1 Marks Question

Question 11 Mark

Find the area of the circle, given that: radius $= 14 mm$ (Take $\pi$ = $\frac{22}{7}$)

Answer

$\mathrm{r}=14 \mathrm{~mm}$
$\therefore \text { Area of the circle }=\pi \mathrm{r}^2$
$=\frac{22}{7}(14)^2=\frac{22}{7} \times 14 \times 14$
$=616 \mathrm{~mm}^2$

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Question 21 Mark

Find the circumference of the circle with the radius:$21\ cm$ (Take $\pi$ = $\frac{22}{7}$)

Answer

$r = 21 \ cm$
$\therefore$ Circumference of the circle
$= 2 \pi r = 2 \times \frac{22}{7} \times$21
$= 132 \ cm$

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Question 31 Mark

Find the circumference of the circle with the radius (Take $\pi = \frac{22}{7}): 28 \ mm$

Answer

$r = 28 mm$
$\therefore$ Circumference of the circle
$= 2 \pi r = 2 \times \frac{22}{7} \times 28$
$= 176 \ mm$

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Question 41 Mark

Find the circumference of the circle with the radius: $14\ cm$ (Take $\pi$ = $\frac{22}{7}$)

Answer

$r = 14 cm$
$\therefore$ Circumference of the circle
$= 2 \pi r = 2 \times \frac{22}{7} \times 14$
$= 88 cm$

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Question 51 Mark

Find the area of triangle

Answer

Area of the triangle
$= \frac{1}{2} \times$ (Base $\times$ Height)
$= \frac{1}{2} \times (3 \ cm \times 4\ cm )$
$= 6 \mathrm{~cm}^2$

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Question 61 Mark

Find the area of the triangle.

Answer

Area of the triangle
$= \frac{1}{2} \times$ (Base $\times$ Height)
$= \frac{1}{2} \times (4 cm \times 3cm )$
$= 6 \mathrm{~cm}^2$

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Question 71 Mark

Find the area of the triangle (Fig).

Answer

We know that:
Area of triangle $=\frac{1}{2} \times b \times h=\frac{1}{2} \times \mathrm{MN} \times \mathrm{LO}$
$=\frac{1}{2} \times 3 \mathrm{cm} \times 2 \mathrm{cm}=3 \mathrm{cm}^{2}$

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Question 81 Mark

Find the area of a triangle (Fig).

Answer

We know that:
Area of triangle $=\frac{1}{2}~\times b~\times h=\frac{1}{2} \times \mathrm{QR} \times \mathrm{PS}$
$=\frac{1}{2} \times 4 \mathrm{cm} \times 2 \mathrm{cm}=4 \mathrm{cm}^{2}$

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Question 91 Mark

Find the height $x$, if the area of the parallelogram is $24 \mathrm{~cm}^2$ and the base is $4 \ cm.$

Answer

Area of parallelogram $= b \times h$
$\Rightarrow 24 = 4 \times x $ (Fig)

$\Rightarrow$ $\frac{24}{4}=x$ or $x = 6 \ cm$
So, the height of the parallelogram is $6 \ cm.$

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Question 101 Mark

One of the sides and the corresponding height of a parallelogram are $4 \ cm$ and $3 \ cm$ respectively. Find the area of the parallelogram (Fig).

Answer

Here, we are given that:
Base $(b) = 4 \ cm$
Height $(h) = 3 \ cm$
$\therefore$ Area of the parallelogram $= b \times h = 4 \ cm \times 3 cm = 12 \mathrm{~cm}^2$

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Question 111 Mark

Anu wants to fence the garden in front of her house (Fig), on three sides with lengths $20 \ m, 12 \ m$ and $12 \ m$. Find the cost of fencing at the rate of $₹ 150$ per metre.

Answer

The total length of the fence required $= 20 m + 12 m + 12 m = 44 m$
Cost of fencing $1 m = ₹ 150$
Therefore, Cost of fencing the whole length $= ₹ 150 × 44 = ₹ 6,600.$

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Question 121 Mark

Two cross roads, each of width $5 m$, run at right angles through the centre of a rectangular park of length $70 m$ and breadth $45 m $and parallel to its sides. Find the area of the roads. Also find the cost of constructing the roads at the rate of $₹ 105$ per $m^2$.

Answer

Clearly, area of the cross roads is the area of shaded portion, i.e., the area of the rectangle $PQRS$ and the area of the rectangle $EFGH$ . But while doing this, the area of the square $KLMN $is taken twice, which is to be subtracted once to get the required area.
Now, $P Q=5 \mathrm{~m}$ and $\mathrm{PS}=45 \mathrm{~m}$
$\mathrm{EH}=5 \mathrm{~m} \text { and } \mathrm{EF}=70 \mathrm{~m}$
$\mathrm{KL}=5 \mathrm{~m} \text { and } \mathrm{KN}=5 \mathrm{~m}$
Therefore, area of the path = Area of the rectangle $PQRS +$ area of the rectangle $EFGH$ - Area of the square $KLMN$
$=\mathrm{PS} \times \mathrm{PQ}+\mathrm{EF} \times \mathrm{EH}-\mathrm{KL} \times \mathrm{KN}$
$=(45 \times 5+70 \times 5-5 \times 5) \mathrm{m}^2$
$=(225+350-25) \mathrm{m}^2=550 \mathrm{~m}^2$
Hence, cost of constructing the path $=₹ 105 \times 550=₹ 57,750$

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Question 131 Mark

Diameter of a circular garden is $9.8\ cm$. What is its area?

Answer

$Area = \pi {r^2}$
$ = \frac{{22}}{7} \times {\left( {\frac{{9.8}}{2}} \right)^2}$
$ \Rightarrow 75.46c{m^2}$

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Question 141 Mark

Find the area of a circle of radius $30 \ cm$ (use $\pi = 3.14).$

Answer

Here, we have
Radius, $\mathrm{r}=30 \mathrm{~cm}$
Therefore, Area of the circle $=\pi \mathrm{r}^2=3.14 \times 30^2=2,826 \mathrm{~cm}^2$

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