We begin by drawing a diagram: [asy]
import three;
triple A = (0,0,0);
triple B = (0,0,44);
triple C = (32,0,44);
triple D = (32,0,0);
triple Ep = (32,24,0);
triple F = (0,24,0);
triple G = (0,24,44);
triple H = (32,24,44);
draw(A--B--C--D--cycle);
draw(A--F--G--B,dashed);
draw(A--D--H--G,dashed);
draw(G--H,dashed);
label("A",A,dir(-140));
label("B",B,dir(-140));
label("C",C,dir(40));
label("D",D,dir(-40));
label("E",Ep,dir(40));
label("F",F,dir(-40));
label("G",G,dir(-40));
label("H",H,dir(40));
label("32",(D+H)/2, dir(0));
[/asy] We see that $\triangle ABC$ and $\triangle AFG$ are similar right triangles, so $\frac{AB}{AF}=\frac{BC}{FG}$. Since $AB=32$ and $BC=24$, we substitute and solve to find $FG = 32 \cdot \frac34 = 24$. Similarly, we can find that $AF = 24 \cdot \frac34 = 18$.
Now we solve for the length of $\overline{BE}$ by using the Pythagorean Theorem on $\triangle BFE$: \begin{align*}
BE^2&=BF^2+FE^2\\
&=18^2+32^2\\
&=324+1024\\
&=1348.
\end{align*}Taking the square root of both sides gives $BE = \sqrt{1348} \approx \boxed{36.7}$ cm.
The length of the prism is 32 cm, the width is 24 cm, the height is 44 cm, and the length of the diagonal of the base, segment BH , is 40 cm. Find the length of the diagonal of the rectangular prism, segment BE . Round the answer to the nearest tenth.
2 answers
Chem Tech makes paper cones used to filter liquids in lab experiments. To make the paper cones, the company starts with the net of the shape, where the slant height is needed. The company wants the height of the cones to be 6 cm. If they make the diameter 13 cm, what must the slant height be? Use the Pythagorean Theorem and round the answer to the nearest tenth.