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Maths 1 Marks Question

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

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Question 11 Mark
All parallelograms having equal areas have same perimeters. Observe all the four triangles $FAB, EAB, DAB$ and $CAB$ as shown in Fig. :
Answer
It is not necessary that all parallelograms having equal areas have same perimeters as their base and height may be different.
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Question 21 Mark
Area of the square $MNOP$ of is $144\ cm^2$. Area of each triangle is ___________. 
Answer
Given, area of square $MNOP = 144\ cm^2$ 
Since, there are $8$ identical triangles in the given square $MNOP$
Hence, area of each triangle $=\frac{1}{8}\times\text{Area of square MNOP}$
$=\frac{1}{8}\times144=18\text{cm}^2$
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Question 31 Mark
$1m^2$ = ______ $mm^2$.
Answer
$1 \mathrm{~m}^2=\underline{100} \mathrm{~mm}^2$.
Solution:
We know that, $1 \mathrm{~cm}=10 \mathrm{~mm}$
$\therefore 1 \mathrm{~cm}^2=(10)^2 \mathrm{~mm}^2=100 \mathrm{~mm}^2$
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Question 41 Mark
In Fig. ratio of the area of triangle $ABC$ to the area of triangle $ACD$ is the same as the ratio of base $BC$ of triangle $ABC$ to the base $CD$ of triangle $ACD. $
Answer
$\because\text{Area of } \triangle\text{ABC}\text{ of }\triangle\text{ACD}=\frac{1}{2}\times\text{BC}\times\text{AC}\frac{1}{2}\times\text{CD}\times\text{AC}=\text{BC}:\text{CD}$
$[\because$ area of tringle $=$ base $\times $ height$]$
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Question 51 Mark
Triangles having the same base have equal area.
Answer
False.
Solution:
$\because$ Area of triangle $= \frac{1}{2}$ × Base × Height
So, area of triangle does not only depend on base, it also depends on height. Hence, if triangles have equal base and equal height, then only their areas are equal.
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Question 61 Mark
Find the area of the shaded portion in question $92. $
Answer
Area of shaded portion $=$ Area of circle $-$ Area of square
$=\pi\text{r}^2-98$ $=\frac{22}{7}\times 7\times7-98=154-98\text{cm}^2$
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Question 71 Mark
If area of a triangular piece of cardboard is $90\ cm^2$, then the length of altitude corresponding to $20\ cm$ long base is ___ $cm.$
Answer
$\text{Area of tringale}=\frac{1}{2}\times\text{Base}\times\text{Height}$
$\Rightarrow90=\frac{1}{2}\times20\times\text{h}\Rightarrow\text{h}=9\text{cm}$
$\therefore\text{Altitude or height}=9\text{cm.}$
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Question 81 Mark
Perimeter of a regular polygon = length of one side × ___________.
Answer
Perimeter of regular polygon = length of one side × Number of sides.
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Question 91 Mark
All parallelograms having equal areas have same perimeters. Observe all the four triangles $FAB, EAB, DAB$ and $CAB$ as shown in Fig. :
All triangles have the same base and the same altitude.
Answer
It is clear from the figure that all triangles have same base $AB$ and all the vertices lie on the same line, so the distance between vertex and base of triangle (i.e. length of altitude) are equal.
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Question 101 Mark
In Fig.
$a.$ Area of $(i)$ is the same as the area of $(ii).$
$b.$ Perimeter of $(ii)$ is the same as $(i).$
$c.$ If $(ii)$ is divided into squares of unit length, then its area is $13$ unit squares.
$d.$ Perimeter of $(ii)$ is $18$ units.
Answer
$a.$ True.
Area of both figures is same, because in both number of blocks are same.
$b.$ False.
Because $2$ new sides are added in $(ii). $ So, the perimeter of $(ii)$ is greater than $(i).$
$c.$ False.
$\therefore$ Area of $1$ square $= 1 \times 1 = 1$ unit squares
$\because$ Number of squares $= 12$ So, total area $= 12 \times 1 = 12$ unit squares
$d.$ True.
$\because$ Perimeter is the sum of all sides. So, it is $18$ units.

 

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Question 111 Mark
$1 \mathrm{~km}^2=$ _________$\mathrm{m}^2$.
Answer
$1 \mathrm{~km}^2=\underline{1000000} \mathrm{~m}^2$.
Solution:
We know that, $1 \mathrm{~km}=1000$
$\therefore 1 \mathrm{~km}^2=(1000)^2 \mathrm{~m}^2=1000000 \mathrm{~m}^2$
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Question 121 Mark
Circumference $‘C’$ of a circle is equal to $2 \cdot \pi \times $ _______ .
Answer
Circumference $= 2\pi \times r$
Hence, $r$ is the answer.
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Question 131 Mark
All parallelograms having equal areas have same perimeters. Observe all the four triangles $FAB, EAB, DAB$ and $CAB$ as shown in Fig. :
All triangles are congruent.
Answer
It is clear from the figure that all triangles have only base line is equal and no such other lines are equal to each other.
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Question 141 Mark
Out of two figures if one has larger area, then its perimeter need not to be larger than the other figure.
Answer
True.

Solution:

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Question 151 Mark
$5$ hectare $=500 \mathrm{~m}^2$
Answer
As we know that, $1$ hectare $=10000 \mathrm{~m}^2$
So, $5$ hectare $=5 \times 10000 \mathrm{~m}^2=50000 \mathrm{~m}^2$
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Question 161 Mark
Area of a triangle $ =\frac{1}{2}$ base × ______ .
Answer
Area of a triangle $ =\frac{1}{2}$ base × Height .
Solution:
Area of triangle $ = \frac{1}{2}$ × Base × Height.
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Question 171 Mark
If perimeter of two parallelograms are equal, then their areas are also equal.
Answer
False. Solution: Their corresponding sides and height may be different. So, area cannot be equal.
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Question 181 Mark
Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding _____________of the base.
Answer
Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding height/ altitude of the base.
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Question 191 Mark
Area of a square of side $6\ m$ is equal to the area of _______ squares of each side $1\ cm.$
Answer
Let number of squares having side $1 \mathrm{~cm}=\mathrm{x}$
According to the question,
Area of side $6\ m$ square $=$ Area of side $1\ cm$ square.
 $\left[\therefore\right.$ area of square $\left.=(\text { side })^2\right]$
$\therefore(6 \mathrm{~m})^2=\times x(1 \mathrm{~cm})^2[\therefore 1 \mathrm{~m}=100 \mathrm{~cm}] $
$\Rightarrow(600 \mathrm{~cm})^2=x \times(1 \mathrm{~cm})^2$
$\Rightarrow 360000 \mathrm{~cm}^2=x \mathrm{~cm}^2$
$\Rightarrow x=360000$
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Question 201 Mark
Ratio of the circumference of a circle to its diameter is denoted by symbol ____________.
Answer
$\because\ \text{Circumference}=2\pi\text{r}$
$\Rightarrow\text{C}=\pi\text{d}$
$\big[\because\text{d}=2\text{r}\big]$
$\Rightarrow\frac{\text{C}}{\text{d}}=\pi$
$\Rightarrow\pi=\text{C}:\text{d}$
Hence, $\pi$ is the answer.
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Question 211 Mark
A circle with radius 16cm is cut into four equal parts and rearranged to form another shape as shown in Fig.  Does the perimeter change? If it does change, by how much does it increase or decrease?
Answer
Yes, the perimeter changes. The perimeter is increased by $2r = 2 \times 16= 32\ cm.$
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Question 241 Mark
The distance around a circle is its ___________.
Answer
The distance around a circle is its circumference.Solution:
The distance around a circle is its circumference. In case of circle, perimeter is known as circumference.
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Question 251 Mark
To find area, any side of a parallelogram can be chosen as ___________ of the parallelogram.
Answer
To find area, any side of a parallelogram can be chosen as base of the parallelogram.Solution:
While calculating the area of the parallelogram, we can choose any side as base.
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Question 261 Mark
In Fig. perimeter of $(ii)$ is greater than that of $(i),$ but its area is smaller than that of $(i). $
Answer
Perimeter is the sum of sides of any polygon and area is space that the polygon required. So, by observing the figures we can say that, perimeter of $(ii)$ is greater than $(i)$ and area is less than that of $(i).$
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Question 271 Mark
An increase in perimeter of a figure always increases the area of the figure.
Answer
Perimeter is the sum of sides of any polygon and area is space that the polygon required. So, by observing the figures we can say that, perimeter of $(ii)$ is greater than $(i)$ and area is less than that of $(i).$
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Question 281 Mark
$1m^2$ = ______ $cm^2$.
Answer
$1 \mathrm{~m}^2= \underline{10000} \mathrm{~cm}^2$
Solution:
We know that, $1 \mathrm{~m}=100 \mathrm{~cm}$
$\therefore 1 \mathrm{~m}^2=(100)^2 \mathrm{~cm}^2$
$\Rightarrow 1 \mathrm{~m}^2=10000 \mathrm{~cm}^2$
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Question 291 Mark
All parallelograms having equal areas have same perimeters. Observe all the four triangles $FAB, EAB, DAB$ and $CAB$ as shown in Fig. :
All triangles may not have the same perimeter.
Answer
It is clear from the figure that all triangles may not have the same perimeter.
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Question 301 Mark
All congruent triangles are equal in area.
Answer
True. Solution:Congruent triangles have equal shape and size. Hence, their areas are also equal.
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Question 311 Mark
If a wire in the shape of a square is rebent into a rectangle, then the ___________ of both shapes remain ___________ same, but may varry.
Answer
If a wire in the shape of a square is rebent into a rectangle, then the perimeter of both shapes remain area same, but may varry.Solution:
When we change the shape, then the perimeter remains same as the length of wire is fixed, but area changes as shape changes.
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Question 321 Mark
Ratio of circumference of a circle to its radius is always $2\pi:1.$
Answer
True. Solution: $\because$ Circumference : Radius $= 2\pi\text{r}: \text{r} = 2\pi : 1$
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Question 331 Mark
All parallelograms having equal areas have same perimeters. Observe all the four triangles $FAB, EAB, DAB$ and $CAB$ as shown in Fig. :
All triangles are equal in area.
Answer
Because the triangles on same base and between same parallel lines have equal in area.
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Question 341 Mark
Circumference $‘C’$ of a circle can be found by multiplying diameter $‘d’$ with _________ .
Answer
$\therefore$ Circumference $= 2\pi r$
Since, diameter $(d) = 2r$
So, $C = \pi \times d$
Hence, $\pi $ is the answer.
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