Question

10
A circular flower bed is surrounded by a path 7 m wide path. The diameter of the flowe
bed is 28 m. What is the area of this path?

Answer 1

Answer :-

• Area of the path = 770m².

Step by step explanation:

To find :-

Area of the path

Solution:

Given that,

A circular flower bed is surrounded by a path 7 m wide path.

The diameter of the flower bed is 28 m.

We know,

$\bigstar \boxed{\bf{Radius \: of \: the \: flower \: bed = \dfrac{Diameter}{2}}}$

Where,

Diameter = 28m

Therefore,

$\longrightarrow \bf \dfrac{28}{2}$

$\longrightarrow \bf 14$

Now,

• Radius of larger circle = Radius of flower bed + width of the path.

So,

$\longrightarrow \bf 14 + 7$

$\longrightarrow \bf 21$

Area of larger circle = πr²

$\longrightarrow \bf \pi {(21)}^{2}$

$\longrightarrow \bf 441\pi \: {cm}^{2}$

Area of smaller circle = πr²

$\longrightarrow \bf \pi {(14)}^{2}$

$\longrightarrow \bf 196\pi \: {cm}^{2}$

Now, Area of path = Area of larger path - Area of smaller flower bed.

Where,

• Area of larger path = 441π cm²
• Area of smaller flower bed = 196π cm².

Therefore,

$\longrightarrow \bf 441\pi - 196\pi$

$\longrightarrow \bf (441 - 196)\pi$

$\longrightarrow \bf (245) \times \pi$

$\longrightarrow \bf (245) \times \dfrac{22}{7}$

$\longrightarrow \bf 35\times 22$

$\longrightarrow \bf 770 {m}^{2}$

Therefore, Area of the path = 770m².

$\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}$