
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
threefourth 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: 
r^{2}
=
(r
 8)^{2}
+ (c/2)^{2}

r^{2} = (r  8)^{2}
+ 12^{2}

r^{2}
=
r^{2}
 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)

h^{2} =
(h
+ 4) ·
(h  3)

h^{2} = h^{2}
+ 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 cm^{2}. 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)b^{2
}  7/5b
 4
= (3/5)b^{2}
 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
nsided 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 nsided
polygon, d_{n}
= n
· (n  3)
/ 2.


By substituting the given condition


(n
+ 3) · n
= n
· (n  3)
+ 36

n^{2
}
+ 3n = n^{2
}
 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 S_{lat} =
2rp
· h = 2rp(r
+ 8),

and the surface area S =
2B + S_{lat} =
2r^{2}p
+ 2rp(r
+ 8).


given
S_{lat}
: S = 5 :
8

2rp(r
+ 8)
:
[2r^{2}p
+ 2rp(r
+ 8)] =
5 :
8


8r
+ 64 = 10r
+ 40

2r
= 24 => r
= 12












Intermediate
algebra contents 



Copyright
© 2004  2020, Nabla Ltd. All rights reserved. 