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BECE Mathematics 2011 Paper

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Question : 19

Total: 40

If

R=\;\frac{h}{2}+\;\frac{d^2}{8h}

R

d=8

h=6

$3\frac{1}{6}$
$4\frac{1}{3}$
$4\frac{3}{4}$
$4\frac{9}{16}$

Solution:

Concept:

Substitute the given values of

d

and

h

into the formula and simplify the expression.

Explanation:

The formula is

R = \frac{h}{2} + \frac{d^2}{8h}

.

Given

d = 8

and

h = 6

.

Substitute these values:

R = \frac{6}{2} + \frac{8^2}{8 \times 6}

Simplify the first term:

\frac{6}{2} = 3

Simplify the second term:

\frac{8^2}{8 \times 6} = \frac{64}{48}

Reduce

\frac{64}{48}

by dividing numerator and denominator by

16

:

\frac{64}{48} = \frac{4}{3}

Now add:

R = 3 + \frac{4}{3}

Write

\frac{4}{3}

as a mixed number:

\frac{4}{3} = 1\frac{1}{3}

So

R = 3 + 1\frac{1}{3} = 4\frac{1}{3}

Answer:

R = 4\frac{1}{3}

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