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# Height and Distance Questions

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A tree of height 36m is on one edge of a road broke at a certain height. It fell in such a way that the top of the tree touches the other edge of the road. If the breadth of the road is 12m, then what is the height at which the tree broke?

A16

B24

C12

D18

Explanation:

If tree is broken then becomes a right angled triangle.
Lets assume the height of broken tree =y
and length of broken part=x
now, x + y = 36 ------------ (1)
x^2 - y^2=144
(x+y)(x-y)=144
x-y = 4 ----------- (2)

By solving eq. (1) & (2)
=>x=20 & y=16

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Two vertical ladders length of 6 m and 11 m are kept vertically at a distance of 12 m. Find the top distance of both ladders?

A13

B8

C11

D15

Explanation:

So distance between the top points = AD = sqrt(12^2+5^2) = 13

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Two poles of height 6 meters and 11 meters stand on a plane ground. If the distance between their feet is 12 meters then find the difference in the distance between their tops:

A12m.

B5m.

C13m.

D11m.

Explanation:

Distance between their tops = sqrt( 12^2+(11-6)^2)
= sqrt 169
= 13m

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If a ladder is 100 m long and distance between bottom of ladder and wall is 60 m. What is the maximum size of cube that can be placed between the ladder and wall.

A34.28

B24.28

C21.42

D28.56

Explanation:

8/6 = tan(y)
x/6*10 -x =tan(y)

8/6=x/60-x
x=34.28

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A ramp makes an angle of 60 degrees with the ground. If I walk up 200 metres on the ramp, at what height from the ground will I reach?

A100 m

B200 m

C150 m

D173.2 m

E120 m

Explanation:

Since 200 metres is travelled along the ramp,
Hypotenuse =200
Angle given = 60 degree
Therefore applying trigonometry,
sine x = Hypotenuse/height [Formula]
sin 60 = 200/h
Therefore, h=(sqrt(3)/2)*200=173.2 metres.

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An ant smartly moves across a staircase taking the shortest distance. Calculate the distance it takes to reach top to B from A. Given that staircase consist of 2 steps. It also known that the length,breadth and height is 6cm,2cm and 1cm respectively.

A3

B6 _/3

C6 _/2

DNone of these

Explanation:

Assume there is a staircase with 2 steps length = 6cm and breadth and height are 1 and 2
so the shortest distance is from left end of the 1st step to right end of the 2nd step which forms a right angled triangle, triangle side are AB, AC and BC.
By using pythagoras theorem,
AC^2 = AB^2+BC^2
AC^2 = 6^2+(2+1+2+1)^2 [height + breath of 2 steps]
AC^2 = 36+36
AC = 6 _/ 2

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The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length. The perpendicular distance between the parallel sides of the trapezium is 12 cm. If the height of the pillar is 20 cm, then the total area, in sq cm, of all six surfaces of the pillar is

A1300

B1340

C1480

D1520

Explanation:

Here is no explanation for this answer

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