RD Sharma solution class 8 chapter 20 Mensuration I(Area of Trapezium and Polygon) Exercise 20.1

Exercise 20.1

Page-20.13

Question 1:

A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²?

Answer 1:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Base} \mathrm{of} \mathrm{a} \mathrm{flooring} \mathrm{tile} \mathrm{that} \mathrm{is} \mathrm{in} \mathrm{the} \mathrm{shape} \mathrm{of} \mathrm{a} \mathrm{parallelogram} = \mathrm{b}=24 \mathrm{cm} \\ \mathrm{Corresponding} \mathrm{height} = \mathrm{h}=10 \mathrm{cm} \\ \mathrm{Now},\mathrm{in} \mathrm{a} \mathrm{parallelogram}: \\ \mathrm{Area}\left(\mathrm{A}\right)= \mathrm{Base} \left(\mathrm{b}\right)\times \mathrm{Height} \left(\mathrm{h}\right) \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{a} \mathrm{tile}= 24 \mathrm{cm}\times 10 \mathrm{cm}=240 {\mathrm{cm}}^2 \\ \mathrm{Now}, \mathrm{observe} \mathrm{that} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{floor} \mathrm{is} 1080 {\mathrm{m}}^2. \\ 1080 {\mathrm{m}}^2=1080\times 1\mathrm{m}\times 1\mathrm{m} \\ =1080\times 100 \mathrm{cm}\times 100 \mathrm{cm} ( \mathrm{Because} 1 \mathrm{m}= 100 \mathrm{cm}) \\ =1080\times 100\times 100\times \mathrm{cm}\times \mathrm{cm} \\ =10800000 {\mathrm{cm}}^2 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{required} \mathrm{tiles}=\frac{10800000}{240}=45000 \\ \mathrm{Hence}, \mathrm{we} \mathrm{need} 45000 \mathrm{tiles} \mathrm{to} \mathrm{cover} \mathrm{the} \mathrm{floor}. \end{gathered}$$

Question 2:

A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. 20.23. If AB = 60 m and BC = 28 m, find the area of the plot.

Answer 2:

$\text{The given figure has a rectangle with a semicircle on one of its sides. }$



$$\begin{gathered} \mathrm{Total} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{plot}=\mathrm{Area} \mathrm{of} \mathrm{rectangle} \mathrm{ABCD} + \mathrm{Area} \mathrm{of} \mathrm{semicircle} \mathrm{with} \mathrm{radius} (\mathrm{r} = \frac{28}{2}=14\mathrm{m}) \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{plot} \mathrm{with} \mathrm{sides} 60\mathrm{m} \mathrm{and} 28\mathrm{m} =60\times 28=1680 {\mathrm{m}}^2 ...\left(\mathrm{i}\right) \\ \mathrm{And}, \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{semicircle} \mathrm{with} \mathrm{radius} 14\mathrm{m} =\frac{1}{2}\mathrm{\pi }\times (14{)}^2=\frac{1}{2}\times \frac{22}{7}\times 14\times 14=308{\mathrm{m}}^2 ...(\mathrm{ii}) \\ \therefore \mathrm{Total} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{plot} =1680+308=1988{\mathrm{m}}^2 ...(\mathrm{from} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right)) \end{gathered}$$

Question 3:

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 24.5 m, find the area of the playground. (Take π = 22/7).

Answer 3:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{playground} \mathrm{is} \mathrm{in} \mathrm{the} \mathrm{shape} \mathrm{of} \mathrm{a} \mathrm{rectangle} \mathrm{with} \mathrm{two} \mathrm{semicircles} \mathrm{on} \mathrm{its} \mathrm{smaller} \mathrm{sides}. \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{portion} \mathrm{is} 36 \mathrm{m} \mathrm{and} \mathrm{its} \mathrm{width} \mathrm{is} 24.5 \mathrm{m} \mathrm{as} \mathrm{shown} \mathrm{in} \mathrm{the} \mathrm{figure} \mathrm{below}. \end{gathered}$$


$$\begin{gathered} \mathrm{Thus}, \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{playground} \mathrm{will} \mathrm{be} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{rectangle} \mathrm{and} \mathrm{the} \mathrm{areas} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{semicircles} \mathrm{with} \mathrm{equal} \mathrm{diameter} 24.5 \mathrm{m}. \\ \mathrm{Now}, \mathrm{area} \mathrm{of} \mathrm{rectangle} \mathrm{with} \mathrm{length} 36\mathrm{m} \mathrm{and} \mathrm{width} 24.5\mathrm{m}: \\ \mathrm{Area} \mathrm{of} \mathrm{rectangle}=\mathrm{length}\times \mathrm{width} \\ =36\mathrm{m}\times 24.5 \mathrm{m}\mathrm{ } \\ =882 {\mathrm{m}}^2 \\ \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{semicircle} = \mathrm{r} = \frac{\mathrm{diameter}}{2}=\frac{24.5}{2}=12.25\mathrm{m} \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{semicircle}=\frac{1}{2}{\mathrm{\pi r}}^2 \\ =\frac{1}{2}\times \frac{22}{7}\times (12.25{)}^2 \\ =235.8 {\mathrm{m}}^2 \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{complete} \mathrm{playground}= \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{ground}+ 2\times \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{semicircle} \\ =882+2\times 235.8 \\ =1353.6 {\mathrm{m}}^2 \end{gathered}$$

Question 4:

A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.

Answer 4:

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



$$\begin{gathered} \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{remaining} \mathrm{part}=\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{piece}-(4\times \mathrm{Area} \mathrm{of} \mathrm{a} \mathrm{quadrant} \mathrm{of} \mathrm{radius} 3.5\mathrm{m}) \\ \mathrm{Now}, \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{piece}=20\times 15=300 {\mathrm{m}}^2 \\ \mathrm{And}, \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{quadrant} \mathrm{with} \mathrm{radius} 3.5 \mathrm{m} \mathrm{=}\frac{1}{4}{\mathrm{\pi r}}^2=\frac{1}{4}\times \frac{22}{7}\times (3.5{)}^2 \\ =\frac{1}{4}\times \frac{22}{7}\times 3.5\times 3.5 \\ =9.625 {\mathrm{m}}^2 \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{remaining} \mathrm{part}=300-(4\times 9.625)=261.5 {\mathrm{m}}^2 \end{gathered}$$

Page-20.14

Question 5:

The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.

Answer 5:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{inside} \mathrm{perimeter} \mathrm{of} \mathrm{the} \mathrm{running} \mathrm{track} \mathrm{is} 400 \mathrm{m}. \mathrm{It} \mathrm{means} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{track} \mathrm{is} 400 \mathrm{m}. \\ \mathrm{Let} \mathrm{r} \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{semicircles}. \\ \mathrm{Observe}: \mathrm{Perimeter} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{track}=\mathrm{Length} \mathrm{of} \mathrm{two} \mathrm{straight} \mathrm{portions} \mathrm{of} 90 \mathrm{m}+\mathrm{Length} \mathrm{of} \mathrm{two} \mathrm{semicircles} \\ \therefore 400=(2\times 90)+(2\times \mathrm{Perimiter} \mathrm{of} \mathrm{a} \mathrm{semicircle}) \\ 400 = 180+(2\times \frac{22}{7}\times \mathrm{r}) \\ 400-180=(\frac{44}{7}\times \mathrm{r}) \\ \frac{44}{7}\times \mathrm{r}=220 \\ \mathrm{r}=\frac{220\times 7}{44}=35 \mathrm{m} \\ \therefore \mathrm{Width} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{track}=2\mathrm{r}=2\times 35=70 \mathrm{m} \\ \mathrm{Since} \mathrm{the} \mathrm{track} \mathrm{is} 14 \mathrm{m} \mathrm{wide} \mathrm{at} \mathrm{all} \mathrm{places}, \mathrm{so} \mathrm{the} \mathrm{width} \mathrm{of} \mathrm{the} \mathrm{outer} \mathrm{track}: 70+(2\times 14)=98 \mathrm{m} \\ \therefore \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{outer} \mathrm{track} \mathrm{semicircles}=\frac{98}{2}=49 \mathrm{m} \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{outer} \mathrm{track}=(\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{portion} \mathrm{with} \mathrm{sides} 90 \mathrm{m} \mathrm{and} 98 \mathrm{m})+ (2\times \mathrm{Area} \mathrm{of} \mathrm{two} \mathrm{semicircles} \mathrm{with} \mathrm{radius} 49 \mathrm{m}) \\ =(98\times 90)+(2\times \frac{1}{2}\times \frac{22}{7}\times {49}^2) \\ =\left(8820\right)+\left(7546\right) \\ =16366 {\mathrm{m}}^2 \\ \mathrm{And}, \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{track}=(\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{portion} \mathrm{with} \mathrm{sides} 90 \mathrm{m} \mathrm{and} 70 \mathrm{m})+ (2\times \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{semicircle} \mathrm{with} \mathrm{radius} 35 \mathrm{m}\text{)} \\ =(70\times 90)+(2\times \frac{1}{2}\times \frac{22}{7}\times {35}^2) \\ =(6300)+(3850) \\ =10150 {\mathrm{m}}^2 \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{running} \mathrm{track}=\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{outer} \mathrm{track}-\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{track} \\ =16366-10150 \\ =6216 {\mathrm{m}}^2 \\ \mathrm{And}, \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{outer} \mathrm{track}=(2\times \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{straight} \mathrm{portion})+(2\times \mathrm{perimeter} \mathrm{of} \mathrm{the} \mathrm{semicircles} \mathrm{with} \mathrm{radius} 49 \mathrm{m}\mathrm{)} \\ =(2\times 90)+(2\times \frac{22}{7}\times 49) \\ =180+308 \\ =488 \mathrm{m} \end{gathered}$$

Question 6:

Find the area of Fig. 20.25, in square cm, correct to one place of decimal. (Take π = 22/7)

Answer 6:

$\text{The given figure is: }$



$$\begin{gathered} \mathrm{Construction} : \mathrm{Connect} \mathrm{A} \mathrm{to} \mathrm{D}. \\ \mathrm{Then}, \mathrm{we} \mathrm{have}: \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{figure} = (\mathrm{Area} \mathrm{of} \mathrm{rectangle} \mathrm{ABCD} + \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{semicircle}) - (\mathrm{Area} \mathrm{of} ∆ \mathrm{AED}). \\ \therefore \mathrm{Total} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{figure}=(\mathrm{Area} \mathrm{of} \mathrm{rectangle} \mathrm{with} \mathrm{sides} 10 \mathrm{cm} \mathrm{and} 10 \mathrm{cm})+(\mathrm{Area} \mathrm{of} \mathrm{semicircle} \mathrm{with} \mathrm{radius}=\frac{10}{2}=5 \mathrm{cm})-(\mathrm{Area} \mathrm{of} \mathrm{triangle} \mathrm{AED} \mathrm{with} \mathrm{base} 6 \mathrm{cm} \mathrm{and} \mathrm{height} 8 \mathrm{cm}\mathrm{)} \\ =(10\times 10)+(\frac{1}{2}\times \frac{22}{7}\times 5^2)-(\frac{1}{2}\times 6\times 8) \\ =100+39.3-24 \\ =115.3 {\mathrm{cm}}^2 \end{gathered}$$

Question 7:

The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. [Use π = 22/7]

Answer 7:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{diameter} \mathrm{of} \mathrm{the} \mathrm{wheel} \mathrm{is} 90 \mathrm{cm}. \\ \therefore \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{circular} \mathrm{wheel}, \mathrm{r}=\frac{90}{2}=45 \mathrm{cm}. \\ \therefore \mathrm{Perimeter} \mathrm{of} \mathrm{the} \mathrm{wheel} =2\times \mathrm{\pi }\times \mathrm{r}=2\times \frac{22}{7}\times 45=282.857 \mathrm{cm} \\ \mathrm{It} \mathrm{means} \mathrm{the} \mathrm{wheel} \mathrm{travels} 282.857 \mathrm{cm} \mathrm{in} \mathrm{a} \mathrm{revolution}. \\ \mathrm{Now}, \mathrm{it} \mathrm{makes} 315 \mathrm{revolutions} \mathrm{per} \mathrm{minute}. \\ \therefore \mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{the} \mathrm{wheel} \mathrm{in} \mathrm{one} \mathrm{minute}=315\times 282.857=89100 \mathrm{cm} \\ \therefore \mathrm{Speed} =89100 \mathrm{cm} \mathrm{per} \mathrm{minute}=\frac{89100 \mathrm{cm}}{1 \mathrm{minute}} \\ \mathrm{Now}, \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{convert} \mathrm{it} \mathrm{into} \mathrm{kilometers} \mathrm{per} \mathrm{hour}. \\ \therefore \frac{89100 \mathrm{cm}}{1 \mathrm{minute}} =\frac{89100\times {\displaystyle \frac{1}{100000}}\mathrm{kilometer}}{{\displaystyle \frac{1}{60}}\mathrm{hour}} \\ =\frac{89100}{100000}\times \frac{60}{1}\times \frac{\mathrm{kilometer}}{\mathrm{hour}} \\ =53.46 \mathrm{kilometers} \mathrm{per} \mathrm{hour} \end{gathered}$$

Question 8:

The area of a rhombus is 240 cm² and one of the diagonal is 16 cm. Find another diagonal.

Answer 8:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rhombus}=240 \mathrm{cm} \\ \mathrm{Length} \mathrm{of} \mathrm{one} \mathrm{of} \mathrm{its} \mathrm{diagonals}=16 \mathrm{cm} \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{if} \mathrm{the} \mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{rhombus} \mathrm{are} {\mathrm{d}}_1 \mathrm{and} {\mathrm{d}}_2, \mathrm{then} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{rhombus} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{Area}=\frac{1}{2}({\mathrm{d}}_1\times {\mathrm{d}}_2) \\ \mathrm{Putting} \mathrm{the} \mathrm{given} \mathrm{values}: \\ 240=\frac{1}{2}(16\times {\mathrm{d}}_2) \\ 240\times 2=16\times {\mathrm{d}}_2 \\ \mathrm{This} \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as} \mathrm{follows}: \\ 16\times {\mathrm{d}}_2=480 \\ {\mathrm{d}}_2=\frac{480}{16} \\ {\mathrm{d}}_2=30 \mathrm{cm} \\ \mathrm{Thus}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{diagonal} \mathrm{of} \mathrm{the} \mathrm{rhombus} \mathrm{is} 30 \mathrm{cm}\mathrm{.} \end{gathered}$$

Question 9:

The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Answer 9:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Lengths} \mathrm{of} \mathrm{the} \mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{rhombus} \mathrm{are} 7.5 \mathrm{cm} \mathrm{and} 12 \mathrm{cm}. \\ \mathrm{Now}, \mathrm{we} \mathrm{know}: \mathrm{Area}=\frac{1}{2}({\mathrm{d}}_1\times {\mathrm{d}}_2) \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{rhombus}=\frac{1}{2}(7.5\times 12)=45 {\mathrm{cm}}^2 \end{gathered}$$

Question 10:

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 10:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Diagonal} \mathrm{of} \mathrm{a} \mathrm{quadrilateral} \mathrm{shaped} \mathrm{field}=24 \mathrm{m} \\ \mathrm{Perpendiculars} \mathrm{dropped} \mathrm{on} \mathrm{it} \mathrm{from} \mathrm{the} \mathrm{remaining} \mathrm{opposite} \mathrm{vertices} \mathrm{are} 8 \mathrm{m} \mathrm{and} 13 \mathrm{m}. \\ \mathrm{Now}, \mathrm{we} \mathrm{know}: \\ \mathrm{Area}=\frac{1}{2}\times \mathrm{d}\times ({\mathrm{h}}_1+{\mathrm{h}}_2) \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{field}=\frac{1}{2}\times 24\times (8+13) \\ =12\times 21 \\ =252 {\mathrm{m}}^2 \end{gathered}$$

Question 11:

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 11:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Side} \mathrm{of} \mathrm{the} \mathrm{rhombus}=6 \mathrm{cm} \\ \mathrm{Altitude}=4 \mathrm{cm} \\ \mathrm{One} \mathrm{of} \mathrm{the} \mathrm{diagonals} = 8 \mathrm{cm} \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rhombus} = \mathrm{Side} \times \mathrm{Altitude} = 6\times 4 = 24 {\mathrm{cm}}^2 ........\left(\mathrm{i}\right) \\ \mathrm{We} \mathrm{know}: \mathrm{Area} \mathrm{of} \mathrm{rhombus} = \frac{1}{2}\times {\mathrm{d}}_1\times {\mathrm{d}}_2 \\ \mathrm{Using} \left(\mathrm{i}\right): \\ 24 = \frac{1}{2}\times {\mathrm{d}}_1\times {\mathrm{d}}_2 \\ 24 = \frac{1}{2}\times 8\times {\mathrm{d}}_2 \\ {\mathrm{d}}_2 = 6 \mathrm{cm} \end{gathered}$$

Question 12:

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 m² is Rs 4.

Answer 12:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{The} \mathrm{floor} \mathrm{consist} \mathrm{of} 3000 \mathrm{rhombus} \mathrm{shaped} \mathrm{tiles}. \\ \mathrm{The} \mathrm{lengths} \mathrm{of} \mathrm{the} \mathrm{diagonals} \mathrm{of} \mathrm{each} \mathrm{tile} \mathrm{are} 45 \mathrm{cm} \mathrm{and} 30 \mathrm{cm}\text{.} \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{a} \mathrm{rhombus} \mathrm{shaped} \mathrm{tile} =\frac{1}{2}\times (45\times 30)=675 {\mathrm{cm}}^2 \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{complete} \mathrm{floor}=3000\times 675=2025000 {\mathrm{cm}}^2 \\ \mathrm{Now}, \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{convert} \mathrm{this} \mathrm{area} \mathrm{into} {\mathrm{m}}^2 \mathrm{because} \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{polishing} \mathrm{is} \mathrm{given} \mathrm{as} \mathrm{per} {\mathrm{m}}^2. \\ \therefore 2025000 {\mathrm{cm}}^2=2025000\times \mathrm{cm}\times \mathrm{cm} \\ =2025000\times \frac{1}{100} \mathrm{m}\times \frac{1}{100} \mathrm{m} \\ =202.5 {\mathrm{m}}^2 \\ \mathrm{Now}, \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{polishing} 1 {\mathrm{m}}^2 \mathrm{is} \mathrm{Rs} 4. \\ \therefore \mathrm{Total} \mathrm{cost} \mathrm{of} \mathrm{polishing} \mathrm{the} \mathrm{complete} \mathrm{floor}=202.5\times 4=810 \\ \mathrm{Thus}, \mathrm{the} \mathrm{total} \mathrm{cost} \mathrm{of} \mathrm{polishing} \mathrm{the} \mathrm{floor} \mathrm{is} \mathrm{Rs} 810. \end{gathered}$$

Question 13:

A rectangular grassy plot is 112 m long and 78 m broad. It has a gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs 4.50 per square metre.

Answer 13:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{The} \mathrm{length} \mathrm{of} \mathrm{a} \mathrm{rectangular} \mathrm{grassy} \mathrm{plot} \mathrm{is} 112 \mathrm{m} \mathrm{and} \mathrm{its} \mathrm{width} \mathrm{is} 78 \mathrm{m}. \\ \mathrm{Also}, \mathrm{it} \mathrm{has} \mathrm{a} \mathrm{gravel} \mathrm{path} \mathrm{of} \mathrm{width} 2.5 \mathrm{m} \mathrm{around} \mathrm{it} \mathrm{on} \mathrm{the} \mathrm{sides} . \\ \mathrm{Its} \mathrm{rough} \mathrm{diagram} \mathrm{is} \mathrm{given} \mathrm{below}: \end{gathered}$$



$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{rectangular} \mathrm{field}=112-(2\times 2.5)=107 \mathrm{m} \\ \mathrm{The} \mathrm{width} \mathrm{of} \mathrm{the} \mathrm{inner} \mathrm{rectangular} \mathrm{field}=78-(2\times 2.5)=73 \mathrm{m} \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{path}=(\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{with} \mathrm{sides} 112 \mathrm{m} \mathrm{and} 78 \mathrm{m})-(\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{with} \mathrm{sides} 107 \mathrm{m} \mathrm{and} 73 \mathrm{m}) \\ =(112\times 78)-(107\times 73) \\ =8736-7811 \\ =925 {\mathrm{m}}^2 \\ \mathrm{Now}, \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{constructing} \mathrm{the} \mathrm{path} \mathrm{is} \mathrm{Rs} 4.50 \mathrm{per} \mathrm{square} \mathrm{meter}. \\ \therefore \mathrm{Cost} \mathrm{of} \mathrm{constructing} \mathrm{the} \mathrm{complete} \mathrm{path}=925\times 4.50=\mathrm{Rs} 4162.5 \\ \mathrm{Thus}, \mathrm{the} \mathrm{total} \mathrm{cost} \mathrm{of} \mathrm{constructing} \mathrm{the} \mathrm{path} \mathrm{is} \mathrm{Rs} 4162.5 \end{gathered}$$

Question 14:

Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.

Answer 14:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Side} \mathrm{of} \mathrm{the} \mathrm{rhombus}=20 \mathrm{cm} \\ \mathrm{Length} \mathrm{of} \mathrm{a} \mathrm{diagonal}=24 \mathrm{cm} \\ \mathrm{We} \mathrm{know}: \mathrm{If} {\mathrm{d}}_1 \mathrm{and} {\mathrm{d}}_2 \mathrm{are} \mathrm{the} \mathrm{lengths} \mathrm{of} \mathrm{the} \mathrm{diagonals} \mathrm{of} \mathrm{the} \mathrm{rhombus}, \mathrm{then} \\ \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{rhombus}=\frac{1}{2}\sqrt{{\mathrm{d}}_1^2+{\mathrm{d}}_2^2} \\ \mathrm{So}, \mathrm{using} \mathrm{the} \mathrm{given} \mathrm{data} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{diagonal} \mathrm{of} \mathrm{the} \mathrm{rhombus}: \\ 20=\frac{1}{2}\sqrt{{24}^2+{\mathrm{d}}_2^2} \\ 40=\sqrt{{24}^2+{\mathrm{d}}_2^2} \\ \mathrm{Squ}ari\mathrm{ng} \mathrm{both} \mathrm{sides} \mathrm{to} \mathrm{get} \mathrm{rid} \mathrm{of} \mathrm{the} \mathrm{square} \mathrm{root} \mathrm{sign}: \\ {40}^2={24}^2+{\mathrm{d}}_2^2 \\ {\mathrm{d}}_2^2\text{=1600-576=1024} \\ {\mathrm{d}}_2\text{=}\sqrt{1024}\text{=32 cm} \\ \therefore \mathrm{A}\text{rea of the rhombus=}\frac{1}{2}(24\times 32)=384{\text{ cm}}^2 \end{gathered}$$

Question 15:

The length of a side of a square field is 4 m. what will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonal is 2 m?

Answer 15:

$$\begin{gathered} \text{Given:} \\ \text{Length of the square field = 4 m} \\ \therefore \mathrm{A}{\text{rea of the square field = 4×4 = 16 m}}^2 \\ \mathrm{Given}\text{: Area of the rhombus = Area of the square field} \\ \text{Length of one diagonal of the rhombus = 2 m} \\ \therefore \mathrm{S}\text{ide of the rhombus=}\frac{1}{2}\sqrt{d_1^2+d_2^2} \\ \text{And, area of the rhombus=}\frac{1}{2}\times (d_1\times d_2) \\ \therefore A\text{rea:} \\ 16=\frac{1}{2}(2\times d_2) \\ {d}_2\text{=16 m} \\ \text{Now, we need to find the length of the side of the rhombus. } \\ \therefore \text{Side of the rhombus = }\frac{1}{2}\sqrt{2^2+{16}^2 }\text{= }\frac{1}{2}\sqrt{260 }\text{= }\frac{1}{2}\sqrt{4\times 65 }\text{= }\frac{1}{2}\text{×2}\sqrt{65 }\text{= }\sqrt{65}\text{ m} \\ \text{Also, we know: Area of the rhombus = Side × Altitude} \\ \therefore \text{16=}\sqrt{65}\text{×Altitude} \\ \text{Altitude=}\frac{16}{\sqrt{65}}\text{ m} \end{gathered}$$

Question 16:

Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm.

Answer 16:

$$\begin{gathered} \text{Given:} \\ \text{Length of each side of a field in the shape of a rhombus = 14 cm} \\ \text{Altitude = 16 cm} \\ \text{Now, we know: Area of the rhombus = Side×Altitude} \\ \therefore {\text{Area of the field = 14×16 = 224 cm}}^2 \end{gathered}$$

Question 17:

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 17:

$$\begin{gathered} \text{Given:} \\ \mathrm{C}\text{ost of fencing 1 metre of a square field = 60 paise} \\ \text{And, the total cost of fencing the entire field = Rs 1200 = 1,20,000 paise} \\ \text{∴ Perimeter of the square field = }\frac{120000}{60}=2000 \text{metres} \\ \text{Now, perimeter of a square = 4×side } \\ \mathrm{For} \mathrm{the} \mathrm{given} \mathrm{square} \mathrm{field}: \\ \text{4×Side = 2000 m} \\ \text{Side = }\frac{2000}{4}= 500 \text{metres} \\ \therefore \mathrm{A}{\text{rea of the square field = 500×500=250000 m}}^2 \\ {\text{Again, given: Cost of reaping per 100 m}}^2 \text{= 50 paise} \\ \therefore \mathrm{C}{\text{ost of reaping per 1 m}}^2 =\frac{50}{100}\text{paise} \\ \therefore \mathrm{C}{\text{ost of reaping 250000 m}}^2=\frac{50}{100}\times 250000=125000\text{ paise} \\ \text{Thus, the total cost of reaping the complete square field is 125000 paise, i.e. Rs 1250.} \end{gathered}$$

Question 18:

In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot.

Answer 18:

$$\begin{gathered} \text{Given:} \\ \text{Side of the square plot = 84 m} \\ \text{Now, the man wants to exchange it with a rectangular plot of the same area with length 144.} \\ \mathrm{A}{\text{rea of the square plot = 84×84 = 7056 m}}^2 \\ \therefore \text{Area of the rectangular plot = Length×Width} \\ 7056=144\times \text{Width} \\ \Rightarrow \text{Width=}\frac{7056}{144}\text{=49 m} \\ \text{Hence, the width of the rectangular plot is 49 m.} \end{gathered}$$

Question 19:

The area of a rhombus is 84 m². If its perimeter is 40 m, then find its altitude.

Answer 19:

$$\begin{gathered} \text{Given:} \\ {\text{Area of the rhombus = 84 m}}^2 \\ \mathrm{P}\text{erimeter = 40 m} \\ \text{Now, we know: Perimeter of the rhombus = 4×Side} \\ \therefore 40=4\times \text{Side} \\ \text{Side=}\frac{40}{4}\text{=10 m} \\ \text{Again, we know: Area of the rhombus = Side×Altitude} \\ \Rightarrow 84=10\times \text{Altitude} \\ \text{Altitude = }\frac{84}{10}\text{ = 8.4 m} \\ \text{Hence, the altitude of the rhombus is 8.4 m.} \end{gathered}$$

Question 20:

A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m².

Answer 20:

$$\begin{gathered} \text{Given:} \\ \text{Side of the rhombus shaped garden = 30 m } \\ \text{Altitude = 16 m} \\ \text{Now, area of a rhombus = side×altitude} \\ {\text{∴ Area of the given garden=30×16=480 m}}^2 \\ {\text{Also, it is given that the rate of levelling the garden is Rs 2 per 1m}}^2\text{.} \\ \therefore \mathrm{T}{\text{otal cost of levelling the complete garden of area 480 m}}^2\text{=480×2= Rs 960} \end{gathered}$$

Question 21:

A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus?

Answer 21:

$$\begin{gathered} \text{Given:} \\ \text{Each side of a rhombus shaped field = 64 m} \\ \text{Altitude = 16 m} \\ \text{We know: Area of rhombus = Side×Altitude} \\ \therefore \mathrm{A}{\text{rea of the field = 64×16=1024 m}}^2 \\ \text{Given: Area of the square field = Area of the rhombus} \\ {\text{We know: Area of a square=(Side)}}^2 \\ \therefore {\text{1024=(Side)}}^2 \\ \Rightarrow \text{Side=}\sqrt{1024}\text{=32 m} \\ \text{Thus, the side of the square field is 32 m.} \end{gathered}$$

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Question 22:

The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.

Answer 22:

$$\begin{gathered} \text{Given: } \\ \text{Area of the rhombus = Area of the triangle with base 24.8 cm and altitude 16.5 cm} \\ \text{Area of the triangle = }\frac{1}{2}\text{×base×altitude = }\frac{1}{2}{\text{×24.8×16.5=204.6 cm}}^2 \\ {\text{∴ Area of the rhombus = 204.6 cm}}^2 \\ \text{Also, length of one of the diagonals of the rhombus=22 cm} \\ \mathrm{We} \mathrm{know}: \text{Area of rhombus=}\frac{1}{2}(d_1\times d_2) \\ 204.6=\frac{1}{2}(22\times d_2) \\ 22\times d_2\text{=409.2} \\ {d}_2\text{=}\frac{409.2}{22}\text{=18.6 cm} \\ \text{Hence, the length of the other diagonal of the rhombus is 18.6 cm.} \end{gathered}$$