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sherxon
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15 Jun 2021, 00:35
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (03:01) correct
62%
(02:25)
wrong
based on 52
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In the middle of a vast grassland, there is a building with length of 11 meters and width of 9 meters. A horse is tied by a 17-meter-long rope at exactly one corner of the building. What is the total area of grassland that the horse can feed on?
A. \(216.75 \pi\)
B. \(241.75 \pi\)
C. \(289 \pi - 99\)
D. \(300 \pi\)
E. \(215 \pi\)
A. \(216.75 \pi\)
B. \(241.75 \pi\)
C. \(289 \pi - 99\)
D. \(300 \pi\)
E. \(215 \pi\)
15 Jun 2021, 15:58
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Total area of grassland horse can feed on
\(= (\frac{3}{4})*\pi*17^2 + (\frac{1}{4})*\pi*6^2+ (\frac{1}{4})*\pi*8^2\)
\(= 216.75 \pi+25\pi\)
\(= 241.75\pi\)
\(= (\frac{3}{4})*\pi*17^2 + (\frac{1}{4})*\pi*6^2+ (\frac{1}{4})*\pi*8^2\)
\(= 216.75 \pi+25\pi\)
\(= 241.75\pi\)
Attachments
Untitled.png [ 8.15 KiB | Viewed 1600 times ]
16 Jun 2021, 01:02
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Solution:
Look at the diagram for reference.
The area which the horse will feed on is = Area of section 1 + Area of section 2 + Area of section 3
Area 1 is 3/4 of circle with radius 17m. Thus its area \(= \frac{3}{4} \times \pi \times 17^2\)
Area 2 is a quadrant with radius 8m. Thus its area \(= \frac{1}{4} \times \pi \times 8^2\)
Area 3 is a quadrant with radius 6m. Thus its area \(= \frac{1}{4} \times \pi \times 6^2\)
So, area the horse will feed on \(= \frac{3}{4} \times \pi \times 17^2 + \frac{1}{4} \times \pi \times 8^2 + \frac{1}{4} \times \pi \times 6^2\)
\(⇒ \frac{1}{4} \pi (3*17^2 + 8^2 + 6^2)\)
\(⇒ \frac{1}{4} \pi (867+64+36)\)
\(⇒ \frac{967 \pi }{ 4}\)
\(⇒ 241.75 \pi\)
Hence the right answer is Option B.
Look at the diagram for reference.
The area which the horse will feed on is = Area of section 1 + Area of section 2 + Area of section 3
Area 1 is 3/4 of circle with radius 17m. Thus its area \(= \frac{3}{4} \times \pi \times 17^2\)
Area 2 is a quadrant with radius 8m. Thus its area \(= \frac{1}{4} \times \pi \times 8^2\)
Area 3 is a quadrant with radius 6m. Thus its area \(= \frac{1}{4} \times \pi \times 6^2\)
So, area the horse will feed on \(= \frac{3}{4} \times \pi \times 17^2 + \frac{1}{4} \times \pi \times 8^2 + \frac{1}{4} \times \pi \times 6^2\)
\(⇒ \frac{1}{4} \pi (3*17^2 + 8^2 + 6^2)\)
\(⇒ \frac{1}{4} \pi (867+64+36)\)
\(⇒ \frac{967 \pi }{ 4}\)
\(⇒ 241.75 \pi\)
Hence the right answer is Option B.
Attachments
horse.png [ 22.42 KiB | Viewed 1471 times ]
