3 Marks Question

Question 13 Marks

The length of the fence of a trapezium-shaped field $ABCD$ is $130m$ and side $AB$ is perpendicular to each of the parallel sides $AD$ and $BC$. If $BC = 54\ m, CD = 19\ m$ and $AD = 42\ m$, find the area of the field.

Answer

Perimeter of trapezium $ABCD = 130\ m$
$\text{AB}\perp\text{AD}$ and $BC$
$BC = 54\ m$, $CD = 19\ m$ $AD = 42\ m$

$\therefore\text{AB}=130-(54+19+42)$
$=130-115=15\text{m}$
Now area of $ABCD$ $=\frac{\text{AD}+\text{BC}}{2}\times\text{AB}$
$=\frac{(42+54)}{2}\times15\text{m}^2$
$=\frac{96}{2}\times15=48\times15\text{m}^2=720\text{m}^2$

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Question 23 Marks

The shape of the top surface of a table is trapezium. Its parallel sides are $1\ m$ and $1.4\ m$ and the perpendicular distance between them is $0.9\ m$. Find its area.

Answer

Parallel sides of the trapezium $= 1\ m, 1.4\ m$
Perpendicular distance between them $= 0.9\ m$.
$\therefore\text{Area}=\frac{\text{Sum of perallel sides}}{2}\times\text{altitude}$
$=\frac{1.4+1}{2}\times0.9=\frac{2.4}{2}\times0.9\text{m}^2$
$=1.2\times0.9=1.08\text{m}^2$

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Question 33 Marks

Find the area of a trapezium whose parallel sides are $24\ cm$ and $20\ cm$ and the distance between them is $15\ cm$.

Answer

In trapezium $ABCD$. Length of parallel sides $AB = 24\ cm, DC = 20\ cm$ and
distance between them $= 15\ cm$

$\therefore$ Area $=\frac{1}{2}$ (Sum of parallel sides) $\times $ distance between them,
$=\frac{1}{2}(24+20)\times15\text{cm}^2$
$=\frac{1}{2}\times44\times15\text{cm}^2$
$330\text{cm}^2$

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Question 43 Marks

Find the area enclosed by the given figure $ABCDEF$ as per dimensions given herewith.

Answer

In the figure, $ABCF$ is $0$ square and $CDEF$ is a trapezium,
Now area of square $ABCF$,
$= (side)^2= (20)^2= 400cm^2$
Area of trap. $CDEF$
$=\frac{1}{2}(\text{ED}+\text{FC})\times\text{height}$
$=\frac{1}{2}(6+20)\times8=\frac{1}{2}\times26\times8\text{cm}^2$
$=104\text{cm}^2$
$\therefore$ Total area of the figure $ABCDEF$
$= 400 + 104 = 504cm^2$

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Question 53 Marks

The area of a trapezium is $1080cm^2$. If the lengths of its parallel sides be $55\ cm$ and $35\ cm$, find the distance between them.

Answer

Area of trapezium $= 1080cm^2$
Lengths of parallel sides are $l_1= 55cm$ and $l_2= 35cm,$
Let h be the distance between them,
Now area $=\frac{1}{2}\text{h}(\text{l}_1+\text{l}_2)$
$\Rightarrow1080=\frac{1}{2}\text{h}(55+35)$
$\Rightarrow1080=\frac{1}{2}\times90\Rightarrow45\text{h}=1080$
$\Rightarrow\text{h}=\frac{1080}{45}=24$
$\therefore$ Distance between parallel sides,
$= 24\ cm$.

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Question 63 Marks

Find the area of a trapezium whose parallel sides are $38.7\ cm$ and $22.3\ cm$, and the distance between them is $16\ cm$.

Answer

Parallel sides of a trapezium $ABCD$ are $l_1= 38.7cm$ and $l_2= 22.3cm.$

And distance between them $(h) = 16\ cm.$
$\therefore$ Area of trapezium,
$=\frac{1}{2}\times\text{h}$ (Sum of parallel sides)
$=\frac{1}{2}\text{h}(\text{l}_1+\text{l}_2)=\frac{1}{2}\times16(38.7+22.3)\text{cm}^2$
$=\frac{1}{2}\times16\times61=488\text{cm}^2$

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Question 73 Marks

In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is $9450m^2$ and the perpendicular distance between the two parallel sides is $84\ m$, find the length of the longer of the parallel sides.

Answer

Let one of parallel sides $= x$
Then second sides $= 2x$
Area $= 9450m^2$
Distance between them $= 84\ m$

$\therefore\frac{(\text{x}+2\text{x})}{2}\times84=9450$
$\Rightarrow\frac{3\text{x}}{2}\times84=9450\Rightarrow\text{x}=\frac{9450\times2}{3\times84}=75$
$\therefore$ Lenght of longer side $= 2x = 2 \times 75$
$= 150\ m.$

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Question 83 Marks

A field is in the form of a right triangle with hypotenuse $50\ m$ and one side $30\ m$. Find the area of the field.

Answer

Let the other side of the triangular field be $x\ m$.
$\therefore\text{x}^2=\Big\{(50)^2-(30)^2\Big\}$
$\Rightarrow\text{x}^2=(2500-900)$
$\Rightarrow\text{x}^2=1600$
$\Rightarrow\text{x}=\sqrt{1600}$
$\Rightarrow\text{x}=40$
$\therefore$ Area of the field $=\Big(\frac{1}{2}\times30\times40\Big) \text{m}^2$ $=600\text{m}^2$

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Question 93 Marks

Find the area of an equilateral triangle of side $6\ cm$.

Answer

Area of an equilateral traingle $=\Big(\frac{\sqrt{3}}{4}\times(\text{side})^2\Big)$ square units.
$=\Big(\frac{\sqrt{3}}{4}\times6\times6\Big)\text{cm}^2$
$=\Big(\frac{\sqrt{3}}{4}\times360\Big)\text{cm}^2$
$=9\sqrt{9}\text{cm}^2$
Hence, the area of an equilateral triangle is $9\sqrt{3}\text{cm}^2.$

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Question 103 Marks

A field is in the form of a trapezium. Its area is $1586m^2$ and the distance between its parallel sides is $26\ m$. If one of the parallel sides is $84\ m$, find the other.

Answer

Area of trapezium shaped field $= 1586m^2$.
Distance between parallel sides $= 26\ m$,
Sum of the parallel sides $=\frac{\text{Area}\times2}{\text{Altitude}}$
$=\frac{1586\times2}{26}=122\text{m}$
One side $= 84\ m$
Second side $= 122 - 84$
$= 38\ m$

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