Lets assume other diagonal = 2x cm,
Since halves of diagonals and one side of rhombus form a right angled triangle with side as hypotenuse, we have:
(20)^2 =(12)^2+x^2 or x=?(20)^2-(12)^2 =?256=16 cm.
So, other diagonal = 32 cm.
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Satish was looking at a new building Kohli Towers constructed having height of 200 m. The angle of elevation of the top of building from a point on ground is 30Â°. What is the distance of the point from the foot of the building?
Given towers Angle APB = 30Â°. And AB = 200 m
AB/AP = tan 30Â° = 1/âˆš3
AP = (AB x âˆš3) m
200âˆš3 m = 200 x 1.73 = 346m
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Jaya Towers and Karuna towers are two tall buildings standing 400 metres apart. Two fast flying birds were sitting on top of these two towers and were aiming at catching the grains put up in one place. Jaya Tower's height is 300 metres and the grains 160 metres apart from Jaya Towers. Both the birds were flying down at the same speed and reached the grains spot at the same time. What is the approximate height of Karuna Towers?
Jaya Tower is 300 metre tall
Lets assume Karuna Tower's height be x
Distance of grains spot from Jaya Towers -- 160 metres.
Both the birds were flying down at the same speed and reached the grains spot at the same time
i.e Time taken for bird on Jaya tower to reach grain = Time taken for bird on Karuna tower to reach grain
Or Distance travelled by bird on Jaya tower / Speed of bird on Jaya tower = Distance travelled by bird on Karuna tower / Speed of bird on Karuna tower ->(1)
But as it is given that both the birds travelled with same speed, Speed of bird on Jaya tower = Speed of bird on Karuna tower. Applying this to equation (1) we get,
Distance travelled by bird on Jaya tower = Distance travelled by bird on Karuna tower -> (2)
Apply Pythagoras theorem, we know that (Distance travelled by bird on Jaya tower)2 = (300)2 + (160)2 -> (3)
Again by applying Pythagoras theorem to the other triangle. (Distance travelled by bird on Karuna tower)2 = (400-160)2 + x2 -> (4)
From equations 2,3 and 4 we get
(300)2 + (160)2 = (400-160)2 + x2
115600 = 57600 + x2
58000 = x2
240.83 m = x
Therefore x is approximately 240 m
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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?
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:
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.
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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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
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