Let the side of the square(ABCD) be $x$ metres.
Then, AB + BC = 2$x$ metres.
AC = $ \sqrt{2} x$ = (1.41$x$) m.
Saving on 2$x$ metres = (0.59$x$) m.
| Saving % =$ \left(\dfrac{0.59x}{2x} \times 100\right) $%= 30% (approx.) |
Let original length = $x$ and original breadth = $y$.
| Decrease in area | = $x$$y$ -$ \left(\dfrac{80}{100} x \times \dfrac{90}{100} y\right) $ |
| =$ \left(xy -\dfrac{18}{25} xy\right) $ | |
| =$ \dfrac{7}{25} xy $ |
| $\therefore$ Decrease % =$ \left(\dfrac{7}{25} xy \times\dfrac{1}{xy} \times 100\right) $%= 28%. |
Area = 5.5 × 3.75 sq. metre
Cost for 1 sq. metre. = Rs. 800
Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500
| $ \dfrac{2(l + b)}{b} $=$ \dfrac{5}{1} $ |
$\Rightarrow$ 2l + 2b = 5b
$\Rightarrow$ 3b = 2l
| b =$ \dfrac{2}{3} $l |
Then, Area = 216 cm2
$\Rightarrow$ l x b = 216
| $\Rightarrow$l x$ \dfrac{2}{3} $l= 216 |
$\Rightarrow$ l2 = 324
$\Rightarrow$ l = 18 cm.
Let l = 9 ft.
Then l + 2b = 37
=> 2b = 37 - l = 37 - 9 = 28
=> b = $\dfrac{28}{2}$ = 14 ft.
Area = lb = 9 × 14 = 126 sq. ft.
Let h be the height of the triangle and b be the base of the triangle.
The area of the triangle A=$\dfrac{1}{2}$bh
The length of the base and the length of the height are increased by 2 times.
The new base is $2 \times b $ and the new height is $ 2 \times h.$
The new area of the triangle = $\dfrac{1}{2} \times (2 \times b) \times (2 \times h)$
= 4 $\left(\dfrac{1}{2}bh\right)$
= 4 × original area of the triangle
The area of the triangle increases by 4 times.
The area of a parallelogram = base × height
= 10 × 7 = 70
So, the area of the parallelogram is 70 $in^{2}$
If the width of the hall is doubled, then the new width = 2 × 10 feet = 20 feet
Area of the new hall = 18 feet × 20 feet = 360 square feet.
[Area of a rectangle = height × width.]
So, the new area of the hall is 360 square feet.
The area of a triangle is half the product of any base b and the corresponding height h.
New base of the triangle = 5b
[Base is increased by 5 times.]
New area of the triangle =$ \dfrac{1}{2} × 5b × h = 5\left(\dfrac{1}{2}bh\right)$ = 5(original area of the triangle)
The area of the triangle increases by 5 times.
The area of a parallelogram is 576 $cm^{2}$
Let h be the height of the parallelogram.
The length of the base is four times the height.
The base of the parallelogram = 4 x h
= (4 x h) x h = 4 x $h^{2}$
The area of the parallelogram = base x height
4 x $h^{2}$ = 576
[From step 1.]
$h^{2}$ = 144
[Divide each side by 4.]
h = 12 cm
[Take the square roots on both the sides.]
The height of the parallelogram is 12 cm