Linear Equations and Word Problems
    Word problems that lead to simple linear equations
      Geometry word problems
Geometry word problems
The general procedure to solve a geometry word problem is:
1. Draw a sketch of the geometric figure.
2. Label its elements using information given in the problem.
3. Choose the correct formula for the given problem (ex: perimeter, area, volume etc.).
4. Substitute the data from the problem into the formula.
5. Solve it for the unknown.

Example:   The angle g of a triangle is twice as large as the angle a, and the angle b is three-fourth of the

angle g. What are the angles measures?.
Solution:  
given,   g = 2a  and  b = 3/4g = 3/4 2a = 3/2a
          since,     a + b + g = 180
          then       a + 3/2a + 2a = 180
                       9/2a = 180    =>    a = 40
                       b = 3/2a = 3/2 40 = 60
                       g = 2a = 2 40 = 80
 

Example:  The exterior angles that lie on the hypotenuse of a right triangle are in the ratio of  11 : 16, find the angles.

Solution:
                  2880 - 16x = 990 + 11x
          27x = 1890    =>    x = a = 70,
  180 - x = 110,     90 + x = 160.
 

Example:  In a circle drawn is a chord whose length is 24 cm and its distance from the center of the circle is 8 cm less than the radius of the circle. Find the radius of the circle.

Solution:  
                      r2 = (r - 8)2 + (c/2)2
                      r2 = (r - 8)2 + 122
                      r2 = r2 - 16r + 64 + 144
                   16r = 208
                      r = 13 cm
 

Example:  The height of a right triangle is 3 units longer than the orthogonal projection of the shorter leg on the hypotenuse and 4 units smaller than the projection of the longer leg. Find lengths of the projections (segments).

Solution:  Using similarity of the two right triangles,
                      (h + 4) : h = h : (h - 3)
                      h2 = (h + 4) (h - 3) 
                      h2 = h2 + h - 12    =>   h = 12
 p = h - 3 = 12 - 3 = 9,
q = h + 4 = 12 + 4 = 16,     c = p + q = 25
 

Example:  Sides of a rectangle are in the ratio of 3 : 5. If the length is decreased by 4 cm and the width is increased by 1 cm, its area will be decreased by 39 cm2. Find dimensions of the rectangle.

Solution:        a : b = 3 : 5    =>   a = (3/5) b
                       A = a b = (3/5)b b
      [(3/5)b + 1] (b - 4) = (3/5)b b - 39
         (3/5)b2 - 7/5b - 4 = (3/5)b2 - 39
         7/5b = 35,         a = (3/5) b = (3/5) 35
           b = 25 cm,       a = 15 cm,

Example:  If number of sides of a polygon increases by 3 then the number of its diagonals increases by 18. Which polygon has this properties?

Solution:  Since an n-sided polygon has n vertices, and from each vertex can be drawn (n - 3) diagonals, then the total number of diagonals that can be drawn is n (n - 3). However, as each diagonal joins two vertices, this means that each diagonal will be drawn twice, so the expression must be divided by 2.
Thus, number of diagonals of an n-sided polygon,   dn = n (n - 3) / 2.
By substituting the given condition
        
      (n + 3) n n (n - 3) + 36
         n2 + 3n = n2 - 3n + 36
                6n = 36    =>    n = 6 

Example:  Of which regular polygon, the difference between the interior and exterior angle, is 132.

Solution:  In a regular polygon a = 360/n  and a + b = 180,  therefore a = b', as shows the picture.
                                          180n - 360 - 360 = 132n
                                          48n = 720     =>    n = 15
 

Example:  The ratio of the lateral area of a cylinder to its surface area is 5 : 8. Find the radius of the base if it is 8 cm shorter than the height of the cylinder.

Solution:  The lateral area of the cylinder    Slat = 2rp h = 2rp(r + 8),
                              and the surface area     S = 2B + Slat = 2r2p + 2rp(r + 8).
    given   Slat : S = 5 : 8
   2rp(r + 8) : [2r2p + 2rp(r + 8)] = 5 : 8
                                          
                                         8r + 64 = 10r + 40
                                         2r = 24    =>   r = 12
  
Intermediate algebra contents
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