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# a rectangular park is 260 m by 480 m. Danny usually trains by running 5 circuits around the edge of the park. After heavy rain, two adjacent

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a rectangular park is 260 m by 480 m. Danny usually trains by running 5 circuits around the edge of the park. After heavy rain, two adjacent sides are too muddy to run along, so he runs a triangular path along the other two sides and the diagonal. Danny does 5 circuits of this path for training. Show that danny runs about 970 metres less than his usual training session.

Guest Jun 4, 2015

#1
+26625
+10

Usual training distance = 5*2*(260 + 480) m

Diagonal distance = √(2602 + 4802) m

Heavy rain training distance = 5*(260 + 480 + √(2602 + 4802)) m

Difference between these:

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{260}}{\mathtt{\,\small\textbf+\,}}{\mathtt{480}}\right){\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}\left({\mathtt{260}}{\mathtt{\,\small\textbf+\,}}{\mathtt{480}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{{\mathtt{260}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{480}}}^{{\mathtt{2}}}}}\right) = {\mathtt{970.531\: \!187\: \!208\: \!763\: \!874\: \!4}}$$

difference = 971 m to nearest metre

.

Alan  Jun 4, 2015
Sort:

#1
+26625
+10

Usual training distance = 5*2*(260 + 480) m

Diagonal distance = √(2602 + 4802) m

Heavy rain training distance = 5*(260 + 480 + √(2602 + 4802)) m

Difference between these:

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{260}}{\mathtt{\,\small\textbf+\,}}{\mathtt{480}}\right){\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}\left({\mathtt{260}}{\mathtt{\,\small\textbf+\,}}{\mathtt{480}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{{\mathtt{260}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{480}}}^{{\mathtt{2}}}}}\right) = {\mathtt{970.531\: \!187\: \!208\: \!763\: \!874\: \!4}}$$

difference = 971 m to nearest metre

.

Alan  Jun 4, 2015

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