# How do you find the area of a rectangle whose diagonal measure 12 inches and whose width measures 6 inches?

Apr 28, 2016

Area = width x height

$\implies a r e a = 6 \sqrt{3} \times 6 = 36 \sqrt{3} {\text{ inches}}^{2}$

#### Explanation:

Let the width be $w$
Let the height be $h$
Let the diagonal be $d$

Using Pythagoras

${d}^{2} = {w}^{2} + {h}^{2}$

We need to determine the value of $h$

Subtract ${w}^{2}$ from both sides

${d}^{2} - {w}^{2} = {w}^{2} - {w}^{2} + {h}^{2}$

But ${w}^{2} - {w}^{2} = 0$

${d}^{2} - {w}^{2} = {h}^{2}$

Square root both sides

$\sqrt{{h}^{2}} = \sqrt{{d}^{2} - {w}^{2}}$

But $\sqrt{{h}^{2}} = h$

$h = \sqrt{{d}^{2} - {w}^{2}}$

Substituting known values

$h = \sqrt{{12}^{2} - {6}^{2}}$

$h = \sqrt{108} \approx 10.39$ This is only an approximate value.
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Finding the exact value

$h = \sqrt{{s}^{2} \times {3}^{2} \times 3}$

$h = 2 \times 3 \times \sqrt{3}$

$h = 6 \sqrt{3} \text{ }$ as an exact value
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Area = width x height

$\implies a r e a = 6 \sqrt{3} \times 6 = 36 \sqrt{3} {\text{ inches}}^{2}$