Verbal Reasoning - Direction Sense Test - Discussion

@All.

Let me explain it simply,

As in the answer pic (please see there) if we add the distances of AB and DC it would be 5 km(2+3) and the distances of BC and DE it would be 12km (10+2). Now, just extend AB and DE , we would get a point say F ( on their intersecting point). Now join A to E, we will get a right-angled triangle named triangle AFE.

Now by Pythagoras theorem, we will find out the line segment AE.
Hypotenuse^2= perpendicular^2 + base^2.
AE^2 = AF^2+ FE^2,
AE^2= 5^2 + 12^2,
AE^2= 25+144,
AE^2 = 169,
AE x AE= 13 x 13,
AE= 13.
So, 13 (option B) is the right answer.
Hope it helps

12km
.......................end
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. 5km
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.
.
Start

Now we want to find distance from startin g point i.e start to end.

We know that pythagarus theorem,

H^2 = side^2+side^2.

Therefore,

H = sqrt(25+144).

H = sqrt169.

H = 13km.

12km comes from adding north distance and similarly 5km.

No. Here we have to find the shortest distance from a to e. so that can be find out like this..

Total distance travel towards north 2+3= 5km
Total distance travel towards east 10+2= 12km
Therefore the shortest distance will sqrt of (5*5+12*12) = sqrt(169) = 13
I think u got my point. Have a nice day.

A taxi driver commented his journey from a point, and drove 10km towards north, and turned to his left and drove another 5km. After waiting to meet a friend here, he turned to his right and continued to drive another 10km. He has covered a distance of 5km to East then, in which direction would he be now.

@ Nagendra..
can you explain it in brief...
i think you are wrong
according to the figure given above there are two right angled triangles rite?
so we have find both AC and CE by using formula (hypotys^2=side^2+side^2) and then we have to calculate AC+CE which gives AE which is required solution...

When to use Pythagoras theorem and when to just add the distance. Can some one explain the difference? By simply adding the answer is 12, by the theorem it's 13.

Logically, the distance he travelled towards north will not be considered when calculating the distance covered, am I right?

Using Pythagoras theorem.

AC+CE = AE required solution.

To find AC, CE ?

(AC)^2 = (2)^2+(10)^2 = Square root of 104 = 10.19.

(CE)^2 = (3)^2+(2)^2 = Square root of 13 = 3.6.

So AC+CE = 13.79. But option gave 13.

Hope all of you got it.

If we assume short side be x then hypotenuse is (2x+6) and 3rd side is (2x+4).

By applying Pythagoras.
(2x+6)square = (2x+4)square + x square.
Finally, gives x square -8x-20 = 0.
Which gives x=10 and x= -2.
So, sides are 10, 24 and 26.

Hi this is calculated using Pythogoras theorem,

Traveled towards east = 10+2 = 12Km
Traveled towards north = 2+3 = 5 Km

We need to substitute in the formula.

= 5^2 + 12^2
= root of 25 + 144
= root of 169
= 13.

The hypotenuse of a right angle triangle is 6cm more than twice the shortage side. If the third side is 2cm less than the hypotenuse. Find the sides of the triangle.

Can anyone solve this problem?