Symbiosis National Aptitude Test SNAP 2015 Paper

Question : 9 of 150

Marks: +1, -0

A student moves

√2 x

km East from his residence and then moves x km North. He, then goes x km North-East and finally he takes a turn of 90° towards right and moves a distance x km and reaches his school. What is the shortest distance of the school from his residence?

$(2 √2 +1)x$ km
3x km
$2 √2 x$ km
$3 √2 x$ km

Validate

Solution:

In fig, O is residence and D is school.
In ΔBCD,

BD^2

BC^2 + CD^2

x^2 + x^2

⇒ BD =

√2 x

⇒ BD = AE =

√2 x

∴ OE = OA + AE =

√2 x + √2 x

2 √2 x

Since BA = DE = x
In ΔOED,

OD^2

OE^2 + DE^2

∴ Minimum distance,
OD =

√{(2√2x)^2+x^2}

√{8x^2+x^2}

= 3x km

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