49
similarity & proportionality MPM2D: Principles of Mathematics Properties of Similar Triangles J. Garvin Slide 1/15
s im i l ar i ty & proport i onal i ty
MPM2D: Principles of Mathematics
Properties of Similar Triangles
J. Garvin
Slide 1/15
s im i l ar i ty & proport i onal i ty
Solving Proportions
Recap
Explain why ∆ABC ∼ ∆DEF .
Each side in ∆ABC is twice as long as that in ∆DEF .Therefore, ∆ABC ∼ ∆DEF by SSS∼.
J. Garvin — Properties of Similar Triangles
Slide 2/15
s im i l ar i ty & proport i onal i ty
Solving Proportions
Recap
Explain why ∆ABC ∼ ∆DEF .
Each side in ∆ABC is twice as long as that in ∆DEF .Therefore, ∆ABC ∼ ∆DEF by SSS∼.J. Garvin — Properties of Similar Triangles
Slide 2/15
s im i l ar i ty & proport i onal i ty
Solving Proportions
Since the ratios of any two corresponding sides of two similartriangles are equal, we can use these ratios to solve for thevalues of unknown sides.
Consider the case where ∆ABC ∼ ∆DEF .
If we know the ratio|AB||DE |
, and we know one of |AC | or
|DF |, then we can solve for the other value using the
proportion|AB||DE |
=|AC ||DF |
.
This can be done by inspection, or by usingcross-multiplication.
J. Garvin — Properties of Similar Triangles
Slide 3/15
s im i l ar i ty & proport i onal i ty
Solving Proportions
Since the ratios of any two corresponding sides of two similartriangles are equal, we can use these ratios to solve for thevalues of unknown sides.
Consider the case where ∆ABC ∼ ∆DEF .
If we know the ratio|AB||DE |
, and we know one of |AC | or
|DF |, then we can solve for the other value using the
proportion|AB||DE |
=|AC ||DF |
.
This can be done by inspection, or by usingcross-multiplication.
J. Garvin — Properties of Similar Triangles
Slide 3/15
s im i l ar i ty & proport i onal i ty
Solving Proportions
Since the ratios of any two corresponding sides of two similartriangles are equal, we can use these ratios to solve for thevalues of unknown sides.
Consider the case where ∆ABC ∼ ∆DEF .
If we know the ratio|AB||DE |
, and we know one of |AC | or
|DF |, then we can solve for the other value using the
proportion|AB||DE |
=|AC ||DF |
.
This can be done by inspection, or by usingcross-multiplication.
J. Garvin — Properties of Similar Triangles
Slide 3/15
s im i l ar i ty & proport i onal i ty
Solving Proportions
Since the ratios of any two corresponding sides of two similartriangles are equal, we can use these ratios to solve for thevalues of unknown sides.
Consider the case where ∆ABC ∼ ∆DEF .
If we know the ratio|AB||DE |
, and we know one of |AC | or
|DF |, then we can solve for the other value using the
proportion|AB||DE |
=|AC ||DF |
.
This can be done by inspection, or by usingcross-multiplication.
J. Garvin — Properties of Similar Triangles
Slide 3/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Example
∆ABC ∼ ∆DEF . Determine |AC | and |EF |.
Since we know the values of the corresponding sides AB andDE , we can use their ratio to create proportions involving theunknown sides.
J. Garvin — Properties of Similar Triangles
Slide 4/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Example
∆ABC ∼ ∆DEF . Determine |AC | and |EF |.
Since we know the values of the corresponding sides AB andDE , we can use their ratio to create proportions involving theunknown sides.J. Garvin — Properties of Similar Triangles
Slide 4/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC ||AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC ||AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC ||AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC |
|AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC ||AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC ||AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC ||AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC ||AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |
14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC ||AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AC corresponds to DF , whosevalue is known.
|AB||DE |
=|AC ||DF |
14
8=|AC |12
168 = 8|AC ||AC | = 21
EF corresponds to BC , whosevalue is known.
|AB||DE |
=|BC ||EF |
14
8=
19
|EF |14|EF | = 152
|EF | = 767
J. Garvin — Properties of Similar Triangles
Slide 5/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Example
Determine |AE |.
Since |AE | = |AC |+ |CE |, we can find |AC | and add it to|CE |, which is known.
J. Garvin — Properties of Similar Triangles
Slide 6/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Example
Determine |AE |.
Since |AE | = |AC |+ |CE |, we can find |AC | and add it to|CE |, which is known.
J. Garvin — Properties of Similar Triangles
Slide 6/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AB corresponds with DE , while AC corresponds with CE .
|AB||DE |
=|AC ||CE |
5
9=|AC |13
65 = 9|AC ||AC | = 65
9
Therefore, |AE | = 13 + 659 = 182
9 m.
J. Garvin — Properties of Similar Triangles
Slide 7/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AB corresponds with DE , while AC corresponds with CE .
|AB||DE |
=|AC ||CE |
5
9=|AC |13
65 = 9|AC ||AC | = 65
9
Therefore, |AE | = 13 + 659 = 182
9 m.
J. Garvin — Properties of Similar Triangles
Slide 7/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AB corresponds with DE , while AC corresponds with CE .
|AB||DE |
=|AC ||CE |
5
9=|AC |13
65 = 9|AC ||AC | = 65
9
Therefore, |AE | = 13 + 659 = 182
9 m.
J. Garvin — Properties of Similar Triangles
Slide 7/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AB corresponds with DE , while AC corresponds with CE .
|AB||DE |
=|AC ||CE |
5
9=|AC |13
65 = 9|AC |
|AC | = 659
Therefore, |AE | = 13 + 659 = 182
9 m.
J. Garvin — Properties of Similar Triangles
Slide 7/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AB corresponds with DE , while AC corresponds with CE .
|AB||DE |
=|AC ||CE |
5
9=|AC |13
65 = 9|AC ||AC | = 65
9
Therefore, |AE | = 13 + 659 = 182
9 m.
J. Garvin — Properties of Similar Triangles
Slide 7/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
AB corresponds with DE , while AC corresponds with CE .
|AB||DE |
=|AC ||CE |
5
9=|AC |13
65 = 9|AC ||AC | = 65
9
Therefore, |AE | = 13 + 659 = 182
9 m.
J. Garvin — Properties of Similar Triangles
Slide 7/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Example
Determine |BD|.
Since BD is part of a trapezoid rather than a triangle, wecannot use it directly in a proportion.
J. Garvin — Properties of Similar Triangles
Slide 8/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Example
Determine |BD|.
Since BD is part of a trapezoid rather than a triangle, wecannot use it directly in a proportion.J. Garvin — Properties of Similar Triangles
Slide 8/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Since |BD| = |AD| − |AB|, we can first find |AD| andsubtract |AB|, which is known.
|AD||AB|
=|DE ||BC |
|AD|14
=20
12|AD|
14=
5
33|AD| = 70
|AD| = 703
Therefore, |BD| = 703 − 14 = 28
3 in.
J. Garvin — Properties of Similar Triangles
Slide 9/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Since |BD| = |AD| − |AB|, we can first find |AD| andsubtract |AB|, which is known.
|AD||AB|
=|DE ||BC |
|AD|14
=20
12|AD|
14=
5
33|AD| = 70
|AD| = 703
Therefore, |BD| = 703 − 14 = 28
3 in.
J. Garvin — Properties of Similar Triangles
Slide 9/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Since |BD| = |AD| − |AB|, we can first find |AD| andsubtract |AB|, which is known.
|AD||AB|
=|DE ||BC |
|AD|14
=20
12
|AD|14
=5
33|AD| = 70
|AD| = 703
Therefore, |BD| = 703 − 14 = 28
3 in.
J. Garvin — Properties of Similar Triangles
Slide 9/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Since |BD| = |AD| − |AB|, we can first find |AD| andsubtract |AB|, which is known.
|AD||AB|
=|DE ||BC |
|AD|14
=20
12|AD|
14=
5
3
3|AD| = 70
|AD| = 703
Therefore, |BD| = 703 − 14 = 28
3 in.
J. Garvin — Properties of Similar Triangles
Slide 9/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Since |BD| = |AD| − |AB|, we can first find |AD| andsubtract |AB|, which is known.
|AD||AB|
=|DE ||BC |
|AD|14
=20
12|AD|
14=
5
33|AD| = 70
|AD| = 703
Therefore, |BD| = 703 − 14 = 28
3 in.
J. Garvin — Properties of Similar Triangles
Slide 9/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Since |BD| = |AD| − |AB|, we can first find |AD| andsubtract |AB|, which is known.
|AD||AB|
=|DE ||BC |
|AD|14
=20
12|AD|
14=
5
33|AD| = 70
|AD| = 703
Therefore, |BD| = 703 − 14 = 28
3 in.
J. Garvin — Properties of Similar Triangles
Slide 9/15
s im i l ar i ty & proport i onal i ty
Solving Similar Triangles
Since |BD| = |AD| − |AB|, we can first find |AD| andsubtract |AB|, which is known.
|AD||AB|
=|DE ||BC |
|AD|14
=20
12|AD|
14=
5
33|AD| = 70
|AD| = 703
Therefore, |BD| = 703 − 14 = 28
3 in.
J. Garvin — Properties of Similar Triangles
Slide 9/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
What happens to the area of a triangle when its dimensionsare doubled?
When the dimensions are doubled, the area is quadrupled.
J. Garvin — Properties of Similar Triangles
Slide 10/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
What happens to the area of a triangle when its dimensionsare doubled?
When the dimensions are doubled, the area is quadrupled.
J. Garvin — Properties of Similar Triangles
Slide 10/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
What happens to the area of a triangle when its dimensionsare tripled?
When the dimensions are tripled, the area increases by afactor of nine.
J. Garvin — Properties of Similar Triangles
Slide 11/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
What happens to the area of a triangle when its dimensionsare tripled?
When the dimensions are tripled, the area increases by afactor of nine.
J. Garvin — Properties of Similar Triangles
Slide 11/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
In general, a triangle whose dimensions are enlarged by afactor of k will have an area k2 times greater.
Areas of Similar Triangles
If ∆ABC ∼ ∆DEF , and if |AB| = k |DE |, thenAreaABC = k2 · AreaDEF .
Any ratio of corresponding sides can be used, so choose theone that is easiest to work with.
J. Garvin — Properties of Similar Triangles
Slide 12/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
In general, a triangle whose dimensions are enlarged by afactor of k will have an area k2 times greater.
Areas of Similar Triangles
If ∆ABC ∼ ∆DEF , and if |AB| = k |DE |, thenAreaABC = k2 · AreaDEF .
Any ratio of corresponding sides can be used, so choose theone that is easiest to work with.
J. Garvin — Properties of Similar Triangles
Slide 12/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
In general, a triangle whose dimensions are enlarged by afactor of k will have an area k2 times greater.
Areas of Similar Triangles
If ∆ABC ∼ ∆DEF , and if |AB| = k |DE |, thenAreaABC = k2 · AreaDEF .
Any ratio of corresponding sides can be used, so choose theone that is easiest to work with.
J. Garvin — Properties of Similar Triangles
Slide 12/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
Example
∆ABC ∼ ∆DEF . Determine AreaDEF .
J. Garvin — Properties of Similar Triangles
Slide 13/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
Determine the scale factor, k , of ∆DEF .
12k = 8
k = 23
The area of ∆DEF will be k2 =(23
)2= 4
9 the size of ∆ABC .
Therefore, the area of ∆DEF is 90× 49 = 40 cm2.
J. Garvin — Properties of Similar Triangles
Slide 14/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
Determine the scale factor, k , of ∆DEF .
12k = 8
k = 23
The area of ∆DEF will be k2 =(23
)2= 4
9 the size of ∆ABC .
Therefore, the area of ∆DEF is 90× 49 = 40 cm2.
J. Garvin — Properties of Similar Triangles
Slide 14/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
Determine the scale factor, k , of ∆DEF .
12k = 8
k = 23
The area of ∆DEF will be k2 =(23
)2= 4
9 the size of ∆ABC .
Therefore, the area of ∆DEF is 90× 49 = 40 cm2.
J. Garvin — Properties of Similar Triangles
Slide 14/15
s im i l ar i ty & proport i onal i ty
Areas of Similar Triangles
Determine the scale factor, k , of ∆DEF .
12k = 8
k = 23
The area of ∆DEF will be k2 =(23
)2= 4
9 the size of ∆ABC .
Therefore, the area of ∆DEF is 90× 49 = 40 cm2.
J. Garvin — Properties of Similar Triangles
Slide 14/15
s im i l ar i ty & proport i onal i ty
Questions?
J. Garvin — Properties of Similar Triangles
Slide 15/15
