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# 10.2: Trapezoids, Rhombi, and Kites

Difficulty Level: At Grade Created by: CK-12

## 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.

1. ABCD\begin{align*}ABCD\end{align*} is a square.
2. ABCD\begin{align*}ABCD\end{align*} is a square.

Know What? The Brazilian flag is to the right. The flag has dimensions of 20×14\begin{align*}20 \times 14\end{align*} (units vary depending on the size, so we will not use any here). The vertices of the yellow rhombus in the middle are 1.7 units from the midpoint of each side.

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 180\begin{align*}180^\circ\end{align*}. Now, this is a parallelogram with height h\begin{align*}h\end{align*} and base b1+b2\begin{align*}b_1 + b_2\end{align*}. The area of this shape is A=h(b1+b2)\begin{align*}A=h(b_1+b_2)\end{align*}.

Because the area of this parallelogram is two congruent trapezoids, the area of one trapezoid would be A=12h(b1+b2)\begin{align*}A=\frac{1}{2} h(b_1+b_2)\end{align*}.

Area of a Trapezoid: A=12h(b1+b2)\begin{align*}A=\frac{1}{2} h(b_1+b_2)\end{align*}

h\begin{align*}h\end{align*} is always perpendicular to the bases.

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) A=12(11)(14+8)A=12(11)(22)A=121 units2\begin{align*}A = \frac{1}{2} (11)(14+8)\!\\ A = \frac{1}{2} (11)(22)\!\\ A = 121 \ units^2\end{align*}

b) A=12(9)(15+23)A=12(9)(38)A=171 units2\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 42\begin{align*}4 \sqrt{2}\end{align*} and the other legs are of length 4.

PP=8+42+16+42=24+8235.3 unitsA=12(4)(8+16)A=48 units2\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: A=12d1d2\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: A=12d1d2\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.

122+82144+64sideP=side2=side2=208=413=4(413)=1613A=121624A=192\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 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 73\begin{align*}7 \sqrt{3}\end{align*}.

P=414=56A=1214143=983\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) 62+5236+25s1Shorter sides of kite=s21=s21=61Longer sides of kite122+52=s22144+25=s22s2=169=13\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*}

P=2(61)+2(13)=261+2641.6A=12(10)(18)=90\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) 202+d2sd2sdsSmaller diagonal portion=252=225=15Larger diagonal portion202+d2l=352  d2l=825dl=533\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*}

A=12(15+533)(40)874.5P=2(25)+2(35)=120\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 A(2,8),B(7,9),C(11,2)\begin{align*}A(2, 8), B(7, 9), C(11, 2)\end{align*}, and D(3,3)\begin{align*}D(3, 3)\end{align*}. Show ABCD\begin{align*}ABCD\end{align*} is a kite and find its area.

Solution: After plotting the points, it looks like a kite. AB=AD\begin{align*}AB = AD\end{align*} and BC=DC\begin{align*}BC = DC\end{align*}. The diagonals are perpendicular if the slopes are opposite signs and flipped.

mACmBD=28112=69=23=9373=64=32\begin{align*}m_{AC} &= \frac{2-8}{11-2}=-\frac{6}{9}=-\frac{2}{3}\\ m_{BD} &= \frac{9-3}{7-3}=\frac{6}{4}=\frac{3}{2}\end{align*}

The diagonals are perpendicular, so ABCD\begin{align*}ABCD\end{align*} is a kite. To find the area, we need to find the length of the diagonals.

d1=(211)2+(82)2=(9)2+62=81+36=117=313d2=(73)2+(93)2=42+62=16+36=52=213\begin{align*}d_1 &= \sqrt{(2-11)^2+(8-2)^2} && d_2=\sqrt{(7-3)^2+(9-3)^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. A=12(313)(213)=39 units2\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 201.71.7=16.6\begin{align*}20-1.7-1.7=16.6\end{align*} and the other is 141.71.7=10.6\begin{align*}14-1.7-1.7=10.6\end{align*}. The area is A=12(16.6)(10.6)=87.98 units2\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 2-16 are similar to Examples 1-4.
• Questions 17-23 are similar to Example 5.
• Questions 24-27 use the area formula for a kite and rhombus and factors.
• Questions 28-30 are similar to Example 4.
1. 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 ABCD\begin{align*}ABCD\end{align*} has vertices A(2,0),B(0,2),C(4,2)\begin{align*}A(-2, 0), B(0, 2), C(4, 2)\end{align*}, and D(0,2)\begin{align*}D(0, -2)\end{align*}. Leave your answers in simplest radical form.

1. 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.
2. 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*}?
3. Find \begin{align*}AB, AD\end{align*}, and \begin{align*}DC\end{align*}.
4. 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*}.

1. Find the slopes of all the sides and diagonals. What type of quadrilateral is this? Plotting the points will help you find the answer.
2. Find \begin{align*}HF\end{align*} and \begin{align*}EG\end{align*}.
3. 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*}.

1. What would the product of the diagonals have to be for the area to be \begin{align*}32 \ units^2\end{align*}?
2. 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*}.

1. What would the product of the diagonals have to be for the area to be \begin{align*}54 \ units^2\end{align*}?
2. 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.

1. What are the lengths of the diagonals for one kite?
2. Find the area of one kite.
3. Find the area of the entire logo.

1. \begin{align*}A = 9(8)+ \left [ \frac{1}{2} (9)(8) \right ] = 72 + 36 = 108 \ units^2\end{align*}
2. \begin{align*}A = \frac{1}{2} (6)(12) 2 = 72 \ units^2\end{align*}
3. \begin{align*}A = 4 \left [ \frac{1}{2} (6)(3) \right ] = 36 \ units^2\end{align*}

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