Question

If Anish is moving along the boundary of a triangular field of sides 35m, 53m and 66 m and you are moving

along the boundary of a circular field whose area is double the area of the triangular field, then the radius of

the circular field is: (Take
$\pi \: as \: \frac{22}{7}$






​

Answer 1

Answer:

Here in the question, we just need to go through the formulas for finding the area of the triangle and area of the circle. Afterwards, applying the conditions will help us to get to our final result.

GiveN:

• Sides of the triangle are 35m, 53m and 66 m
• Area of the circular field = Area of triangle.

We have to find the radius of the circular field...?

Here, we can use the Heron's formula for finding the area of the triangle since it's not a right angled triangle.

• Area of ∆ = $\sf{ \sqrt{s(s - a)(s - b)(s - c)} }$
• Area of the circle = πr²

So, let's start solving....

Area of the triangle:

Finding semiperimeter:

• s = 35 m + 53 m + 66 m / 2 = 77 m

Now let a = 35 m, b = 53 m and c = 77 m, Let's plug the values in the formula to find the area of the ∆.

$\sf{ \longrightarrow{area = \sqrt{77(77 - 35)(77 - 53)(77 - 66)} \: {m}^{2} }}$

$\sf{ \longrightarrow{area = \sqrt{77(42)(24)(11)} \:{m}^{2} }}$

Now for easier calculation, let's simplify the numbers inside the square root to form the pairs,

$\sf{ \longrightarrow{area = \sqrt{11 \times 7 \times 7 \times 6 \times 6 \times 2 \times 2 \times 11}\: {m}^{2} }}$

$\sf{ \longrightarrow{area = 11 \times 7 \times 6 \times 2}\: {m}^{2} }$

$\sf{ \longrightarrow{area = 924 \: {m}^{2} }}$

Area of the circle:

Given in the question that area of the circle is double of the area of the triangle. So,

• Area of the circle = 1848 m²

Now plug in the value in area of the circle formula to find the radius of the circular field.

$\sf{ \longrightarrow{\pi {r}^{2} = 1848 \: {m}^{2} }}$

Put the value of π be 22/7,

$\sf{ \longrightarrow{ \dfrac{22}{7} {r}^{2} = 1848 \: {m}^{2} }}$

$\sf{ \longrightarrow{ {r}^{2} = \dfrac{1848 \times 7}{22} \: {m}^{2} }}$

$\sf{ \longrightarrow{ {r}^{2} = 84 \times 7 \: {m}^{2} }}$

Now squaring both sides,

$\sf{ \longrightarrow{r = \sqrt{12 \times 7 \times 7} \: m}}$

$\sf{ \longrightarrow{r = 3 \sqrt{14} \: m}}$

This is around 11.225 m approximately. So the required length of the radius of the circular path:

$\large{ \boxed{ \sf{ \red{ \approx \: 11.225 \: m}}}}$

And we are done !!

Answer 2

Answer:

24.14

Step-by-step explanation:

See image for full explanation...