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in the diagram below, \\( \\overrightarrow{ol} \\) is perpendicular to …

Question

in the diagram below, \\( \overrightarrow{ol} \\) is perpendicular to \\( \overline{dx} \\).
find the length of \\( \overline{ox} \\).
if entering your answer as a decimal, round your final answer to the nearest hundredth.
\\( ox = \\) \\( \square \\) units

Explanation:

Step1: Let \( OX = x \), then \( OD = 13 - x \)

We know \( \triangle BOX \) and \( \triangle POD \) are right triangles. By Pythagoras, \( BX^2 + OX^2 = BO^2 \) and \( PD^2 + OD^2 = PO^2 \). Also, \( BX = PD \) (since \( OL \perp DX \) and \( BX \perp DX \), \( PD \perp DX \), so \( BX \) and \( PD \) are both perpendicular to the same line, hence equal in length). Let \( BX = PD = y \).

Step2: Set up equations

For \( \triangle BOX \): \( y^2 + x^2 = 12^2 = 144 \)
For \( \triangle POD \): \( y^2 + (13 - x)^2 = 10^2 = 100 \)
Subtract the second equation from the first:
\( y^2 + x^2 - (y^2 + (13 - x)^2) = 144 - 100 \)
Simplify: \( x^2 - (169 - 26x + x^2) = 44 \)
\( x^2 - 169 + 26x - x^2 = 44 \)
\( 26x - 169 = 44 \)
\( 26x = 44 + 169 = 213 \)
\( x = \frac{213}{26} \approx 8.19 \)

Answer:

\( 8.19 \)