Question

The side of triangular field are 55m ,300mand 300m its area is eqal to7÷15th area of circular park . what is the perimeter(in m) of the circular park​

Answer 1

★Given:-

• The sides of the triangular field are 55m, 300m, 300m.
• Its area is equal to 7/15 the area of the circular park.

★To find:-

• Perimeter of the circular park​.

★Solution:-

We know,

Heron's formula:-

$\displaystyle \leadsto \underline{\boxed{\sf Area = \sqrt{s(s-a)(s-b)(s-c)} }}$

$\displaystyle \leadsto \underline{\boxed{\sf s = \frac{a+b+c}{2} }}$

Where,

• a,b,c are the sides of triangle
• s is the semi-perimeter

Putting values in the formula,

➜S = (a+b+c)/2

➜S = (55+300+300)/2

➜S = 655/2

➜S = 327.5m

$\implies \sf Area = \sqrt{327.5(327.5-55)(327.5-300)(327.5-300)}$

$\implies \sf \sqrt{327.5\times 272.5 \times 27.5 \times 27.5}$

$\implies \sf 27.5 \times 298.73$

$\implies \sf Area = \bold{8215.075m^2.}$

Therefore,

Area of the triangular field is 8215.075m².

Area of the circular park:-

According to question,

➜Area of the triangular field =7/15(Area of the circular park)

Area of the circular park:

➜Area of triangular field×(15/7)

➜8215.075 × 15/7

➜17603.73m²

Now,

Let the radius of the circular park is "r".

 

Using the formula,

✦Area of circle = πr²

Putting values,

Area of the circular park = 17603.73  

➜πr² = 17603.73

➜r² = 17603.73×(7/22)

➜r = √5601.18

➜r = 74.84

Perimeter of circular park:-  

Using the formula,

✦Perimeter of circle = 2πr

Putting values,

➜Perimeter = 2πr  

➜2(22/7)(74.84)

➜470.42m²

Hence,

The perimeter of the circular park is 470.42m².

______________

Answer 2

★ Given :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• The sides of triangular field are 55m, 300m and 300m. It's area is equal to (7/15)th area of circular park. What is perimeter (in m) of the circular park (corrected to one deciml place) ? (Take pi = 22/7)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ Solution :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ we know that,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{Area of a ∆ with sides a, b , c and semi- perimeter s is √[s * (s - a) * (s - b) * (s - c)] .}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{semi - perimeter = (a + b + c)/2 }$.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{given sides of triangular field are 55m, 300m and 300m.. }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ So,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{→ s = (55 + 300 + 300) / 2 = 327.5m. }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ Than,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\textsf{→ Area of triangular field = √[327.5 * (327.5 - 55) * (327.5 - 300) * (327.5 - 300)] = √[327.5 * 272.5 * 27.5 * 27.5] = 27.5√(327.5 * 272.5) = 27.5 * 298 = 8195 m².}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ Now given that,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{→ Area of triangular field = (7/15)th area of circular park. }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ So,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{→ 8195 = (7/15) * Area of circular park.}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{→ (8195 * 15)/7 = Area of circular park.}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{→ Area of circular park = 17560.7 m². }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ Therefore,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{→ πr² = 17560.7 }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{→ (22/7) * r² = 17560.7 }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{→ r² = (17560.7 * 7) / 22 }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{→ r = 74.74 m. }$

⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ Hence,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

→ $\textsf{Perimeter of the circular Park = 2πr = 2 * (22/7) * 74.74 = (3288.56)/7 = 469.8m. (Ans). }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\pink{\sf \pink{\bigstar} }}$ Final answer $\large\underline{\pink{\sf \pink{\bigstar} }}$

⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\pink{\sf \red{\bigstar} }}$$\textsf{→ 470m ( by round off rule)}$$\large\underline{\pink{\sf \red{\bigstar} }}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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