Lesson Try to solve these nice Geometry problems !

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Try to solve these nice Geometry problems !


Problem 1

Two sides of triangle  ABC  are  AB = 34 cm  and  AC = 25 cm and their included angle measures  62°.
Find the distance of the orthocenter to side  AB.

Solution


        In this problem,  the given data looks like to be disconnected one from the other,
        so,  it seems that it is unsolvable.

        Nevertheless,  the solution does exist and is quite beautiful,  although not obvious,  from the first glance. 


So, we have a triangle ABC with the sides AB = 34 cm and AC = 25 cm.
Their included angle A is 62°, so we can find the length of the third side BC
opposite to angle A.  Use the cosine law

    BC = sqrt%28AB%5E2+%2B+AC%5E2+-+2%2AAB%2AAC%2Acos%28A%29%29 = sqrt%2834%5E2+%2B+25%5E2+-+2%2A34%2A25%2A0.46947156278%29 = 31.3512096 cm.


Now, having the lengths of the three sides of triangle ABC, we can find its area,
using the Heron's formula. In order for do not bother with calculations, I will use 
one of numerous online calculators, 
https://www.omnicalculator.com/math/herons-formula



It gives the area of triangle ABC

    area(ABC) = 375.253 cm^2.


Other online calculators

https://www.inchcalculator.com/herons-formula-calculator/
https://www.wolframalpha.com/widgets/view.jsp?id=7ac490665df1b278eb748160468147bc

give practically the same value.



Now we can determine the radius of the circumscribed circle around triangle ABC

    R = %28a%2Ab%2Ac%29%2F%284%2Aarea%29 = %2834%2A25%2A31.3512096%29%2F%284%2A375.253%29 = 17.7537  cm  (rounded).



Now the distance from the orthocenter to the side AB is the leg of the right angled triangle,
whose hypotenuse is  R = 17.7537 cm  and the other leg is half the length of the side AB.


So, we write

    the distance from the orthocenter to the side AB = sqrt%2817.7537%5E2+-+%2834%2F2%29%5E2%29 = 5.118 cm  (rounded).


ANSWER.  The distance from the orthocenter to the side AB  is  5.118 cm (rounded).

Thus,  all the data was woven into one logical thread that led to a complete solution.

So,  we can celebrate the victory.


Problem 2

The circle circumscribes an equilateral triangle.  The area of the circle is  12pi.
What is the triangle's perimeter ?

Solution 

For any triangle with the side lengths 'a', 'b' and 'c', the radius of the circumscribed circle is

    R = %28a%2Ab%2Ac%29%2F%284%2AA%29,     (1)


where A is the area of the triangle.  In our case, all thee sides are congruent a = b = c,
so the radius of the circumscribed circle is 

    R = a%5E3%2F%284A%29.     (2)


The area of the circumscribed circle is  A = a%5E2%2A%28sqrt%283%29%2F4%29 square units.

Substituting it into the formula (1), we get

    R = a%5E3%2F%284%2Aa%5E2%2A%28sqrt%283%29%2F4%29%29 = a%2Fsqrt%283%29.    (3)
 

We are given that  the area of the circle is 12%2Api, so we can write this equation

    pi%2AR%5E2 = 12%2Api.


Cancel  pi  in both sides to get

    R%5E2 = 12.


Substitute here R from (3).  You will get

    a%5E2%2F3 = 12.


Simplify and find 'a'

    a%5E2 = 3*12 = 36,   a = sqrt%2836%29 = 6 units.


Hence, the perimeter of the triangle is 3*6 = 18 units.


ANSWER.  The perimeter of the triangle is 18 units.

Thus we produced short, straightforward, compact and elegant Geometry solution.


Problem 3

The center of a circle is at  (-3,-2).  If a chord of length  4  is bisected at  (3,1),  find the length of the radius.

Solution 

Let P = (-3,-2) be the center of the circle.

Let C = (3,1) be the midpoint of the chord.

Let point A be one of the intersections of the chord with the circle.


Triangle PCA is a right-angled triangle with right angle at vertex C.


The leg CA has the length of 4/2 = 2 units  (half the length of the chord).


The leg PC has the length  sqrt%28%283-%28-3%29%29%5E2+%2B+%281-%28-2%29%29%5E2%29 = sqrt%286%5E2+%2B+3%5E2%29 = sqrt%2836%2B9%29 = sqrt%2845%29.


Therefore the length of the radius, which is the hypotenuse of this right-angled triangle PCA, is


    r = |PA| = sqrt%28+PC%5E2+%2B+CA%5E2%29 = sqrt%2845+%2B+4%29 = sqrt%2849%29 = 7.


ANSWER.  The radius is 7 units long.


My other additional lessons on miscellaneous  Geometry problems in this site are
    - Find the rate of moving of the tip of a shadow
    - A radio transmitter accessibility area
    - Miscellaneous geometric problems
    - Miscellaneous problems on parallelograms
    - Remarkable properties of triangles into which diagonals divide a quadrilateral
    - A trapezoid divided in four triangles by its diagonals
    - A problem on a regular heptagon
    - The area of a regular octagon
    - The fraction of the area of a regular octagon
    - Find the angle between sides of folded triangle
    - A problem on three spheres
    - A sphere placed in an inverted cone
    - An upper level Geometry problem on special (15°,30°,135°)-triangle
    - A great Math Olympiad level Geometry problem
    - Nice geometry problem of a Math Olympiad level
    - OVERVIEW of my additional lessons on miscellaneous advanced Geometry problems

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


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