Nisha walks 15 metre west and 8m North to reach the opposite corner of a rectangular field Mary walked from the same point diagonally to the opposite corner what is the distance covered by Mary
Answers
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Nisha walks 15 m West and 8 m North to reach the opposite corner of a rectangular field. Mary walked from the same point diagonally to the opposite corner.
Distance covered by Mary.
Firstly, let ABCD be the required rectangle.
Then, Nisha moves from A to B - 15 m and then B to C - 8 m
And, Mary walks from A to C. Then we have to find the distance between A and C.
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Now,
All the angles of a rectangle are 90°
=> angles A = B = C = D = 90°
Hence, in the fig.,
AB = 15 m
BC = 8 m
Angle B = 90°
Now, side opposite to 90° is called the hypotenuse
Hence, AC is the hypotenuse
That makes a right angled triangle ABC at angle B.
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Now, we use the Pythagoras' Theorem; which states that in a right angled triangle, the square of the hypotenuse is equal to the sum of square of other two sides, to solve the question.
AC² = AB² + BC²
=> AC² = 15² + 8²
=> AC² = 225 + 64
=> AC² = 289
=> AC = √289
=> AC = 17 m
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Hence, the distance covered by Mary = AC
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Heya mate!!! Here's your answer
______________________________
Nisha walks 15 m West and 8 m North to reach the opposite corner of a rectangular field. Mary walked from the same point diagonally to the opposite corner.
Distance covered by Mary.
Firstly, let ABCD be the required rectangle.
Then, Nisha moves from A to B - 15 m and then B to C - 8 m
And, Mary walks from A to C. Then we have to find the distance between A and C.
____________________
Now,
All the angles of a rectangle are 90°
=> angles A = B = C = D = 90°
Hence, in the fig.,
AB = 15 m
BC = 8 m
Angle B = 90°
Now, side opposite to 90° is called the hypotenuse
Hence, AC is the hypotenuse
That makes a right angled triangle ABC at angle B.
___________________
Now, we use the Pythagoras' Theorem; which states that in a right angled triangle, the square of the hypotenuse is equal to the sum of square of other two sides, to solve the question.
AC² = AB² + BC²
=> AC² = 15² + 8²
=> AC² = 225 + 64
=> AC² = 289
=> AC = √289
=> AC = 17 m
_____________________
Hence, the distance covered by Mary = AC
\boxed{\bold{\mathcal{\red{=\: 17\: m}}}}
=17m
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