| Other side | = | $ \sqrt{\left(\dfrac{15}{2} \right)^2-\left(\dfrac{9}{2} \right)^2} $ | ft |
| = | $ \sqrt{\dfrac{225}{4} -\dfrac{81}{4} } $ | ft | |
| = | $ \sqrt{\dfrac{144}{4} } $ | ft | |
| = | 6 ft. |
$\therefore$ Area of closet = $\left(6 \times 4.5\right)$ sq. ft = 27 sq. ft.
Let breadth = x metres.
Then, length = (x + 20) metres.
| Perimeter =$ \left(\dfrac{5300}{26.50} \right) $m = 200 m. |
$\therefore$ 2[(x + 20) + x] = 200
$\Rightarrow$ 2x + 20 = 100
$\Rightarrow$ 2x = 80
$\Rightarrow$ x = 40.
Hence, length = x + 20 = 60 m.
Area of the park = (60 x 40) m2 = 2400 m2.
Area of the lawn = 2109 m2.
$\therefore$ Area of the crossroads = (2400 - 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
$\Rightarrow$ x2 - 100x + 291 = 0
$\Rightarrow$ (x - 97)(x - 3) = 0
$\Rightarrow$ x = 3.
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = $(41 \times 41)$ cm2.
| $\therefore$ Required number of tiles =$ \left(\dfrac{1517 \times 902}{41 \times 41} \right) $= 814. |
$lb=460$ m2...(Equation 1)
Let the breadth = b
Then length, $l=b \times \dfrac{(100 + 15)}{100} = \dfrac{115b}{100}$ ...(Equation 2)
From Equation 1 and Equation 2,
$\dfrac{115b}{100} \times b = 460$
$b^2 = \dfrac{46000}{115} = 400$
$\Rightarrow b = \sqrt{400} = 20\text{ m}$
Let original length = $x$ metres and original breadth = $y$ metres.
Original area = $(xy)$ m2.
| New length =$ \left(\dfrac{120}{100} x\right) $m=$ \left(\dfrac{6}{5} x\right) $m. |
| New breadth =$ \left(\dfrac{120}{100} y\right) $m=$ \left(\dfrac{6}{5} y\right) $m. |
| New Area =$ \left(\dfrac{6}{5} x \times\dfrac{6}{5} y\right) $m2=$ \left(\dfrac{36}{25} xy\right) $m2. |
The difference between the original area = $xy$ and new-area 36/25 $xy$ is
= (36/25)$x$$y$ - $x$$y$
= $x$$y$(36/25 - 1)
= $x$$y$(11/25) or (11/25)$x$$y$
| $\therefore$ Increase % =$ \left(\dfrac{11}{25} xy \times\dfrac{1}{25} \times 100\right) $%= 44%. |
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
$\therefore Area = (l \times b) = (63 \times 40)$ m2 = 2520 m2.
Let the areas of the parts be x hectares and (700 - x) hectares
Difference of the areas of the two parts = x - (700 - x) = 2x - 700
one-fifth of the average of the two areas = $\dfrac{1}{5}\dfrac{[\text{x}+(700-\text{x})]}{2}$
$=\dfrac{1}{5} \times \dfrac{700}{2}=\dfrac{350}{5}=70$
Given that difference of the areas of the two parts = one-fifth of the average of the two areas
=> 2x - 700 = 70
=> 2x = 770
$\Rightarrow x = \dfrac{770}{2}= 385$
Hence, area of smaller part = (700 - x) = (700 – 385) = 315 hectares.
$ \sqrt{l^2 + b^2} $ = $ \sqrt{41} $ .
Also, lb = 20.
(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81
$\Rightarrow$ (l + b) = 9.
$\therefore$ Perimeter = 2(l + b) = 18 cm.
Perimeter of a rectangle = 2(l + b)
where l is the length and b is the breadth of the rectangle
Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres
Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which isvery important.
Hence number of poles required = $\dfrac{280}{5}$ = 56
| Area to be plastered= $[2\left(l + b\right) \times h] $+ $\left(l \times b\right)$ = ${[2\left(25 + 12\right) \times 6] + \left(25 \times 12\right)}$ m2 = (444 + 300) m2 = 744 m2. |
| $\therefore$ Cost of plastering = Rs.$ \left(744 \times\dfrac{75}{100} \right) $= Rs. 558. |
Given that area of the field = 680 sq. feet
=> lb = 680 sq. feet
Length(l) = 20 feet
=> 20 × b = 680
=> b $= \dfrac{680}{20} = 34$ feet
Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet
100 cm is read as 102 cm.
$\therefore$ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
| $\therefore$ Percentage error =$ \left(\dfrac{404}{100 \times 100} \times 100\right) $%= 4.04% |
Given that breadth of a rectangular field is 60% of its length
$\Rightarrow b = \dfrac{60l}{100} = \dfrac{3l}{5}$
perimeter of the field = 800 m
=> 2 (l + b) = 800
$\Rightarrow 2\left(l + \dfrac{3l}{5} \right)=800$
$\Rightarrow l + \dfrac{3l}{5} = 400$
$\Rightarrow \dfrac{8l}{5} = 400$
$\Rightarrow \dfrac{l}{5} = 50$
$\Rightarrow l = 5 \times 50 = 250\text{ m}$
$\text{b = }\dfrac{3l}{5} = \dfrac{3 \times 250}{5} = 3 \times 50 = 150\text{ m}$
Area = $\text{lb = }250 \times 150 = 37500\text{ m}^2$
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
$\therefore$ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
let each side of the square be a , then area = a x a
New side = 125a / 100 = 5a / 4
New area =$\left(5a \times 5a\right)$ / $\left(4 \times 4\right)$ = (25a²/16)
increased area== (25a²/16) - a²
Increase %= [(9a²/16 ) x (1/a² ) x 100]% = 56.25%
Area of a square = length × length
= 35 × 35 sq. m.
= 1225 m2
Length of the rectangle = 24 mm.
Breadth of the rectangle = 8 mm.
Area of the rectangle = l × b
= 24 × 8m2.
= 192m2.
8 cm = 2 × radius
8 cm × 2 = radius
radius = 4 cm
Area= $\pi$ × r2
= $\pi$ × 42
= $\pi$ × (4 × 4)
Area = 50.24 cm
Length of the square field = 275 m
Area of the square field = side × side
= 275 m × 275 m
= 75625 m2
Cost of levelling the field for 1 sq m = 10 cent.
Now, we have to find the cost for entire area ie 75625 sq m.
Cost of levelling the field = 75625 × 10 cent
= 756250 cent
= $ 7562.50