Lesson Solved problems on area of regular triangles

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Solved problems on area of regular triangles


Problem 1

Find the area of a regular triangle,  if its side is  5 cm.

Solution
The area of a regular triangle with the side  a  is equal to  a%5E2%2Asqrt%283%29%2F4  (see the lesson  Formulas for area of a triangle  under the topic  Area and surface area  of the section  Geometry  in this site).  Therefore,  the area of the given regular triangle is equal to  5%5E2%2Asqrt%283%29%2F4 =~ 25%2A1.73%2F4 = 10.825 cm%5E2.
Answer.  The area of a regular triangle with the side length of  5 cm  is equal to  10.825 cm%5E2  (approximately).


Problem 2

Find the side of a regular triangle,  if its area is  9 sm%5E2.

Solution

Let  x  be the measure of the side of our triangle.
Since the area of a regular triangle with the side  x  is  x%5E2%2Asqrt%283%29%2F4,  you get the equation
x%5E2%2Asqrt%283%29%2F4 = 9.
Hence,  x%5E2 = 9%2A4%2Fsqrt%283%29   and   x = sqrt%289%2A4%2Fsqrt%283%29%29 =~ sqrt%289%2A4%2F1.73%29 = sqrt%2820.809%29 = 4.562.
Answer.  The side of a regular triangle with the area of  9 cm%5E2  is equal to  4.562 cm  (approximately).


Problem 3

Three regular triangles are constructed in the exterior of a right-angled triangle              
on its sides as shown in the  Figure 1.

Prove that the area of the triangle constructed on the hypotenuse is equal to
the sum of the areas of the two triangles constructed on the legs.

Solution

We are given a right-angled triangle  DELTAABC  (Figure 1).
Let  a  and  b  be the measures of its legs and  c  be the measure of its
hypotenuse. Then

a%5E2 + b%5E2 = c%5E2           (1)



      Figure 1.  To the  Problem 3

in accordance with the  Pythagorean Theorem  (see the lesson  The Pythagorean Theorem  under the topic  Pythagorean Theorem  of the section  Geometry  in this site).

The area of the regular triangle constructed on the leg  a  is  a%5E2%2Asqrt%283%29%2F4.  The area of the regular triangle constructed on the leg  b  is  b%5E2%2Asqrt%283%29%2F4.
The sum of the areas of these two triangles is  a%5E2%2Asqrt%283%29%2F4 + b%5E2%2Asqrt%283%29%2F4 = %28a%5E2+%2B+b%5E2%29%2Asqrt%283%29%2F4.
Due to the equality  (1),  it is exactly equal to  c%5E2%2Asqrt%283%29%2F4,  which is the area of the regular triangle constructed on the hypotenuse  c.
The solution is completed.


Problem 4

An arbitrary point is selected inside a regular triangle with the side measure  a.
Find the sum of distances from this point to the triangle sides and prove that this sum does not depend on the point inside the triangle.

Solution

Let  DELTAABC  be a regular triangle and  P  be an arbitrary point inside the triangle          
(Figure 2a).
The distance from the point  P  to any side of the triangle  DELTAABC  is the length
of the perpendicular drawn from the point  P  to this side.  So,  let us draw the
perpendiculars  PD,  PE  and  PF  from the point  P  to the sides of the triangle,
and let  d%5B1%5D = |PD|,  d%5B2%5D = |PE|,  d%5B3%5D = |PF| be the measures of these perpendiculars.
We need to find the sum of the measures  d%5B1%5D + d%5B2%5D + d%5B3%5D = |PD| + |PE| + |PD|.

Let us connect the point  P  with the vertices of the triangle by the segments


  Figure 2a.  To the  Problem 4          



    Figure 2b. To the solution
            of the Problem 4
 PA,  PB  and  PC  (Figure 2b),  and consider the triangles  DELTAPAB,  DELTAPBC  and  DELTAPAC.
The area of the entire original triangle  DELTAABC  is the sum of the areas of the triangles  DELTAPAB,  DELTAPBC  and  DELTAPAC.  So,  we can write
a%5E2%2Asqrt%283%29%2F4 = a%2Ad%5B1%5D%2F2 + a%2Ad%5B2%5D%2F2 + a%2Ad%5B3%5D%2F2.
Now,  cancel the common factor  a%2F2  in both sides.  You get
a%2Asqrt%283%29%2F2 = d%5B1%5D + d%5B2%5D + d%5B3%5D.
Note that   a%2Asqrt%283%29%2F2   is nothing else as the altitude  h  of the regular triangle with the side  a.  Thus we proved that
d%5B1%5D + d%5B2%5D + d%5B3%5D = h.
In other words,  the sum of distances from the arbitrary point inside the regular triangle to the sides of the triangle is equal to the altitude of the triangle.
This sum is the same for all points inside the triangle and does not depend on the location of the point inside the triangle.

Amazing fact,  isn't?   The solution is completed.


My other lessons on the topic  Area  in this site are
    - WHAT IS area?
    - Formulas for area of a triangle
    - Proof of the Heron's formula for the area of a triangle
    - One more proof of the Heron's formula for the area of a triangle
    - Proof of the formula for the area of a triangle via the radius of the inscribed circle
    - Proof of the formula for the radius of the circumscribed circle
    - Area of a parallelogram
    - Area of a trapezoid
    - Area of a quadrilateral
    - Area of a quadrilateral circumscribed about a circle  and
    - Area of a quadrilateral inscribed in a circle
under the topic  Area and surface area  of the section  Geometry,  and
    - Solved problems on area of triangles
    - Solved problems on area of right-angled triangles
    - Solved problems on the radius of inscribed circles and semicircles
    - Solved problems on the radius of a circumscribed circle
    - A Math circle level problem on area of a triangle
    - Solved problems on area of parallelograms
    - Solved problems on area of rhombis, rectangles and squares
    - Solved problems on area of trapezoids  and
    - Solved problems on area of quadrilaterals
under the topic  Geometry  of the section  Word problems.

For navigation over the lessons on  Area of Triangles  use this file/link  OVERVIEW of lessons on area of triangles.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link  GEOMETRY - YOUR ONLINE TEXTBOOK.

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