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| Linear
Equations and Word Problems |
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Word
problems that lead to simple linear equations |
Geometry word problems |
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Geometry word problems
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The
general procedure to solve a geometry word problem is:
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| 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. |
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Example:
The
angle g
of a triangle is twice as large as the
angle a,
and the angle b is
three-fourth of the
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angle g.
What are the angles measures?.
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| Solution: |
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given, g
= 2a
and b
= 3/4g
= 3/4 ·
2a =
3/2a |
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since,
a
+
b +
g
= 180° |
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then
a
+ 3/2a
+ 2a
= 180° |
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9/2a
= 180°
=> a
= 40° |
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b
= 3/2a
= 3/2 ·
40 =
60° |
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g
= 2a
= 2 · 40°
= 80° |
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Example: The
exterior angles that lie on the hypotenuse of a right triangle are in the
ratio of 11 : 16, find the angles.
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| Solution: |
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| 2880 -
16x = 990
+ 11x
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| 27x = 1890
=> x = a
= 70°,
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| 180° -
x
= 110°,
90°
+ x =
160°. |
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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.
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| Solution: |
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r2
=
(r
- 8)2
+ (c/2)2
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r2 = (r - 8)2
+ 122
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| r2
=
r2
- 16r
+ 64
+ 144
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| 16r
=
208 |
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r =
13 cm |
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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).
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| Solution:
Using similarity of the two right triangles, |
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(h
+ 4)
: h =
h
: (h - 3)
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| h2 =
(h
+ 4) ·
(h - 3)
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| h2 = h2
+ h
- 12
=> h
= 12
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| p
= h - 3
= 12 - 3
= 9, |
| q
= h
+ 4
= 12
+ 4
= 16,
c
= p +
q
= 25 |
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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.
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| Solution:
a
: b = 3
: 5 => a = (3/5) ·
b |
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A = a ·
b = (3/5)b
· b
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| [(3/5)b
+ 1] · (b - 4)
= (3/5)b
· b
- 39
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(3/5)b2
- 7/5b
- 4
= (3/5)b2
- 39 |
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7/5b =
35,
a = (3/5) ·
b = (3/5) · 35 |
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b = 25 cm,
a = 15 cm, |
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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?
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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.
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| Thus, number of diagonals of an n-sided
polygon, dn
= n
· (n - 3)
/ 2.
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| By substituting the given condition
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| (n
+ 3) · n
= n
· (n - 3)
+ 36
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n2
+ 3n = n2
- 3n
+ 36 |
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6n = 36
=> n =
6 |
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Example:
Of which regular polygon, the difference between the interior and
exterior angle, is 132°.
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Solution:
In a regular polygon a
= 360°/n
and a
+
b
= 180°,
therefore a
= b', as shows the picture.
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180n
- 360
- 360 =
132n
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48n
= 720 => n
= 15
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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.
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Solution: The
lateral area of the cylinder Slat =
2rp
· h = 2rp(r
+ 8),
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and the surface area S =
2B + Slat =
2r2p
+ 2rp(r
+ 8).
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given
Slat
: S = 5 :
8
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2rp(r
+ 8)
:
[2r2p
+ 2rp(r
+ 8)] =
5 :
8
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8r
+ 64 = 10r
+ 40
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2r
= 24 => r
= 12
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| Intermediate
algebra contents |
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