Question

a rectangle contains 3 identical circle if each circle has a radius of r then what is the area of rectangle in terms of r​

Answer 1

Answer:

4(2 + √3)r²  or  14.92 r²

Step-by-step explanation:

In the formed triangle, by Pythagoras theorem,

⇒ OD² = (r + r)² - r²

            = (2r)² - r²

            = 4r² - r²

            = 3r²

⇒ OD = r√3

Hence, breadth is r + r√3 + r = r(2 + √3)

length = 2r + 2r = 4r

Hence, area is length * breadth

⇒ 4r * r(2 + √3)

⇒ 4(2 + √3)r²

   Or, 14.92r² or 15r²

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow THE\:RECTANGLE\:OF\:LENGTH\:AND\:BREADTH\:CONTAINING,$

$\sf\dashrightarrow THREE\:CIRCLES\:WITH\:RADIUS\::-r$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow AREA\:OF\:RECTANGLE.$

✯FORMULA IN USE,

$\large{\boxed{\bf{ \star\:\: AREA\:OF\:RECTANGLE= LENGTH \times BREADTH\:\: \star}}}$

$\large\underline\bold{SOLUTION,}$

$\bf\therefore BY\:USING\: PYTHAGORAS\:THEOEM,$

$\sf\therefore OQ^2=AQ^2+AO^2$

$\sf\implies (r+r)^2= (r)^2+OA^2$

$\sf\implies OA^2= (r+r)^2-(r)^2$

$\sf\implies OA^2= (4r^2)-r^2$

$\sf\implies OA^2= 3r^2$

$\sf\implies OA= \sqrt{3r^2}$

$\sf\implies OA= \sqrt{3} \:r$

$\sf\implies OA= r\sqrt{3}$

$\large{\boxed{\bf{ \star\:\: OA= r\sqrt{3}cm\:\: \star}}}$

✯LENGTH,

$\sf\therefore 2r+2r$

$\sf\implies 4r$

$\large{\boxed{\bf{ \star\:\: length= 4r\:cm\:\: \star}}}$

✯BREADTH,

$\sf\therefore r+r+r \sqrt{3}$

$\sf\dashrightarrow 2r+ r\sqrt{3}$

$\sf\dashrightarrow r( 2+ \sqrt{3})$

$\large{\boxed{\bf{ \star\:\: breadth= r(2+ \sqrt{3})cm\:\: \star}}}$

NOW, FINDING AREA OF RECTANGLE,

$\sf\therefore AREA\:OF\:RECTANGLE= LENGTH \times BREADTH$

$\sf\implies (4r) \times r(\sqrt{3} +2)$

$\sf\implies 8r^2+4r^2 \sqrt{3}$

$\sf\implies r^2(8+4\sqrt{3})$

$\sf\implies r^2(8+4(1.73)$

$\sf\implies r^2(8+6.92)$

$\sf\implies r^2(14.92)$

$\sf\implies 14.92r^2$

$\large{\boxed{\bf{ \star\:\: 14.92r^2 cm^2 \:\: \star}}}$

$\large\underline\bold{AREA\:OF\:RECTANGLE\:IS\:14.92r^2cm^2}$

$\red{\text{REFER THE DIAGRAM.}}$

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