Maths Olympiad Prep

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, 2020

Geometry Difficulty 3.1 AMC 10/12 Prove it Canada

For a rectangular prism with length \ell, width ww, and height hh as shown, the surface area is given by the formula A=2w+2h+2whA = 2\ell w + 2\ell h + 2wh and the volume is given by the formula V=whV = \ell wh.

What is the surface area of a rectangular prism with length 22 cm, width 55 cm, and height 99 cm?
A rectangular prism with height 10 cm has a square base. The volume of the prism is 160 cm3^3. What is the side length of the square base?
A rectangular prism has a square base with area 3636 cm2^2. The surface area of the prism is 240240 cm2^2. Determine the volume of the prism.
A rectangular prism has length kk cm, width 2k2k cm, and height 3k3k cm, where k>0k > 0. The volume of the prism is x cm 3\text{x cm 3}. The surface area of the prism is x cm 2\text{x cm 2}. Determine the value of kk.

Solution

The surface area of a rectangular prism is given by the formula A=2w+2h+2whA=2\ell w+2\ell h+2wh.

Thus, the rectangular prism with length 2 cm, width 5 cm, and height 9 cm has surface area 2(2)(5)+2(2)(9)+2(5)(9)=20+36+90=1462(2)(5)+2(2)(9)+2(5)(9)=20+36+90=146 cm2^2.
The volume of a rectangular prism is given by the formula V=whV=\ell wh.

If the rectangular prism has a square base, then =w\ell =w and so V=2hV=\ell ^2h.

Substituting V=160V=160 cm3^3 and h=10h=10 cm, we get 160=2(10)160=\ell ^2(10) or 2=16\ell ^2=16, and so =4\ell =4 cm (since >0\ell >0).

Therefore, the side length of the square base of a rectangular prism with height 10 cm and volume 160 cm3^3 is 4 cm.
If a rectangular prism has a square base, then =w\ell=w.

Since the area of the base is 36 cm2^2, then 36=w=236=\ell \cdot w=\ell^2, and so =w=36=6\ell =w=\sqrt{36}=6 cm (since >0\ell>0).

If the surface area of this prism is 240 cm2^2, then substituting, we get 240=2(6)(6)+2(6)h+2(6)h240=2(6)(6)+2(6)h+2(6)h or 240=72+24h240=72+24h, and so h=2407224=7h=\dfrac{240-72}{24}=7 cm.

Thus, the volume of the prism is wh=(6)(6)(7)=252\ell wh=(6)(6)(7)=252 cm3^3.
Substituting into the formula for volume, we get x=k(2k)(3k)x=k(2k)(3k) or x=6k3x=6k^3.

Substituting into the formula for surface area, we get x=2(k)(2k)+2(k)(3k)+2(2k)(3k)x=2(k)(2k)+2(k)(3k)+2(2k)(3k) or x=4k2+6k2+12k2=22k2x=4k^2+6k^2+12k^2=22k^2.

Equating the two expressions that are each equal to xx and solving, we get 6k3=22k26k322k2=02k2(3k11)=0\begin{aligned} 6k^3& = 22k^2\\ 6k^3-22k^2& = 0\\ 2k^2(3k-11)& = 0\end{aligned} Since k>0k>0, then 3k11=03k-11=0 and so k=113k=\dfrac{11}{3}.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.