Question
Mathematics
In the figure below, $PR=32$PR=32 and $QR=14$QR=14. Find $PQ$PQ.
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From the given information, we can use the Pythagorean theorem to find $PR$PR and $QR$QR:
$P{R}^{2}=P{Q}^{2}+Q{R}^{2}$PR^(2)=PQ^(2)+QR^(2)
${32}^{2}=P{Q}^{2}+{14}^{2}$32^(2)=PQ^(2)+14^(2)
Simplifying this equation, we get:
$1024=P{Q}^{2}+196$1024=PQ^(2)+196
Subtracting 196 from both sides, we get:
$828=P{Q}^{2}$828=PQ^(2)
Taking the square root of both sides, we get:
$PQ=\sqrt{828}\approx \overline{)28.77}$PQ=sqrt828~~28.77
Therefore, $PQ\approx 28.77$PQ~~28.77.
Step-by-step solution:
1. Use the Pythagorean theorem to find $PR$PR: $P{R}^{2}=P{Q}^{2}+Q{R}^{2}$PR^(2)=PQ^(2)+QR^(2)
2. Substitute the given values for $PR$PR and $QR$QR: ${32}^{2}=P{Q}^{2}+{14}^{2}$32^(2)=PQ^(2)+14^(2)
3. Simplify the equation: $1024=P{Q}^{2}+196$1024=PQ^(2)+196
4. Solve for $PQ$PQ: $P{Q}^{2}=828$PQ^(2)=828
5. Take the square root of both sides: $PQ=\sqrt{828}\approx 28.77$PQ=sqrt828~~28.77
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