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A took 15 seconds to cross a rectangular field diagonally walking at the rate of 56 m/min and B took the same time to cross the same field along its sides walking at the rate of 72 m/min. What is the area of the field?
(A) 46 m²
(B) 50 m²
(C) 54 m²
(D) 64 m²
(E) 72 m²
(A) 46 m²
(B) 50 m²
(C) 54 m²
(D) 64 m²
(E) 72 m²
18 Mar 2021, 08:46
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Converting metre/minute to metre/second.
Let sides be a and b.
Diagonal be c.
Apply Pythagorean. Can isolate 2ab. Values of all other terms are known.
Area= ab
Posted from my mobile device
Let sides be a and b.
Diagonal be c.
Apply Pythagorean. Can isolate 2ab. Values of all other terms are known.
Area= ab
Posted from my mobile device
20 Mar 2021, 22:25
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A took 15 seconds to cross a rectangular field diagonally walking at the rate of 56 m/min and B took the same time to cross the same field along its sides walking at the rate of 72 m/min. What is the area of the field?
A
Distance=Speed·Time
\(Distance=\frac{56m}{min}·15 seconds=\frac{56m}{min}·\frac{1·min}{4}=\frac{56·m·1·min}{min·4}=\frac{56m}{4}=14m\)
B
Distance=Speed·Time
\(Distance=\frac{72m}{min}·15 seconds=\frac{72m}{min}·\frac{1·min}{4}=\frac{72·m·1·min}{min·4}=\frac{72m}{4}=18m\)
A and B formed a triangle with their walking: A the diagonal and B the two sides. A field is a right angle triangle, therefore the sides and diagonal are related using \(a^2+b^2=c^2\)
We don't have \(a^2+b^2\), but \((a+b)^2=a^2+b^2+2ab\) and we have a+b — the combined length of the two sides of the field.
\((a+b)^2-2ab=c^2\);
\((18)^2-2ab=(14)^2\);
\(324-2ab=196\);
\(324-196=2ab\);
\(128=2ab\);
\(64=ab\)
length·width=area
ab=area
A
Distance=Speed·Time
\(Distance=\frac{56m}{min}·15 seconds=\frac{56m}{min}·\frac{1·min}{4}=\frac{56·m·1·min}{min·4}=\frac{56m}{4}=14m\)
B
Distance=Speed·Time
\(Distance=\frac{72m}{min}·15 seconds=\frac{72m}{min}·\frac{1·min}{4}=\frac{72·m·1·min}{min·4}=\frac{72m}{4}=18m\)
A and B formed a triangle with their walking: A the diagonal and B the two sides. A field is a right angle triangle, therefore the sides and diagonal are related using \(a^2+b^2=c^2\)
We don't have \(a^2+b^2\), but \((a+b)^2=a^2+b^2+2ab\) and we have a+b — the combined length of the two sides of the field.
\((a+b)^2-2ab=c^2\);
\((18)^2-2ab=(14)^2\);
\(324-2ab=196\);
\(324-196=2ab\);
\(128=2ab\);
\(64=ab\)
length·width=area
ab=area
