~mm}^2\left[\because\right.$ area of square $\left.=(\text { side })^2\right]$
Area of square of side $1 \mathrm{~cm}=1 \times 1 \mathrm{~cm}=1 \mathrm{~cm}^2$
According to the question,
Area of squares of side $1 \mathrm{~mm}=$ Area of square side $1\ cm$
$\Rightarrow p \times 1 \mathrm{~mm}^2=1 \mathrm{~cm}^2$
$\Rightarrow \mathrm{pmm}^2=(10 \mathrm{~mm})^2[\because 1 \mathrm{~cm}=10 \mathrm{~mm}]$
$\Rightarrow \mathrm{pmm}^2=10 \mathrm{~mm}^2$
So, $p=100$

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MCQ 51 Mark
$12m^2$ is the area of:
  • A
    A square with side $12m.$
  • ✓
    $12$ squares with side $1m$ each.
  • C
    $3$ squares with side $4m$ each.
  • D
    $4$ squares with side $3m$ each.
Answer
Correct option: B.
$12$ squares with side $1m$ each.
For option $(a)$, Area of a square with side $12 \mathrm{~cm}=12 \times 12=144 \mathrm{~m}^2\left[\because\right.$ area of square $=\left(\right.$ side $\left.\left.^2\right)\right]$
For option $(b)$, Area of $12$ square with side $1\ m$ each $=12 \times 1 \times 1=12 \mathrm{~m}^2$
For option $(c)$, Area of $3$ squares with side $4\ m$ each $=3 \times$ Area of square of side $4\ m$
$=3 \times 4 \times 4=48 \mathrm{~m}^2$
For option $(d)$, Area of $4$ square with side $3 \mathrm{~m}=4 \times$ Area of square of side $3\ m$
$=4 \times 3 \times 3=36 \mathrm{~m}^2$
Hence, option $(b)$ is correct.
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MCQ 61 Mark
Ratio of area of $\triangle\text{MNO},\ \triangle\text{MOP},\text{ and }\triangle\text{MPQ}$ in is:
  • ✓
    $2 : 1 : 3$
  • B
    $1 : 3 : 2$
  • C
    $2 : 3 : 1$
  • D
    $1 : 2 : 3$
Answer
Correct option: A.
$2 : 1 : 3$
$ \text{Area of }\triangle\text{MNO}=\frac{1}{2}\times\text{NO}\times\text{MO}$
$=\frac{1}{2}\times4\times5=10\text{cm}^2$ $\big[\because\text{area of triangle}=\frac{1}{2}\times\text{base}\times\text{height}\big]$
$ \text{Area of }\triangle\text{MOP}=\frac{1}{2}\times\text{OP}\times\text{MO}$
$=\frac{1}{2}\times2\times5=5\text{cm}^2$
$ \text{Area of }\triangle\text{MPQ}=\frac{1}{2}\times\text{PQ}\times\text{MO}$
$=\frac{1}{2}\times6\times5=15\text{cm}^2$
Hence, required ratio $= 10 : 5 : 15 = 2 : 1 : 3$
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MCQ 71 Mark
In $\triangle\text{MNO}$ is a right-angled triangle. Its legs are $6\ cm$ and $8\ cm$ long. Length of perpendicular $NP$ on the side $MO$ is:
  • ✓
    $4.8\ cm$
  • B
    $3.6\ cm$
  • C
    $2.4\ cm$
  • D
    $1.2\ cm$
Answer
Correct option: A.
$4.8\ cm$
Given, $\triangle\text{MNO}$ is a right angled triangle.
So, according to Pythagoras theorem,
$\text{MO}^2=\text{MN}^2+\text{NO}^2$
$\Rightarrow\text{MO}^2=6^2+8^2$
$\Rightarrow\text{MO}^2=36+64$
$\Rightarrow\text{MO}^2=100$
$\Rightarrow\text{MO}=\sqrt{100}$
$\Rightarrow\text{MO}=10\text{cm}$
$\therefore\text{ Area of }\triangle\text{MNO}=\frac{1}{2}\times\text{Base}\times\text{Height}$
$\Rightarrow\frac{1}{2}\times\text{MN}\times\text{NO}=\frac{1}{2}\times\text{MO}\times\text{NP}$
$\Rightarrow\frac{1}{2}\times6\times8=\frac{1}{2}\times10\times\text{NP}$
$\Rightarrow\text{NP}=\frac{24}{5}$
$\Rightarrow\text{NP}=4.8\text{cm}$
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MCQ 81 Mark
Area of a right-angled triangle is $30\ cm^2$. If its smallest side is $5\ cm$, then its hypotenuse is:
  • A
    $14\ cm$
  • ✓
    $13\ cm$
  • C
    $12\ cm$
  • D
    $11\ cm$
Answer
Correct option: B.
$13\ cm$
Given, area of a right angled triangle = $30cm^2$
and smallest side i.e., base $= 5\ cm$
We know that,
Area of right angled triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$
$\therefore30=\frac{1}{2}\times5\times\text{Height}$
$\Rightarrow\text{Height}=\frac{30\ \times\ 2}{5}$
$\Rightarrow\text{Height}=12\text{cm}$
Now, according to Pythagoras theorem,
$\text{(Hypotenuse)}^2=\text{(Perpendicular)}^2+\text{(Base)}^2$
$\Rightarrow\text{(Hypotenuse)}^2=(12)^2+(5)^2$ $\big[\because\text{height}=\text{perpendicular}\big]$
$\Rightarrow\text{(Hypotenuse)}^2=144+25$
$\Rightarrow\text{(Hypotenuse)}^2=169$
$\Rightarrow\text{Hypotenuse}=\sqrt{169}$
$\Rightarrow\text{Hypotenuse}=13\text{cm}$
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MCQ 91 Mark
Ratio of area of $\triangle\text{MNO}$ to the area of parallelogram $MNOP$ in the same is:
  • A
    $2 : 3$
  • B
    $1 : 1$
  • ✓
    $1 : 2$
  • D
    $2 : 1$
Answer
Correct option: C.
$1 : 2$
$\text{Area of }\triangle\text{MNO}:\text{Area of parallelogram MNOP}$
$=\frac{1}{2}\times\text{Base}\times\text{Height}:\text{Base}\times\text{Corresponding Height}$
$=\frac{1}{2}\text{NO}\times\text{OQ}:\text{MP}\times\text{OQ}$
$=\frac{1}{2}\times\text{NO}:\times\text{NO}$ $\big[\because\text{NO}=\text{MP}\big]$
$=1:2$
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MCQ 101 Mark
The circumference of a circle whose area is $81\pi\text{r}^2,$ is:
  • A
    $9\pi\text{r}$
  • ✓
    $18\pi\text{r}$
  • C
    $3\pi\text{r}$
  • D
    $81\pi\text{r}$
Answer
Correct option: B.
$18\pi\text{r}$
Let the radius of circle be $R$
$\therefore$ Area of circle $=\pi\text{R}^2$
$\Rightarrow81\pi\text{r}^2=\pi\text{R}^2$
$\Rightarrow\text{R}=\sqrt{81\text{r}^2}$
$\Rightarrow\text{R}=9\text{r}$
Now, circumference of circle $=2\pi\text{R}=2\pi(9\text{r})=18\pi\text{r}$
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MCQ 111 Mark
If $1 \mathrm{~m}^2=x \mathrm{~mm}^2$, then the value of $x$ is:
  • A
    $1000$
  • B
    $10000$
  • C
    $100000$
  • ✓
    $1000000$
Answer
Correct option: D.
$1000000$
$\text { Given, } 1 \mathrm{~m}^2=\mathrm{x} \mathrm{mm}^2$
$\therefore(1000 \mathrm{~mm})^2=\mathrm{x} \mathrm{mm}^2[\because 1 \mathrm{~m}=1000 \mathrm{~mm}]$
$\Rightarrow 1000000 \mathrm{~mm}^2=\mathrm{xmm}^2$
$\Rightarrow \mathrm{x}=1000000$
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MCQ 121 Mark
The perimeter of the figure $ABCDEFGHIJ$ is:
  • ✓
    $60\ cm$
  • B
    $30\ cm$
  • C
    $40\ cm$
  • D
    $50\ cm$
Answer
Correct option: A.
$60\ cm$
Perimeter = Sum of all sides
So, $AJ + JI + IH + HG + GF + FE + ED + CD + BC + AB$
$= (AJ + IH + GF + BC) + 3 + 5 + 2 + 20 + 4 + 6$
$= DE + 40$
$= 20 + 40$
$= 60\ cm$
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MCQ 131 Mark
If the sides of a parallelogram are increased to twice its original lengths, how much will the perimeter of the new parallelogram?
  • A
    $1.5$ times.
  • ✓
    $2$ times.
  • C
    $3$ times.
  • D
    $4$ times.
Answer
Correct option: B.
$2$ times.
Let the length and breadth of the parallelogram be l and b, respectively.
Then, perimeter $= 2(1 + b)$ $[\because$ perimeter of parallelogram $= 2 \times $ (length + breadth)$]$
If both sides are increased twice, then new length and breadth will be $21$ and $2b$, respectively.
Now, new perimeter $= 2(21 + 2b) = 2 × 2(l + b) = 2$ times of original perimeter.
Hence, the perimeter of parallelogram will be increased $2$ times.
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MCQ 141 Mark
Length of tape required to cover the edges of a semicircular disc of radius $10\ cm$ is:
  • A
    $62.8\ cm$
  • B
    $51.4\ cm$
  • ✓
    $31.4\ cm$
  • D
    $15.7\ cm$
Answer
Correct option: C.
$31.4\ cm$
In order to find the length of tape required to cover the edges of a semi-circular disc, we have to find the perimeter of semi-circle.

From the above figure it is clear that,
Perimeter of semi-circle = Circumference of semi-circle + Diameter
$\therefore$ Circumference of semi-circle $=\frac{2\pi\text{r}}{2}=\pi\times\text{r}=\frac{22}{7}\times10=\frac{270}{7}=31.4\text{cm}$
$\therefore$ Total tape required $= 31.4 + 2 \times 10 = 51.4\ cm$ $\big[\because\text{diameter}=2\times\text{radius}\big]$
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MCQ 151 Mark
A wire is bent to form a square of side $22\ cm$. If the wire is rebent to form a circle, its radius is:
  • A
    $22\ cm$
  • ✓
    $14\ cm$
  • C
    $11\ cm$
  • D
    $7\ cm$
Answer
Correct option: B.
$14\ cm$
Given, side of a square $= 22\ cm$
Perimeter of square and circumference of circle are equal, because the wire has same length.
According to the question,
Perimeter of square = Circumference of circle
$\Rightarrow4\times(\text{Side})=2\times\pi\times\text{r}$
$\Rightarrow4\times22=2\times\frac{22}{7}\times\text{r}$ $\Big[\because\ \pi=\frac{22}{7}\Big]$
$\Rightarrow\text{r}=\frac{4\ \times\ 22\ \times\ 7}{2\ \times\ 22}$
$\Rightarrow\text{r}=14\text{cm}$
Hence, the redius is $14\ cm$
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MCQ 161 Mark
What is the radius of the largest circle that can be cut out of the rectangle measuring $10\ cm$ in length and $8\ cm$ in breadth?
  • ✓
    $4\ cm$
  • B
    $5\ cm$
  • C
    $8\ cm$
  • D
    $10\ cm$
Answer
Correct option: A.
$4\ cm$


From the above it is clear that largest circle will have diameter equals smaller side i.e., $8\ cm$
So, diameter $= 8\ cm$
$\therefore\text{ Radius}=\frac{\text{Diameter}}{2}=4\text{cm}$
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MCQ 171 Mark
A table top is semicircular in shape with diameter $2.8m$. Area of this table top is:
  • ✓
    $3.08 \mathrm{~m}^2$
  • B
    $6.16 \mathrm{~m}^2$
  • C
    $12.32 \mathrm{~m}^2$
  • D
    $24.64 \mathrm{~m}^2$
Answer
Correct option: A.
$3.08 \mathrm{~m}^2$
Given, diameter $= 2.8m$
Now, radius $=\frac{2.8}{2}\text{m}=1.4\text{m}$ $\big[\because\text{radius}=\frac{\text{diameter}}{2}\big]$
$\therefore$ Area of table top = Area of semi-circle
$=\frac{\pi\text{r}^2}{2}=\frac{22}{7}\times\frac{1.4}{2}\times1.4=3.08\text{m}^2$
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MCQ 181 Mark
Area of a circle with diameter $‘m’$ radius $‘n’$ and circumference $‘p’$ is
  • A
    $2\pi\text{n}$
  • B
    $\pi\text{m}^2$
  • C
    $\pi\text{p}^2$
  • ✓
    $\pi\text{n}^2$
Answer
Correct option: D.
$\pi\text{n}^2$
Given, diameter $= m$, radius $= n$ and circumference $= p$
$\therefore$ Area of circle $\pi\text{r}^2=\pi\text{n}^2$
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MCQ 191 Mark
If each side of a rhombus is doubled, how much will its area increase?
  • A
    $1.5$ times.
  • ✓
    $2$ times.
  • C
    $3$ times.
  • D
    $4$ times.
Answer
Correct option: B.
$2$ times.
Let $b$ be the side and $h$ be the height of a rhombus.
$\therefore$ Area of rhombus $= b \times h$ $\big[\because$ area of rhombus = base $x$ corresponding height$]$
If each side of rhombus is doubled, then side of rhombus $= 2b$
Now, area of rhombus $= 2b \times h = 2(b \times h) = 2$ times of original
Hence, the area of rhombus will be increased by $2$ times.
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MCQ 201 Mark
Circumference of a circle disc is $88\ cm$. Its radius is:
  • A
    $8\ cm$
  • B
    $11\ cm$
  • ✓
    $14\ cm$
  • D
    $44\ cm$
Answer
Correct option: C.
$14\ cm$
We know that, circumference $=2\pi\text{r}$
$\Rightarrow88=2\times\frac{22}{7}\times\text{r}$
$\Rightarrow\text{r}=\frac{88\ \times\ 7}{2\ \times\ 22}$ $\big[\because\text{circumference}=88\text{cm},\text{ give}\big]$
$\Rightarrow\text{r}=14\text{cm}$
Hence, the radius is $14\ cm$.
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MCQ 211 Mark
If radius of a circle is increased to twice its original length, how much will the area of the circle increase?
  • A
    $1.4$ times.
  • B
    $2$ times.
  • C
    $3$ times.
  • ✓
    $4$ times.
Answer
Correct option: D.
$4$ times.
Let $r$ be the radius of the circle.
$\therefore$ Area of circle $=\pi^2$
If radius is increased to twice its original length, then radius will be $2r.$
Now, area of new circle $=\pi(2\text{r})^2=4\pi^2=4$ times of original area
Hence, the area of circle will be increased by $4$ times.
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MCQ 221 Mark
The area of a square is $100cm^2$. The circumference (in cm) of the largest circle cut of it is:
  • A
    $5\pi$
  • ✓
    $10\pi$
  • C
    $15\pi$
  • D
    $20\pi$
Answer
Correct option: B.
$10\pi$
Let the side of square be $a\ cm$.

Given, area of square = $100cm^2$
$\therefore$ Area of square = $a^2$
$\Rightarrow\text{a}^2=100\text{cm}^2$ $\big[\because\text{area of square }=(\text{side})^2\big]$
$\Rightarrow\text{a}=\sqrt{100}$
$\Rightarrow\text{a}=10\text{cm}$
Now, for the largest circle in the square, diameter of the circle must be equal to the side of square.
$\therefore$ Diameter = side of a square $= 10\ cm$
$\Rightarrow2\text{r}=10\text{cm}$ $\big[\because\text{diameter}=2\times\text{radius}\big]$
$\Rightarrow\text{r}=5\text{cm}$
$\therefore$ Circumference of the circle $=2\pi\text{r}=2\times\pi\times5=10\pi\text{cm}$
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MCQ 231 Mark
Area of circular garden with diameter $8m$ is:
  • A
    $12.56 \mathrm{~m}^2$
  • B
    $25.12 \mathrm{~m}^2$
  • ✓
    $50.24 \mathrm{~m}^2$
  • D
    $200.96 \mathrm{~m}^2$
Answer
Correct option: C.
$50.24 \mathrm{~m}^2$
Given, diameter $= 8m$
So, radius $=\frac{8}{2}\text{m}=4\text{m}$ $\big[\because\text{ radius}=\frac{\text{diameter}}{2}\big]$
$\therefore$ Area of circle garden $=\pi\text{r}^2=\frac{22}{7}\times4\times4=50.24\text{m}^2$
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MCQ 241 Mark
Area of parallelogram $ABCD$ is not equal to:
  • ✓
    $DE \times DC$
  • B
    $BE \times AD$
  • C
    $BF \times DC$
  • D
    $BE \times BC$
Answer
Correct option: A.
$DE \times DC$
We know that,
Area of parallelogram = Base $\times $ Corresponding Height
So, area of parallelogram $ABCD = AD \times BE = BC \times BE$ $\big[\because\text{AD}=\text{BC}\big]$
or area of parallelogram $ABCD = DC \times BF$
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MCQ 251 Mark
In if $PR = 12\ cm, QR = 6\ cm$ and $PL = 8\ cm$, then $QM$ is:
  • A
    $6\ cm$
  • B
    $9\ cm$
  • ✓
    $4\ cm$
  • D
    $2\ cm$
Answer
Correct option: C.
$4\ cm$
Given that, $PR = 12\ cm, QR = 6\ cm$ and $PL = 8\ cm$
Now, in right angled $APLR$, using Pythagoras theorem,
$\text{(Hypotenuse)}^2=\text{(Perpendicular)}^2+\text{(Base)}^2$
$\Rightarrow\text{PR}^2=\text{PL}^2+\text{LR}^2$
$\Rightarrow\text{LR}^2=\text{PR}^2-\text{PL}^2$
$\Rightarrow\text{LR}^2=(12)^2-(8)^2$
$\Rightarrow\text{LR}^2=144-64=80$
$\Rightarrow\text{LR}=\sqrt{80}=4\sqrt{5}\text{cm}$
$\therefore\ \text{LR}=\text{LQ}+\text{QR}$
$\Rightarrow\text{LQ}=\text{LR}-\text{QR}$
$\Rightarrow\text{LQ}=\big(4\sqrt{5}-6\big)\text{cm}$
Now, $\text{Area of }\triangle\text{PLR},$
$\text{A}_1=\frac{1}{2}\times\text{LR}\times\text{PL}$
$\text{A}_1=\frac{1}{2}\times\big(4\sqrt{5}\big)\times8$
$\text{A}_1=16\sqrt{5}\text{cm}^2$
Again, $\text{Area of }\triangle\text{PLQ},$
$\text{A}_2=\frac{1}{2}\times\text{LQ}\times\text{PL}$
$\text{A}_2=\frac{1}{2}\times\big(4\sqrt{5}-6\big)\times8$
$\text{A}_2=\big(6\sqrt{5}-24\big)\text{cm}^2$
$\therefore\ \text{Area of }\triangle\text{PLR}=\text{Area of }\triangle\text{PLQ}+\text{Area of }\triangle\text{PQR}$
$\Rightarrow16\sqrt{5}=\big(16\sqrt{5}-24\big)+\text{Area of }\triangle\text{PQR}$
$\Rightarrow\text{Area of }\triangle\text{PQR}=24\text{cm}^2$
$\Rightarrow\frac{1}{2}\times\text{PR}\times\text{QM}=24$
$\Rightarrow\frac{1}{2}\times12\times\text{QM}=24$
$\therefore\ \text{QM}=4\text{cm}$
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MCQ 261 Mark
What will be the area of the largest square that can be cut out of a circle of radius $10\ cm$?
  • A
    $100 \mathrm{~cm}^2$
  • ✓
    $200 \mathrm{~cm}^2$
  • C
    $300 \mathrm{~cm}^2$
  • D
    $400 \mathrm{~cm}^2$
Answer
Correct option: B.
$200 \mathrm{~cm}^2$
Given, radius of circle $= 10\ cm$
The largest square that can be cut-out of a circle of radius $10\ cm$ will have its diagonal equal to the diameter of the circle.
Let the side of a square be $x$.
Then, area of the square = $x+x=x^2 \mathrm{~cm}^2$ $\big[\because\ \text{area of square}=(\text{side}^2)\big]$

Now, in right angled $\triangle\text{DAB},$
$(\text{BD})^2=(\text{AD})^2+(\text{AB})^2$ [by Pythagora theorem]
$\therefore(20)^2=\text{x}^2+\text{x}^2$
$\big[\because$ diagonal = diameter and diameter $= 2 \times $ radius $= 2 \times 10 = 20\ cm$$]$
$\Rightarrow2\text{x}^2=400$
$\Rightarrow\text{x}^2=200$
$\therefore(\text{side})^2=200$
Hence, the area of the largest square is $200cm^2$
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MCQ 271 Mark
The area of a semicircle of radius $4r$ is:
  • ✓
    $8\pi\text{r}^2$
  • B
    $4\pi\text{r}^2$
  • C
    $12\pi\text{r}^2$
  • D
    $2\pi\text{r}^2$
Answer
Correct option: A.
$8\pi\text{r}^2$
Given, radius of semi-circle $= 4r$
$\because$ Area of semi-circle $=\frac{1}{2}\times\pi\text{r}^2$
$=\frac{1}{12}\times\pi\times(4\text{r})^2$
$=\frac{16}{2}\pi\text{r}^2$
$=8\pi\text{r}^2$
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MCQ 281 Mark
In reference to a circle the value of $\pi$ is equal to:
  • A
    $\frac{\text{area}}{\text{circumference}}$
  • B
    $\frac{\text{area}}{\text{dinmeter}}$
  • ✓
    $\frac{\text{circumference}}{\text{diameter}}$
  • D
    $\frac{\text{circumference}}{\text{radius}}$
Answer
Correct option: C.
$\frac{\text{circumference}}{\text{diameter}}$
$\frac{\text{circumference}}{\text{diameter}}$
We know that,
Circumference of a circle $=2\pi\text{r}$
Circumference $=\pi\times\text{Diameter}$ $\big[\because\text{diameter}=2\text{r}\big]$
$\Rightarrow\pi=\frac{\text{circumference}}{\text{diameter}}$
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MCQ 291 Mark
Area of triangle $MNO$ of is:
  • A
    $\frac{1}{2}\text{MN}\times\text{NO}$
  • B
    $\frac{1}{2}\text{NO}\times\text{MO}$
  • C
    $\frac{1}{2}\text{MN}\times\text{OQ}$
  • ✓
    $\frac{1}{2}\text{NO}\times\text{OQ}$
Answer
Correct option: D.
$\frac{1}{2}\text{NO}\times\text{OQ}$
$\frac{1}{2}\text{NO}\times\text{OQ}$
We know that,
Area of triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$
$=\frac{1}{2}\text{NO}\times\text{OQ}$
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MCQ 301 Mark
Area of the circle obtained in A wire is bent to form a square of side $22\ cm$. If the wire is rebent to form a circle, its radius is:
  • A
    $196 \mathrm{~cm}^2$
  • B
    $212 \mathrm{~cm}^2$
  • ✓
    $616 \mathrm{~cm}^2$
  • D
    $644 \mathrm{~cm}^2$
Answer
Correct option: C.
$616 \mathrm{~cm}^2$

Area of the circle $=\pi\text{r}^2=\frac{22}{7}\times14\times14=616\text{cm}^2$
$\big[\because\text{ r}=14\text{cm},\text{ find above}\big]$

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MCQ 311 Mark
$36$ unit squares are joined to form a rectangle with the least perimeter. Perimeter of the rectangle is:
  • A
    $12$ units
  • ✓
    $26$ units
  • C
    $24$ units
  • D
    $36$ units
Answer
Correct option: B.
$26$ units
Area of rectangle formed $= 36$ units
We have, $36 = 6 \times 6$
$= 2 \times 3 \times 2 \times 3$
$= 22 \times 32$
$= 4 \times 9$
So, the sides of a rectangle are $4\ cm$ and $9\ cm$
$\therefore$ Perimeter $= 2(l + b)$
$= 2(4 + 9)$
$= 2 \times 13$
$= 26$ units
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MCQ 321 Mark
A rectangular piece of dimensions $3\ cm \times 2\ cm$ was cut from a rectangular sheet of paper of dimensions $6\ cm \times 5\ cm$ (see the figure). Area of remaining sheet of paper is:

 
  • A
    $30 \mathrm{~cm}^2$
  • B
    $36 \mathrm{~cm}^2$
  • ✓
    $24 \mathrm{~cm}^2$
  • D
    $22 \mathrm{~cm}^2$
Answer
Correct option: C.
$24 \mathrm{~cm}^2$

Given dimensions of the bigger rectangle are $6\ cm$ and $5\ cm$
$\therefore$ Area of bigger rectangle $=6 \mathrm{~cm} \times 5 \mathrm{~cm}=30 \mathrm{~cm}^2[\because$ area of rectangle $=$ length $\times$ breadth $]$
Also given, dimensions of the smaller rectangle are $3\ cm$ and $2\ cm$
$\therefore$ Area of smaller rectangle $=3 \mathrm{~cm} \times 2 \mathrm{~cm}=6 \mathrm{~cm}^2$
$\therefore$ Area of remaining sheet of paper $=$ Area of bigger rectangle - Area of smaller rectangle $=(30-6) \mathrm{cm}^2=24 \mathrm{~cm}^2$

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MCQ 331 Mark
Circumference of a circle of diameter $5\ cm$ is:
  • A
    $3.14\ cm$
  • B
    $31.4\ cm$
  • ✓
    $15.7\ cm$
  • D
    $1.57\ cm$
Answer
Correct option: C.
$15.7\ cm$
Given, diameter of a circle $= 5\ cm$
$\therefore\ \text{Radius}=\frac{5}{2}\text{cm}$ $\big[\because\text{radius}=\frac{\text{diameter}}{2}\big]$
Now, Circumference $=2\pi\text{r}$
$=2\times\frac{22}{7}\times\frac{5}{2}=\frac{110}{7}=15.7\text{cm}$
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MCQ 341 Mark
If the radius of a circle is tripled, the area becomes:
  • ✓
    $9$ times.
  • B
    $3$ times.
  • C
    $6$ times.
  • D
    $30$ times.
Answer
Correct option: A.
$9$ times.
Let $r$ be the radius of a circle.
$\therefore$ Area of circle $=\pi\text{r}^2$
If radius is tripled, then new radius will be $3r$
$\therefore$ Area of new circle $=\pi(3\text{r}^2)=9\pi^2=9$ times of original
Hence, the area of a cirlce becomes 9 times to the original area.
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