AD, EF and BD are perpendicular to AB.

AD = 3 and BC = 5

What is the length h?

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- May 8th 2006, 01:06 PMNatasha1Symmetry
AD, EF and BD are perpendicular to AB.

AD = 3 and BC = 5

What is the length h? - May 8th 2006, 01:26 PMOReillyQuote:

Originally Posted by**Natasha1**

- May 8th 2006, 01:31 PMCaptainBlackQuote:

Originally Posted by**Natasha1**

$\displaystyle

\frac{h}{3}=\frac{DB}{AB}\ \ \ \dots (1)

$

Also trianglea ABC and ADF are similar:

$\displaystyle

\frac{h}{5}=\frac{AB-DB}{AB}

$

So dividing these:

$\displaystyle

\frac{5}{3}=\frac{DB}{AB}\ \frac{AB}{AB-DB}=\frac{DB}{AB-DB}

$

Hence:

$\displaystyle

DB=\frac{5}{8}AB

$

Substituting this into $\displaystyle (1)$ then gives:

$\displaystyle

\frac{h}{3}=\frac{5}{8}

$,

or:

$\displaystyle

h=\frac{3\times 5}{8}=\frac{15}{8}

$

RonL