http://www.ck12.org Chapter 10. Perimeter and Area
122 + 82 =side^2 A=1
2
· 16 · 24
144 + 64 =side^2 A= 192
side=√
208 = 4
√
13
P= 4
(
4
√
13
)
= 16
√
13
Example B
Find the perimeter and area of the rhombus below.
In a rhombus, all four triangles created by the diagonals are congruent.
Here, each triangle is a 30-60-90 triangle with a hypotenuse of 14. From the special right triangle ratios the short leg
is 7 and the long leg is 7
√
3.
P= 4 · 14 = 56 A=
1
2
· 7 · 7
√
3 =
49
√
3
2
≈ 42. 44
Example C
The vertices of a quadrilateral areA( 2 , 8 ),B( 7 , 9 ),C( 11 , 2 ), andD( 3 , 3 ). Determine the type of quadrilateral and find
its area.
For this problem, it might be helpful to plot the points. From the graph we can see this is probably a kite. Upon
further review of the sides,AB=ADandBC=DC(you can do the distance formula to verify). Let’s see if the
diagonals are perpendicular by calculating their slopes.
mAC=2 − 8
11 − 2
=−
6
9
=−
2
3
mBD=9 − 3
7 − 3
=
6
4
=
3
2
Yes, the diagonals are perpendicular because the slopes are opposite signs and reciprocals.ABCDis a kite. To find
the area, we need to find the length of the diagonals. Use the distance formula.
d 1 =√
( 2 − 11 )^2 +( 8 − 2 )^2 d 2 =√
( 7 − 3 )^2 +( 9 − 3 )^2
=
√
(− 9 )^2 + 62 =
√
42 + 62
=
√
81 + 36 =
√
117 = 3
√
13 =
√
16 + 36 =
√
52 = 2
√
13
Now, plug these lengths into the area formula for a kite.
A=
1
2
(
3
√
13
)(
2
√
13
)
= 39 units^2Watch this video for help with the Examples above.