Question

Ravi walks northward upto 10m he turns left and walks 30m finally he turns left and walks 50m at what distance ravi is now from his starting position​

Answer 1

Given,

Ravi walks 10m in north direction.

Then he turns left and walks 30m.

Then he again turns left and walks 50m.

To find,

The distance between the starting point and the end point of the journey.

Solution,

As described in the diagram, Ravi starts his journey from point A and reaches point B by travelling 10 metres. Then from point B Ravi reaches point C by travelling 30 metres. At last, Ravi reaches point D by travelling 50 metres.

We have to calculate the distance between point A and point D.

For easier mathematical calculation, we dropped a perpendicular from the point A on the CD side and marked the intersection point as point E.

Now, ABCE is a four sided polygon where the angles ABC,BCE and AEC are already 90°.

So, angle BAE will be = 360°-(3×90°) = 90°

And, from visual observation we can say that consecutive sides are not equal.

Hence, we can say that the ABCE is a rectangle.

And, according to the properties of rectangle,

AB = CE = 10 m

BC = AE = 30 m

The CD is a straight line, so angle CED will be 180°.

We previously know that, angle AEC is 90°.

So, angle AED will be = (180°-90°) = 90°

Hence, ∆AED will be a right angled triangle.

In the AED right angled triangle,

AE = 30 m = Base

DE = CD-CE = 50-10 = 40 m = Height

AD = Hypotenuse = Let, x m

According to the Pythagoras theorem,

(x)² = (30)²+(40)²

x² = 900+1600

x² = 2500

x = 50

Hence, Ravi was 50 metres away from his starting point.