Question

In a triangular park, the three corners are used to make play areas for children in the form of sectors of radii 14m each. If the sides of the park are 52m, 56m and 60m, find the area of the remaining portion of the park. use pie 22/7

Answer 1

Area of shaded region is 1036 $m^2$.

Step-by-step explanation:

Given:

Radius = 14 m

As shown in the figure below, BC = 52 m, AB = 56 m and AC = 60 m .

Area of shaded portion = Area of triangular park - Area of all three circular section  

Also we Know Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

Here $s = \frac{a+b+c}{2}$ and Here a = 52, b = 56 and c = 60,

So, $s =\frac{52+56+60}{2}=\frac{168}{2}=84$

So, Area of triangular park = $\sqrt{84(84-52)(84-56)(84-60)}$

                                            = $\sqrt{84\times 32 \times 28 \times 24}$

                                            = $\sqrt{1806336}$

                                            = $1344 m^2$

We know Area of sector of circle = $\frac{Center angle}{360} \times \pi r^2$

So,

Area of circular sections  = $\frac{\angle A}{360} \times \pi r^2 +\frac{\angle B}{360} \times \pi r^2 +\frac{\angle C}{360} \times \pi r^2$

                                          = $\frac{\angle A + \angle B + \angle C}{360} \times \pi r^2$

                                          = $\frac{180}{360}\times \frac{22}{7} \times 14 \times 14$         $[\pi=\frac{22}{7} ]$

                                          = $\frac{1}{2} \times 22 \times 2 \times 14$

                                           = $308 m^2$

Now, area of shaded region = 1344 - 308 = 1036 $m^2$

   

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given:

Radius = 14 m

As shown in the figure below, BC = 52 m, AB = 56 m and AC = 60 m .

Area of shaded portion = Area of triangular park - Area of all three circular section

Also we Know Area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

s(s−a)(s−b)(s−c)

Here s = \frac{a+b+c}{2}s=

2

a+b+c

and Here a = 52, b = 56 and c = 60,

So, s =\frac{52+56+60}{2}=\frac{168}{2}=84s=

2

52+56+60

=

2

168

=84

So, Area of triangular park = \sqrt{84(84-52)(84-56)(84-60)}

84(84−52)(84−56)(84−60)

= \sqrt{84\times 32 \times 28 \times 24}

84×32×28×24

= \sqrt{1806336}

1806336

= 1344 m^21344m

2

We know Area of sector of circle = \frac{Center angle}{360} \times \pi r^2

360

Centerangle

×πr

2

So,

Area of circular sections = \frac{\angle A}{360} \times \pi r^2 +\frac{\angle B}{360} \times \pi r^2 +\frac{\angle C}{360} \times \pi r^2

360

∠A

×πr

2

+

360

∠B

×πr

2

+

360

∠C

×πr

2

= \frac{\angle A + \angle B + \angle C}{360} \times \pi r^2

360

∠A+∠B+∠C

×πr

2

= \frac{180}{360}\times \frac{22}{7} \times 14 \times 14

360

180

×

7

22

×14×14 [\pi=\frac{22}{7} ][π=

7

22

]

= \frac{1}{2} \times 22 \times 2 \times 14

2

1

×22×2×14

= 308 m^2308m

2

Now, area of shaded region = 1344 - 308 = 1036 m^2m

2