10.2: Trapezoids, Rhombi, and Kites
Learning Objectives
 Derive and use the area formulas for trapezoids, rhombi, and kites.
Review Queue
Find the area of the shaded regions in the figures below.

\begin{align*}ABCD\end{align*}
ABCD is a square. 
\begin{align*}ABCD\end{align*}
ABCD is a square.
Know What? The Brazilian flag is to the right. The flag has dimensions of \begin{align*}20 \times 14\end{align*}
Find the area of the rhombus (including the circle). Do not round your answer.
Area of a Trapezoid
Recall that a trapezoid is a quadrilateral with one pair of parallel sides. The lengths of the parallel sides are the bases and the perpendicular distance between the parallel sides is the height of the trapezoid.
To find the area of the trapezoid, make a copy of the trapezoid and then rotate the copy \begin{align*}180^\circ\end{align*}
Because the area of this parallelogram is two congruent trapezoids, the area of one trapezoid would be \begin{align*}A=\frac{1}{2} h(b_1+b_2)\end{align*}
Area of a Trapezoid: \begin{align*}A=\frac{1}{2} h(b_1+b_2)\end{align*}
\begin{align*}h\end{align*}
You could also say the area of a trapezoid is the average of the bases times the height.
Example 1: Find the area of the trapezoids below.
a)
b)
Solution:
a) \begin{align*}A = \frac{1}{2} (11)(14+8)\!\\
A = \frac{1}{2} (11)(22)\!\\
A = 121 \ units^2\end{align*}
b) \begin{align*}A = \frac{1}{2} (9)(15+23)\!\\
A = \frac{1}{2} (9)(38)\!\\
A = 171 \ units^2\end{align*}
Example 2: Find the perimeter and area of the trapezoid.
Solution: Even though we are not told the length of the second base, we can find it using special right triangles. Both triangles at the ends of this trapezoid are isosceles right triangles, so the hypotenuses are \begin{align*}4 \sqrt{2}\end{align*}
\begin{align*}P &= 8+4\sqrt{2}+16+4\sqrt{2} && A=\frac{1}{2}(4)(8+16)\\
P &= 24+8\sqrt{2} \approx 35.3 \ units && A=48 \ units^2\end{align*}
Area of a Rhombus and Kite
Recall that a rhombus is an equilateral quadrilateral and a kite has adjacent congruent sides.
Both of these quadrilaterals have perpendicular diagonals, which is how we are going to find their areas.
Notice that the diagonals divide each quadrilateral into 4 triangles. If we move the two triangles on the bottom of each quadrilateral so that they match up with the triangles above the horizontal diagonal, we would have two rectangles.
So, the height of these rectangles is half of one of the diagonals and the base is the length of the other diagonal.
Area of a Rhombus: \begin{align*}A=\frac{1}{2} d_1 d_2\end{align*}
The area is half the product of the diagonals.
Area of a Kite: \begin{align*}A=\frac{1}{2} d_1 d_2\end{align*}
Example 3: Find the perimeter and area of the rhombi below.
a)
b)
Solution: In a rhombus, all four triangles created by the diagonals are congruent.
a) To find the perimeter, you must find the length of each side, which would be the hypotenuse of one of the four triangles. Use the Pythagorean Theorem.
\begin{align*}12^2+8^2 &=side^2 && A=\frac{1}{2} \cdot 16 \cdot 24\\ 144+64 &= side^2 && A=192\\ side &= \sqrt{208}=4 \sqrt{13}\\ P &= 4 \left( 4\sqrt{13} \right)=16 \sqrt{13}\end{align*}
b) Here, each triangle is a 306090 triangle with a hypotenuse of 14. From the special right triangle ratios the short leg is 7 and the long leg is \begin{align*}7 \sqrt{3}\end{align*}.
\begin{align*}P &= 4 \cdot 14=56 && A=\frac{1}{2} \cdot 14 \cdot 14\sqrt{3}=98\sqrt{3}\end{align*}
Example 4: Find the perimeter and area of the kites below.
a)
b)
Solution: In a kite, there are two pairs of congruent triangles. Use the Pythagorean Theorem in both problems to find the length of sides or diagonals.
a) \begin{align*}&\text{Shorter sides of kite} && \text{Longer sides of kite}\\ 6^2+5^2 &= s_1^2 && 12^2+5^2=s_2^2\\ 36+25 &= s_1^2 && 144+25=s_2^2\\ s_1 &= \sqrt{61} && \qquad \quad s_2=\sqrt{169}=13\end{align*}
\begin{align*}P = 2 \left( \sqrt{61} \right)+2(13)=2\sqrt{61}+26 \approx 41.6 && A=\frac{1}{2} (10)(18)=90\end{align*}
b) \begin{align*}&\text{Smaller diagonal portion} && \text{Larger diagonal portion}\\ 20^2+d_s^2&=25^2 && 20^2+d_l^2=35^2\\ d_s^2&=225 && \qquad \ \ d_l^2=825\\ d_s&=15 && \qquad \quad d_l=5\sqrt{33}\end{align*}
\begin{align*}A=\frac{1}{2} \left(15+5 \sqrt{33} \right)(40) \approx 874.5 && P=2(25)+2(35)=120\end{align*}
Example 5: The vertices of a quadrilateral are \begin{align*}A(2, 8), B(7, 9), C(11, 2)\end{align*}, and \begin{align*}D(3, 3)\end{align*}. Show \begin{align*}ABCD\end{align*} is a kite and find its area.
Solution: After plotting the points, it looks like a kite. \begin{align*}AB = AD\end{align*} and \begin{align*}BC = DC\end{align*}. The diagonals are perpendicular if the slopes are opposite signs and flipped.
\begin{align*}m_{AC} &= \frac{28}{112}=\frac{6}{9}=\frac{2}{3}\\ m_{BD} &= \frac{93}{73}=\frac{6}{4}=\frac{3}{2}\end{align*}
The diagonals are perpendicular, so \begin{align*}ABCD\end{align*} is a kite. To find the area, we need to find the length of the diagonals.
\begin{align*}d_1 &= \sqrt{(211)^2+(82)^2} && d_2=\sqrt{(73)^2+(93)^2}\\ &= \sqrt{(9)^2+6^2} && \quad =\sqrt{4^2+6^2}\\ &= \sqrt{81+36}=\sqrt{117}=3\sqrt{13} && \quad =\sqrt{16+36}=\sqrt{52}=2\sqrt{13}\end{align*}
Plug these lengths into the area formula for a kite. \begin{align*}A=\frac{1}{2} \left(3 \sqrt{13} \right)\left( 2\sqrt{13} \right)=39 \ units^2\end{align*}
Know What? Revisited To find the area of the rhombus, we need to find the length of the diagonals. One diagonal is \begin{align*}201.71.7=16.6\end{align*} and the other is \begin{align*}141.71.7=10.6\end{align*}. The area is \begin{align*}A=\frac{1}{2} (16.6)(10.6)=87.98 \ units^2\end{align*}.
Review Questions
 Question 1 uses the formula of the area of a kite and rhombus.
 Questions 216 are similar to Examples 14.
 Questions 1723 are similar to Example 5.
 Questions 2427 use the area formula for a kite and rhombus and factors.
 Questions 2830 are similar to Example 4.
 Do you think all rhombi and kites with the same diagonal lengths have the same area? Explain your answer.
Find the area of the following shapes. Round your answers to the nearest hundredth.
Find the area and perimeter of the following shapes. Round your answers to the nearest hundredth.
Quadrilateral \begin{align*}ABCD\end{align*} has vertices \begin{align*}A(2, 0), B(0, 2), C(4, 2)\end{align*}, and \begin{align*}D(0, 2)\end{align*}. Leave your answers in simplest radical form.
 Find the slopes of \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{DC}\end{align*}. What type of quadrilateral is this? Plotting the points will help you find the answer.
 Find the slope of \begin{align*}\overline{AD}\end{align*}. Is it perpendicular to \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{DC}\end{align*}?
 Find \begin{align*}AB, AD\end{align*}, and \begin{align*}DC\end{align*}.
 Use #19 to find the area of the shape.
Quadrilateral \begin{align*}EFGH\end{align*} has vertices \begin{align*}E(2, 1), F(6, 4), G(2, 7)\end{align*}, and \begin{align*}H(2, 4)\end{align*}.
 Find the slopes of all the sides and diagonals. What type of quadrilateral is this? Plotting the points will help you find the answer.
 Find \begin{align*}HF\end{align*} and \begin{align*}EG\end{align*}.
 Use #22 to find the area of the shape.
For Questions 24 and 25, the area of a rhombus is \begin{align*}32 \ units^2\end{align*}.
 What would the product of the diagonals have to be for the area to be \begin{align*}32 \ units^2\end{align*}?
 List two possibilities for the length of the diagonals, based on your answer from #24.
For Questions 26 and 27, the area of a kite is \begin{align*}54 \ units^2\end{align*}.
 What would the product of the diagonals have to be for the area to be \begin{align*}54 \ units^2\end{align*}?
 List two possibilities for the length of the diagonals, based on your answer from #26.
Sherry designed the logo for a new company, made up of 3 congruent kites.
 What are the lengths of the diagonals for one kite?
 Find the area of one kite.
 Find the area of the entire logo.
Review Queue Answers
 \begin{align*}A = 9(8)+ \left [ \frac{1}{2} (9)(8) \right ] = 72 + 36 = 108 \ units^2\end{align*}
 \begin{align*}A = \frac{1}{2} (6)(12) 2 = 72 \ units^2\end{align*}
 \begin{align*}A = 4 \left [ \frac{1}{2} (6)(3) \right ] = 36 \ units^2\end{align*}
Notes/Highlights Having trouble? Report an issue.
Color  Highlighted Text  Notes  

Please Sign In to create your own Highlights / Notes  
Show More 