Lesson Miscellaneous geometric problems

Miscellaneous geometric problems

Problem 1

At 1:35 pm, the angle between the minute hand and vertical direction up is  

     =  =   radians

(since the minute hand makes a full rotation in 60 minutes).


The angle between the hour hand and vertical direction up is

      +  =  +  = .

(since the hour hand makes a full rotation in 12 hours = 12*60 minutes).


The difference between these angles is

     -  =  -  = .


Thus the angle between the two hands at 1:35 pm is    radians.

It is more than the right angle  = , but less than the straight angle of  = .



Now, we have an obtuse triangle with the sides of 14 cm and 9 cm and the angle of   radians between them.


To find the distance between the tips of the hands, apply the cosine law equation

    d =  = 22.745 cm.


ANSWER.  The distance between the tips is about  22.745 cm.

Problem 2

Draw the radii of the circles from their centers to the intersection point C.
Connect the centers by the straight line AB.


You will get a triangle ABC with the sides 10 in, 16 in and 22 in.


Find its area using the Heron's formula

    area = ,

where s =  = 24 is the semi-perimeter.


Substituting this number into the formula, you will get

    area =  =  =  = 

         =  =  = 73.32121112... square inches.


Write the formula for the area of triangle ABC in other way, using the base AB = 22 in and the height h
from point C to the base

    area =  = 11*h  square inches.


You will get an equation

    11*h = .

Hence,  

    h = .


This value of h is half of the length CD;  so

    CD =  = 13.331 inches  (rounded).


At this point, the problem is just solved.


ANSWER.  The length CD is   = 13.331 inches  (rounded).

Problem 3

When we flatten out a cone into the sector of a circle,
the radius of this circle is the slant height of the cone,
while the arc of this sector is the circumference of the base circle of the cone.


So, we should find the slant height and the circumference of the base circle.


The slant height is the hypotenuse of the right triangle, whose legs are
the radius of the base circle and the height of the cone.

So, in our case, the slant height is  

    h =  =  =  =  = 13 cm.


The circumference of the base circle is

    C =  =  = 31.4159265 cm  (approximately).


Now the angle of the sector    is connected with the slant height and the arc length
by equation

    C = ,  

where    is in radians.  Here "h"  plays the role of the radius of the arc.


So, we find

     =  =  = 2.416609731 radians.


If you want to have it in degrees, convert radians to degrees

     =  =  = 138.4615 degrees (approximately).


ANSWER.  The angle of the sector is about 2.416609731 radians or 138.4615 degrees.

Problem 4

Make a sketch. In the sketch, you will see the circle representing the Earth,
with the radius of 3,958.8 miles.

Also, you will see the right-angled triangle with the hypotenuse length of

      R + h = 3958.8 +  = 3958.8 + 0.06856 = 3958.87 miles

and one leg of the length R = 3958.8 miles.


So, you can find the other leg of this triangle, which is a tangent to the Earth surface

    L =  =  = 23.5 miles (rounded).


It is the distance to the horizon from the top of the lighthouse.

Problem 5

The area of the regular hexagon ABCDEF is 6 times the area of an equilateral triangle 
with the side length 2.


So, the area of ABCDEF is   =  square units.


To get the area of the regular hexagon PQRSTU, we should subtract from the area ABCDEF 
the area of triangle PBQ 6 times.


PBQ is an isosceles triangle with the lateral side length of 1 and the concluded angle of 120 degrees.
Therefore, its area is

     = .


Thus the area of PQRSTU is  

     -  =  =  = 7.794228634 square units.    ANSWER

Problem 6

We have one circle and four rings: in all, 1 + 4 = 5 shapes of equal areas.


The area of the outer circle is   = .


The area of each shape is 1/5 of the total area,
since the partial areas of all shapes are equal to each other.


They want you find the width of the band for the ring between the third and the second circles 
(counting from the center).


For the area of the second circle you have

     = ,


where    is the radius of the second circle.  So, from this equation

     = ,

     =  =  meters.



For the area of the third circle you have

     = ,


where    is the radius of the third circle.  So, from this equation

     = ,

     =  =  meters.


Now the width of the band  " a "  is  the difference

    a =  -  =  = 1.706 meters  (rounded).    ANSWER

Problem 7

Let A, C and P be the given points A = (-4,0), C = (4,0).
Let P = (1,y) be the point, for which we want to find its coordinate 'y' such that

    |AP| + |CP| = 9.    (1)


Notice that line AC is horizontal (lies on x-axis of the (x,y) coordinate system).


Draw a perpendicular PD from vertex P of triangle APC to its base AC, 
so point D = (1,0) is the intersection of the perpendicular with AC.


Thus we have now triangle APC and two right-angled triangles ADP and CDP.

Let 'a' be the length CP and 'c' be the length AP:  

    a = |CP|,  c = |AP|.    (2)


We have  |AD| = 1 - (-4) = 5;  |CD| = 4 - 1 = 3.   (3)


Perpendicular PD is the common leg of triangles ADP and CDP, so we can write, using Pythagorean equation 

   |AP|^2 - |AD|^2 = |CP|^2 - |CD|^2.



Substituting here from relations (2) and (3), we get this equation

   c^2 - 5^2 = a^2 - 3^2,


which implies 

    c^2 - a^2 = 5^2 - 3^2,

    c^2 - a^2 = 16.    (4)



So, now we have this system of equations (1) and (4)

    c   + a   =  9.    (1')

    c^2 - a^2 = 16,    (4')


Now we are on a finish line to complete the solution.



In equation (4'),  factor left side as  c^2 - a^2 = (c+a)*(c-a) and replace (c+a) by 9,
based on equation (1').  Then instead the system (1'), (4') you will get the system

    a  + c =  9.    (1'')

    9(c-a) = 16.    (4'')



Open parentheses in (4'')  and multiply equation (1'')  by 9  (both sides)

    9c + 9a = 81.

    9c - 9a = 16,


Add      the two last equations and get  18c = 81 + 16 = 97,  c = 97/18.

Subtract the two last equations and get  18a = 81 - 16 = 65,  c = 65/18.



Now we can find 'y'

    y^2 = |PD|^2 = |CP|^2 - |AD|^2 = a^2 - 3^2 =  =  = ,

    y =  = 2.010005835,  or  y = 2.01, approximately.


ANSWER.  y =  = 2.010005835,  or  y = 2.01, approximately.