Question

two small circular Park of diameter 16 and 12 are to be replaced by a bigger circular Park what would be the radius of the new park if the two new park occupies the same space as the two small park​

Answer 1

AnswEr :

• Diameter of Park ( D₁ ) = 16 unit
• Radius of Park ( R₁ ) = 8 unit
• Diameter of Park ( D₂ ) = 12 unit
• Radius of Park ( R₂ ) = 6 unit

A new Bigger Circular Park replace those two Smaller Circular Parks, and occupies the same space as of two small parks.

• Let's Head to the Question Now :

$:\implies \sf{Area \:of \:Park_1 + Area \:of \:Park_2 = Area \:of \:Park_3}\\\\\\:\implies \sf \pi {(R_1)}^{2} + \pi {(R_2)}^{2} = \pi {(R_3)}^{2}\\\\\\:\implies \sf \cancel\pi [{(R_1)}^{2} + {(R_2)}^{2}] = \cancel\pi {(R_3)}^{2}\\\\\\:\implies \sf[{(R_1)}^{2} + {(R_2)}^{2}] ={(R_3)}^{2}\\\\\\:\implies \sf[{(8 \:unit)}^{2} + {(6 \:unit)}^{2}] ={(R_3)}^{2}\\\\\\:\implies \sf[64 \:unit^{2} + 36 \:unit^{2}] ={(R_3)}^{2}\\\\\\:\implies \sf100\:unit^{2} ={(R_3)}^{2}\\\\\\:\implies \sf \sqrt{100 \:unit^{2}}=R_3\\\\\\:\implies \sf \sqrt{10 \:unit \times 10 \:unit}=R_3 \\\\\\:\implies\boxed{\sf R_3 = 10 \:unit}$

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∴ Radius of Bigger Circular Park is 10 unit.

Answer 2

Answer:

$\huge{\boxed{\mathfrak\green{Answer:10\:unit}}}$

______________________

$\bf\green{Given\::}$

• Diameter of park 1 = 16 unit.
• Diameter of park 2 = 12 unit.
• Radius of new park if the new park occupies the same place as the two small parks = ?

As the new park occupies the place of the two smaller two parks,so the area of the bigger park will be as equal to the sum of areas of the two smaller parks.

$\tt Now\:radius\:of\:park 1 =\frac{Dimeter(1)}{2}\\ \leadsto\tt Radius(1) =\frac{16}{2}\:unit.\\ \leadsto\tt Radius(1) = 8\:unit.$

$\tt Radius(2)=\frac{Diameter(2)}{2}\\ \leadsto\tt Radius(2) =\frac{12}{2}\:unit.\\ \leadsto\tt Radius(2) = 6\:unit.$

$\tt Area\:of\:new\:park = Area\:of \:park\:1 + Area\:of\:park\:2$

$\leadsto\tt Area(3) =\pi({r1})^{2} + \pi({r2})^{2}$

$\leadsto\tt \pi({r3})^{2} = \pi\times {8}^{2} + \pi \times {6}^{2}$

$\leadsto\tt \pi({r3})^{2} = \pi \times 64 + \pi \times 36$

$\leadsto\tt \pi({r3})^{2} = \pi(64+36)$

$\leadsto\tt ({r3})^{2} = 100$

$\leadsto\tt r3 =\sqrt{100}\:unit$

$\leadsto\tt r3 = 10\:unit$

$\therefore\tt {\underline{Radius\:of\:new\:park=10\:unit.}}$