Case study (4 Marks)

Question 14 Marks

The area of a trapezium is $91cm^2$ and its height is $7\ cm$. If one of the parallel sides is longer than the other by $8\ cm$, find the two parallel sides.

Answer

Area of trapezium = $91cm^2$
Height $(h) = 7cm$
Sum of parallel sides.
$=\frac{\text{Area}\times2}{\text{heght}}=\frac{91\times2}{7}=26\text{cm}$
Let shorter side $= x + 8$
Then longer side $= x + 8$
$\therefore\text{x}+\text{x}+8=26$
$\Rightarrow2\text{x}=26-8=18$
$\Rightarrow\text{x}=\frac{18}{2}=9$
One parallel side $= 9cm$
and second side $= 9 + 8 = 17\ cm$
Hence parallel sides are $17\ cm, 9\ cm$

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Question 24 Marks

A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are $36\ m$ and $24.5m$, find the area of the playground. $\Big(\text{Take }\pi=\frac{22}{7}\Big )$

Answer

Length of rectangular portion $(l) = 36m$
and breath $(b) = 24.5m$

$\therefore$ Area $= l × b = 36 × 24.5m^2 = 882m^2$
Radius of each semicircle (r)
$=\frac{24.5}{2}=12.25\text{m}$
$\therefore$ Area of two semicircles
$=2\times\frac{1}{2}\pi\text{ r}^2=\frac{22}{7}\times(12.25)^2\text{m}^2$
$=\frac{22}{7}\times150.0625\text{m}^2$
$=471.625\text{m}^2$
Total area of the playground = 471.625 + 882 = $1353.625 m^2$

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Question 34 Marks

Find the area of a rhombus whose side is $6\ cm$ and whose altitude is $4\ cm$. If one of its diagonals is $8\ cm$ long, find the length of the other diagonal.

Answer

Side of rhombus $(b) = 6\ cm$
Altitude $(h) = 4\ cm$

$\therefore \text { Area }=b \times h=6 \times 4 cm^2=24 cm^2$
Now one diagonal $\left( d _1\right)=8 cm$
$\therefore$ Second diagonal $\left( d _2\right)=\frac{\text { Area } \times 2}{d_1}$
$=\frac{24 \times 2}{8}=6 cm$

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Question 44 Marks

Find the area of Fig. as the sum of the areas of two trapezium and a rectangle.

Answer

In the figure,
One rectangle is $ABCD$ whose sides are $50\ cm$ and $10\ cm.$
Two trapezium of equal size in which parallel sides are $30\ cm$ and $10\ cm$ and height.
$=\frac{70-50}{2}=\frac{20}{2}=10\text{cm}$
Now area of the figure

= Area of rectangle + area of two trapezium
$= 50×10+2\times\frac{1}{2}(30+10)\times10\text{cm}^2$
$=500+40\times10\text{cm}^2=500+400$
$=900\text{cm}^2$

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Question 54 Marks

Find the area enclosed by the following figures as the sum of the areas of a rectangle and a trapezium:

Answer

In the figure $ABCDEF,$
Join $CF$, then, the figure consists ine square and one trapezium $ABCF$ is a square whose side $= 18cm$

Area of the square $=18 \times 18 cm^2=324 cm^2$
Area of trapezium $FCDE=\frac{1}{2}(CF+ED) \times 8 cm^2$
$=\frac{1}{2}(18+7) \times 8$
$=\frac{1}{1} \times 25 \times 8 cm^2$
$=100 cm^2$
Total area of figure $A B C D E F=324+100=424 cm^2$

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Question 64 Marks

The cost of fencing a square field at $60$ paise per metre is Rs. $1200$. Find the cost of reaping the field at the rate of $50$ paise per $100$ sq. metres.

Answer

Cost of fencing the square field $= Rs. 1200$
Rate $= 60$ pise per m.
$\therefore$ Perimeter $=\frac{1200\times100}{60}=2000\text{m}$
Area of square field $= a^2 = (500)^2$
Rate of reaping the field = 50p per $100m^2$.
$\therefore$ Total cost $=\text{Rs. }\frac{250000\times50}{100\times100}$
$=\text{Rs. }1,250$

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Question 74 Marks

Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is $10500m^2$ and the perpendicular distance between the two parallel sides is $100\ m$, find the length of the side along the river.

Answer

Area of the trapezium shaped field = $10500m^2$
Perpendicular distance between them $(h) = 100m$
$\therefore$ Sum of parallel cides $=\frac{\text{Area}\times2}{\text{h}}$
$=\frac{10500\times2}{100}=210\text{m}$
Let one side $= x$
Then second side $= 2x.$
$\therefore\text{x}+2\text{x}=210$
$\Rightarrow3\text{x}=210$
$\Rightarrow\text{x}=\frac{210}{3}=70$
$\therefore$ Lenght of parallel side along river $= 2x$
$=2\times70=140\text{m}$

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Question 84 Marks

The diagonal of a quadrilateral shaped field is $24\ m$ and the perpendiculars dropped on it from the remaining opposite vertices are $8 \ m$ and $13\ m$. Find the area of the field.

Answer

In quadriateral shaped field $ABCD.$
diagonal $AC = 24m$
and perpendicular $BL = 13m$
and perpendicular $DM$ on $AC = 8m$
Area of the filed $ABCD$ $=\frac{1}{2}\times\text{Ac}\times(\text{BL}+\text{DM})$
$=\frac{1}{2}\times24\times(13+8)\text{m}^2$
$=12\times21=252\text{m}^2$

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Question 94 Marks

It is given that the length of the rectangular piece is $20\ m$ and its width is $15\ m$.And, from each corner a quadrant each of radius $3.5\ m$ has been cut out. A rough figure for this is given below:

Answer

Length of rectangular portion $(l) = 20m$
breadth $(b) = 15m$

Area of the rectangular piece $= l \times b = 20 \times 15 = 300m^2$
Radius of semicircular portion $(r) = 3.5m$
Total area of $4$ quadrants
$=4\times\frac{1}{4}\times\pi\text{r}^2=\frac{22}{7}\times(3.5)^2\text{m}^2$
$=\frac{22}{7}\times12.25\text{m}^2=22\times1.75\text{m}^2=38.5\text{m}^2$

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Question 104 Marks

The floor of a building consists of $3000$ tiles which are rhombus shaped and each of its diagonals are $45\ cm$ and $30\ cm$ in length. Find the total cost of polishing the floor, if the cost per $m^2$ is Rs. $4.$

Answer

Number of rhombus shaped tiles $= 300$
Diagonals of each tile $= 45\ cm$ and $130\ cm$
$\therefore$ Area of one tile $=\frac{\text{d}_1\times\text{d}_2}{2}$
$=\frac{45\times30}{2}=675\text{cm}^2$
$\therefore$ Area of 3000 tiles $= 675\times3000\text{cm}^2$
$=2025000\text{cm}^2$
$=\frac{2025000}{100\times100}=202.5\text{m}^2$
Rate of polishing the tiles = Rs. $4$ per $m^2$
Total cost $= 202.5 \times 4 = Rs. 810$

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MATHS Case study (4 Marks)

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