 # Similar Triangles Sample Problems Lecture Notes page 1

```Lecture Notes
Similar Triangles
page 1
Sample Problems
1. The triangles shown below are similar. Find the exact values of a and b shown on the picture below.
2. Consider the picture shown below.
a) Use the Pythagorean Theorem to …nd the value of a.
b) Prove that the triangles ABE and ACD are similar.
c) Use similar triangles to …nd the value of x.
d) Find the value of b.
3. a) A person is standing 40 ft away from a street light that is 30 ft tall. How tall is he if his shadow
is 10 ft long?
b) A 6 ft tall person is standing 24 ft away from a street light that is 15 ft tall. How long is her
4. Prove the following statement. Let ABC be any right triangle, the right angle at point C. The
altitude drawn from C to the hypotenuse splits the triangle into two right triangles that are similar
to each other and to the original triangle.
5. The picture below shows a right triangle. Find the length of h; the height drawn to the hypotenuse.
Last revised: January 6, 2013
Lecture Notes
Similar Triangles
page 2
6. Find x, y, and h based on the picture below.
Practice Problems
1. The picture below shows two similar right triangles. Find the exact values of x and y.
2. The picture below shows two similar right triangles. Find the exact value of a.
3. Find the value of x based on the …gures below.
Last revised: January 6, 2013
Lecture Notes
Similar Triangles
page 3
4. a) A person is standing 24 ft away from a street light that is 25 ft tall. How tall is he if his shadow
is 6 ft long?
b) A 5:2 ft tall person is standing 20 ft away from a street light that is 15:6 ft tall. How long is her
5. Find the exact value of x, y, and z, based on the …gures shown below.
6. Find the exact value of a, b, and h, based on the picture shown below.
Last revised: January 6, 2013
Similar Triangles
Lecture Notes
page 4
1. a =
40
28
in; b =
in
7
5
2. a) 13 ft
b) see solutions c) 10 ft
3. a) 6 ft
b) 16 ft
d) 13 ft
4. see solutions
5. 30 units
6. h =
420
400
441
; x=
; y=
29
29
29
1. x =
15
39
= 3: 75; y =
= 9: 75
4
4
2. 105 in
3. a)
35
= 8: 75
4
4. a) 5 ft
5. a) x =
b)
60
7
b) 10 ft
24
32
18
= 4: 8; y =
= 6: 4; z =
= 3: 6
5
5
5
6. a) h = 12; a = 15; b = 20
b) x = 60; y = 25; z = 144
b) h = 168; a = 175; b = 600
Last revised: January 6, 2013
Similar Triangles
Lecture Notes
page 5
Sample Problems - Solutions
1. The triangles shown below are similar. Find the exact values of a and b shown on the picture below.
Solution: In similar triangles, the ratios of corresponding sides are preserved. To …nd a; we write
the ratio
side opposite angle
side opposite angle
for both triangles.
side opposite angle
side opposite angle
=
a
8
=
5
7
We now solve the equation for a.
a
8
=
5
7
7a = 40
40
a =
7
Similarly, we can …nd b by writing the ratio
multiply both sides by 35
divide by 7
side opposite angle
side opposite angle
side opposite angle
side opposite angle
=
for both triangles.
4
b
=
5
7
We now solve the equation for b.
4
b
=
5
7
28 = 5b
28
b =
5
Thus a =
multiply both sides by 35
divide by 5
40
28
in and b =
in.
7
5
Last revised: January 6, 2013
Similar Triangles
Lecture Notes
page 6
2. Consider the picture shown below.
a) Use the Pythagorean Theorem to …nd the value of a.
Solution: The shorter sides are 5 ft and 12 ft long. The hypotenuse is a: We state the Pythagorean
Theorem for this triangle and solve the equation for a.
52 + 122
25 + 144
169
13
Since distances can never be negative, a =
=
=
=
=
a2
a2
a2
a
13 is ruled out. Thus a = 13 ft.
b) Prove that the triangles ABE and ACD are similar.
Solution: First, angles ABE and ACD are both right angles. Second, the two triangles literally
share angle EAB (or angle DAC). Finally, if two triangles agree in the measure of two of their
angles, the third angles must be equal since in every triangle, the three angles add up to 180 . The
two triangles are similar because they have identical angles.
c) Use similar triangles to …nd the value of x.
Solution: The triangles 4ABE and 4ACD
side opposite point A
for both triangles.
are similar.
To …nd x; we write the ratio
horizontal side
side opposite point A
5
x
=
=
horizontal side
12
24
and solve the equation for x.
5
x
=
12
24
10 = x
multiply both sides by 24
Thus x is 10 ft. Indeed, once we established that the triangles are similar, and noticed that the
horizontal side was doubled from 12 ft to 24 ft; we could easily predict this answer.
d) Find the value of b.
Solution: We can either use similar triangles or the Pythagorean Theorem to …nd the side AD.
Either way, we easily get that 26 ft. However, the length of side AD is not b, but a + b: From part
a), we know that a = 13 ft.
13 + b = 26
b = 13
Thus b = 13 ft.
Last revised: January 6, 2013
Similar Triangles
Lecture Notes
page 7
3. a) A person is standing 40 ft away from a street light that is 30 ft tall. How tall is he if his shadow
is 10 ft long?
Solution: After we draw a picture, we see that this problem is very similar to the previous one.
Triangles 4ADE and 4ABC are similar. We use the ratio
x
30
=
10
50
5x = 30
x = 6
DE
BC
=
and solve for x.
AB
multiply both sides by 50
divide by 5
Thus the person is 6 ft tall. Notice that the number 40 did not occur in the equation. It is a common
error to use 40 instead of 50.
b) A 6 ft tall person is standing 24 ft away from a street light that is 15 ft tall. How long is her
Solution: After we draw a picture, write an equation expressing that triangles ADE and ABC are
similar.
We can use the same ratio as before,
6
x
6 (x + 24)
6x + 144
144
16
=
=
=
=
=
DE
BC
=
and solve for x.
AB
15
x + 24
15x
15x
9x
x
multiply both sides by x (x + 24)
distribute
subtract 6x
divide by 9
Thus her shadow is 16 ft long.
Last revised: January 6, 2013
Lecture Notes
Similar Triangles
page 8
Note: If the …rst step, multiplying by x (x + 24) (same as cross-multiplying) is confusing, here is the
break-down:
6
15
=
multiply by x (x + 24)
x
x + 24
6
15
x (x + 24)
=
x (x + 24)
expressing everything as a fraction
x
x + 24
x (x + 24) 6
15
x (x + 24)
=
1
x
x + 24
1
x (x + 24) 6
15x (x + 24)
=
cancel
x
x + 24
(x + 24) 6
15x
=
simplify
1
1
(x + 24) 6 = 15x
4. Prove the following statement. Let ABC be any right triangle, the right angle at point C. The
altitude drawn from C to the hypotenuse splits the triangle into two right triangles that are similar
to each other and to the original triangle.
Solution: Let us draw a picture and use standard labeling of points.
The two triangles created, M ADC and M DBC are both right triangles. 4ADC is similar to the
original triangle, because they agree in two angles: the right angle and . 4DBC is similar to the
original triangle, because they agree in two angles: the right angle and . Thus all three triangles
are similar. Also, this will be very useful later: \ACD =
and \BCD = .
5. The picture below shows a right triangle. Find the length of h; the height drawn to the hypotenuse.
Solution: Let us …rst label the points, angles and sides in the triangle. As we proved it in the
previous problem, the two new triangles are similar to the original triangle.
Last revised: January 6, 2013
Similar Triangles
Lecture Notes
Consider now the ratio
side opposite
side opposite
similar, this ratio is preserved.
page 9
in triangles 4AP C and M P BC. Since these triangles are
side opposite
side opposite
=
50
h
=
h
18
We solve this equation for h.
50
h
50 18
900
h
h=
h
18
= h2
= h2
=
30
=
30 is ruled out since distances can not be negative. Thus h =
p
18 50 = 30.
6. Find x, y, and h based on the picture below.
Solution: We can easily
p …nd the hypotenuse of this triangle via the Pythagorean Theorem. The
hypotenuse, x + y is 202 + 212 = 29 units long. Next, let us …rst label the points, angles and sides
in the triangle.
Last revised: January 6, 2013
Similar Triangles
Lecture Notes
page 10
We now re-draw the three similar triangles in a separate …gure, all three of them rotated and re‡ected
into the same direction. This way, it is easy to realize what sides correspond to each other. (Hint:
start with the anges, they are in the same location. Then identify the points, and …nally the sides.)
We can …nd y using the following ratio in the …rst two triangles
side opposite
hypotenuse
=
21
y
=
29
21
21
y
=
multiply both sides by 21 29
29
21
441 = 29y
divide by 29
441
= y
29
A di¤erent ratio in the same triangles can be used to obtain
side opposite
hypotenuse
20
h
=
29
21
420 = 29h
420
= h
29
=
20
h
=
29
21
multiply both sides by 21 29
divide by 29
For x; we can simply use the fact that x + y = 29 and we already computed y =
x+
441
.
29
441
= 29
29
x = 29
441
400
=
29
29
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