
ABC is an equilateral triangular plot. An electric pole stands at the vertex and makes an angle of \[{60^o}\] at either of the two vertices. If the same height of the triangle is 100m, then the height of the pole is
A) 100m
B) 150m
C) 200m
D) 300m

Image: A equilateral triangular plot ABC,
Answer
216k+ views
Hint: An equilateral triangle is also known as a regular triangle. It is a type of triangle in which all three sides are equal or they have the same length. An equilateral triangle also has equal angles on all three sides and is equal to \[{60^o}\] .
Formula used:
Let us assume that AP is the electric pole of height ‘h’
Therefore, to calculate the height of the pole formula to be used is:
\[\dfrac{{AB}}{h} = \cot {60^o}\]
Complete step-by-step solution:
Let us assume that the height of the electric pole is of height ‘h’
Then, \[\dfrac{{AB}}{h} = \cot {60^o}\]
\[ = AB = \dfrac{h}{{\sqrt 3 }} = AC = BC\]
Since \[AD = 100m\]
\[B{D^2} = A{B^2} - A{D^2}\]
\[(\dfrac{1}{2}) \times {(\dfrac{h}{{\sqrt 3 }})^2} = {(\dfrac{h}{{\sqrt 3 }})^2} - {100^2}\]
\[(\dfrac{{{h^2}}}{3}) = (\dfrac{{{h^2}}}{{12}}) = {100^2}\]
\[(\dfrac{{{h^2}}}{4}) = {100^2}\]
\[h = 200m\]
Note: Equilateral triangle exhibit certain properties such as:
1) All the three sides of an equilateral triangle are of the same length.
2) All the three angles that are present in an equilateral triangle are equal to \[{60^o}\] and are congruent to each other.
3) When a perpendicular is drawn from the vertex of the equilateral triangle to its other side, it bisects the triangle into two equal halves.
4) The ortho-center and the centroid in an equilateral triangle are present at the same point.
5) The perimeter of an equilateral triangle is equal to 3a.
6) The area of an equilateral triangle is equal to \[\dfrac{{\sqrt {3{a^2}} }}{4}\] .
7) An equilateral triangle is nothing but a polygon that has three sides.
Formula used:
Let us assume that AP is the electric pole of height ‘h’
Therefore, to calculate the height of the pole formula to be used is:
\[\dfrac{{AB}}{h} = \cot {60^o}\]
Complete step-by-step solution:
Let us assume that the height of the electric pole is of height ‘h’
Then, \[\dfrac{{AB}}{h} = \cot {60^o}\]
\[ = AB = \dfrac{h}{{\sqrt 3 }} = AC = BC\]
Since \[AD = 100m\]
\[B{D^2} = A{B^2} - A{D^2}\]
\[(\dfrac{1}{2}) \times {(\dfrac{h}{{\sqrt 3 }})^2} = {(\dfrac{h}{{\sqrt 3 }})^2} - {100^2}\]
\[(\dfrac{{{h^2}}}{3}) = (\dfrac{{{h^2}}}{{12}}) = {100^2}\]
\[(\dfrac{{{h^2}}}{4}) = {100^2}\]
\[h = 200m\]
Note: Equilateral triangle exhibit certain properties such as:
1) All the three sides of an equilateral triangle are of the same length.
2) All the three angles that are present in an equilateral triangle are equal to \[{60^o}\] and are congruent to each other.
3) When a perpendicular is drawn from the vertex of the equilateral triangle to its other side, it bisects the triangle into two equal halves.
4) The ortho-center and the centroid in an equilateral triangle are present at the same point.
5) The perimeter of an equilateral triangle is equal to 3a.
6) The area of an equilateral triangle is equal to \[\dfrac{{\sqrt {3{a^2}} }}{4}\] .
7) An equilateral triangle is nothing but a polygon that has three sides.
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