KoKo Chocolatiers sells their truffes in rectangular boxes. They are going to start ordering circular boxes, too. The rectangular box is made from metal that is 0.3 cm thick. The circular box will be constructed from laminated cardboard that is 0.2 cm thick. The height of the circular box will be the same as the rectangular box. If the volumes of the two boxes are equal, then what is the radius (r) of the circular box?

r = 9.61 cm

Step-by-step explanation:

Volume of a rectangular box = length × width × height

= 21.5 cm × 13.5 cm × 6 cm

= 1,741.5 cm³

Volume of a rectangular box = 1,741.5 cm³

Since,

Volume of a rectangular box = Volume of a circular box

Then,

1,741.5 cm³ = Volume of a circular box

Volume of a circular box = πr²h

1,741.5 cm³ = 3.14 * r² * 6 cm

1,741.5 = 18.84r²

r² = 1,741.5/18.84

r² = 92.436305732484

r = √92.436305732484

r = 9.6143801533163

Approximately,

r = 9.61 cm

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)