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प्रश्न
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.
उत्तर
Given: Direction of north = 5 m i.e. AC Direction of east = 12 m i.e. AB
To find: BC
According to Pythagoras Theorem,
In right angled Δ ABC
(BC)2 = (AC)2 + (AB)2
(BC)2 = (5)2 + (12)2
(BC)2 = 25 + 144
(BC)2= 169
∴ BC = `sqrt169=sqrt(13xx13)` = 13 m
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From given figure, In ∆ABC, If AC = 12 cm. then AB =?
Activity: From given figure, In ∆ABC, ∠ABC = 90°, ∠ACB = 30°
∴ ∠BAC = `square`
∴ ∆ABC is 30° – 60° – 90° triangle
∴ In ∆ABC by property of 30° – 60° – 90° triangle.
∴ AB = `1/2` AC and `square` = `sqrt(3)/2` AC
∴ `square` = `1/2 xx 12` and BC = `sqrt(3)/2 xx 12`
∴ `square` = 6 and BC = `6sqrt(3)`
