Karan’s house is south facing. From the front door of his house Karan started walking and after walking 20 metres straight he turned to his left and walked 50 metres. After this he turned to his right and walked 80 metres. Find the distance between the point from where he finished his walk and the door of his house.
Answers
Answer:
50.6449510225
Step-by-step explanation:
it will be a sum of two triangles Hypotenuse .
and the triangles are of dimensions 20,10
and 40,80
- now let's find hypotenuse for 20,10 √20² + 10²
- now let's find hypotenuse for 80,40 √40²+80²
- now total sum is √500+√8000=50.6449510225
Answer:
The distance between the point where he finished his walk and the door of his house is 50√5 m.
Solution:
Let A be the door of Karan's house. After walking 20 meters straight, he reached point B. From B he turned left and walked 50 meters to reach point C. From C, he turned right and reached point D after walking 80 meters.
The diagram for the same is shown below.
From the figure, we have
AB = 20 m
BC = 50 m
CD = 80 m
Let us construct a rectangle ABCE. As we know, in a rectangle opposite sides are equal.
∴ AE = BC = 50 m
CE = AB = 20 m
We have to find the distance between the starting point and ending point i.e., between A and D. So, we have to find the length AD.
From the figure,
DE = CD + CE
= 80 m + 20 m
= 100 m
In ΔADE, applying Pythagoras theorem, we get
(Hypotenuse)² = (Perpendicular)² + (Base)²
AD² = DE² + AE²
AD² = (100)² + (50)²
AD² = 10000 + 2500
AD² = 12500
AD = √12500
AD = 50√5 m
Hence, the distance between the point where he finished his walk and the door of his house, AD = 50√5 m.
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