An isoceles triangle is to be cut from one edge of a square lamina such that the remaining portion when suspended from the apex P of the cut will remain in equilibrium in any position. The value of h is ?
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See the diagram.
The isosceles triangle is EFG. It is symmetrical around Line L2. If the remaining piece GDABCFE is suspended along L2 (EK), it will be balanced in equilibrium.
If it is suspended along L1 (IJ) also, it should be in balance. Then the mass of the rectangular piece ABIJ = mass of JICFEGD.
As the mass is proportional to the area, (density constant), we equate the areas.
Ar(ΔEFG) = 1/2 * (a-h) * x
Ar(ABIJ) = a * h
Ar(JICFEGD) = (a-h) a - (a-h)x/2
= a² - a h - a x/2 + x h/2
=> a² - a h + x h/2 - a x/2 = a h
=> h (4 a - x) = a (2a - x)
=> h = (2a - x) a /(4a - x)
If the base of isosceles triangle x = CD = a, then
h = a * a / (4a-a) = a/3
The isosceles triangle is EFG. It is symmetrical around Line L2. If the remaining piece GDABCFE is suspended along L2 (EK), it will be balanced in equilibrium.
If it is suspended along L1 (IJ) also, it should be in balance. Then the mass of the rectangular piece ABIJ = mass of JICFEGD.
As the mass is proportional to the area, (density constant), we equate the areas.
Ar(ΔEFG) = 1/2 * (a-h) * x
Ar(ABIJ) = a * h
Ar(JICFEGD) = (a-h) a - (a-h)x/2
= a² - a h - a x/2 + x h/2
=> a² - a h + x h/2 - a x/2 = a h
=> h (4 a - x) = a (2a - x)
=> h = (2a - x) a /(4a - x)
If the base of isosceles triangle x = CD = a, then
h = a * a / (4a-a) = a/3
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