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Problem 829

# In the rectangular solid shown, $$\underline{D A}=4$$ in. $$\underline{D C}=3$$, and $$\underline{\mathrm{GC}}=12$$. (a) Find the length of $$\underline{\mathrm{CA}}$$, a diagonal of the base $$\mathrm{ABCD}$$. (b) Using the result found in part (a), find the length of $$\underline{G A}$$. a diagonal of the solid. (c) If $$\underline{\mathrm{DA}}=\ell, \underline{\mathrm{DC}}=\mathrm{w}$$, and $$\underline{\mathrm{GC}}=\mathrm{h}$$, represent the length of

Expert verified
In summary, we found: (a) The length of the diagonal of the base, $$\underline{\mathrm{CA}} = 5$$. (b) The length of the diagonal of the solid, $$\underline{G A} = 13$$. (c) The length of the diagonal of the solid, $$\underline{G A}$$, can be represented in terms of $$\ell, \mathrm{w}$$, and $$\mathrm{h}$$ as: $$\underline{G A} = \sqrt{h^2 + \ell^2 + w^2}$$.
See the step by step solution

## Step 1: Finding the length of $$\underline{\mathrm{CA}}$$ (diagonal of the base)

Using the Pythagorean Theorem, we can find the length of the diagonal of the base, $$\underline{\mathrm{CA}}$$, on triangle $$\triangle DAC$$. $\mathrm{CA^2 = DA^2 + DC^2}$ Substitute values $$\underline{D A}=4$$ and $$\underline{D C}=3$$: $\mathrm{CA^2 = 4^2 + 3^2 = 16 + 9 = 25}$ So, taking the square root of both sides: $\underline{\mathrm{CA}} = \sqrt{25} = 5$ The length of the diagonal of the base, $$\underline{\mathrm{CA}}$$, is 5 inches.

## Step 2: Finding the length of $$\underline{G A}$$ (diagonal of the solid)

We can find the length of the diagonal of the solid, $$\underline{G A}$$, by using the Pythagorean Theorem again on triangle $$\triangle GCA$$. $GA^2 = GC^2 + CA^2$ Substitute values $$\underline{\mathrm{GC}}=12$$ and the result from part (a), $$\underline{\mathrm{CA}}=5$$: $GA^2 = 12^2 + 5^2 = 144 + 25 = 169$ So, taking the square root of both sides: $\underline{G A} = \sqrt{169} = 13$ The length of the diagonal of the solid, $$\underline{G A}$$, is 13 inches.

## Step 3: Generalizing the length of $$\underline{G A}$$ in terms of $$\ell, \mathrm{w}$$, and $$\mathrm{h}$$

Using our work from step 2, we can generalize our expression for the diagonal of the solid, $$\underline{G A}$$: $GA^2 = GC^2 + CA^2$ Now, we go back to step 1 and generalize our expression for $$\underline{\mathrm{CA}}^2$$ in terms of $$\ell$$ and $$w$$: $CA^2 = DA^2 + DC^2 \Rightarrow CA^2 = \ell^2 + w^2$ Substituting the values $$\underline{D A} = \ell$$, $$\underline{D C}=w$$, and $$\underline{\mathrm{GC}}=h$$ back into our expression, we get: $GA^2 = h^2 + (\ell^2 + w^2)$ Taking the square root of both sides, we find: $\underline{G A} = \sqrt{h^2 + \ell^2 + w^2}$ The length of the diagonal of the solid, $$\underline{G A}$$, can be represented in terms of $$\ell, \mathrm{w}$$, and $$\mathrm{h}$$ as: $\underline{G A} = \sqrt{h^2 + \ell^2 + w^2}$

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