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प्रश्न
Rahul walks 12 m north from his house and turns west to walk 35 m to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk while returning?
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उत्तर
Let Rahul walked x m, while returning home.
In ΔABC, by using Pythagoras theorem, we get
AC2 = AB2 + BC2
⇒ AC2 = (12)2 + (35)2
⇒ AC2 = 144 + 1225 = 1369
⇒ AC = `sqrt(1369)`
⇒ AC = 37 m
| 37 | |
| 3 | `bar13` `bar69` |
| 9 | |
| 67 | 469 |
| 469 | |
| 0 |
Hence, Rahul walked 37 m distance for returning to his house diagonally.
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संबंधित प्रश्न
Find the square of the given number.
93
Write a Pythagorean triplet whose one member is 18.
Observe the following pattern
22 − 12 = 2 + 1
32 − 22 = 3 + 2
42 − 32 = 4 + 3
52 − 42 = 5 + 4
and find the value of
1112 − 1092
Observe the following pattern \[1^2 = \frac{1}{6}\left[ 1 \times \left( 1 + 1 \right) \times \left( 2 \times 1 + 1 \right) \right]\]
\[ 1^2 + 2^2 = \frac{1}{6}\left[ 2 \times \left( 2 + 1 \right) \times \left( 2 \times 2 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 = \frac{1}{6}\left[ 3 \times \left( 3 + 1 \right) \times \left( 2 \times 3 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 + 4^2 = \frac{1}{6}\left[ 4 \times \left( 4 + 1 \right) \times \left( 2 \times 4 + 1 \right) \right]\] and find the values :
12 + 22 + 32 + 42 + ... + 102
Observe the following pattern \[1^2 = \frac{1}{6}\left[ 1 \times \left( 1 + 1 \right) \times \left( 2 \times 1 + 1 \right) \right]\]
\[ 1^2 + 2^2 = \frac{1}{6}\left[ 2 \times \left( 2 + 1 \right) \times \left( 2 \times 2 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 = \frac{1}{6}\left[ 3 \times \left( 3 + 1 \right) \times \left( 2 \times 3 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 + 4^2 = \frac{1}{6}\left[ 4 \times \left( 4 + 1 \right) \times \left( 2 \times 4 + 1 \right) \right]\] and find the values :
52 + 62 + 72 + 82 + 92 + 102 + 112 + 122
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