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