Areas of Trapezoids, Rhombuses, and Kites

Properties of Trapezoid

• Only one pair of opposite sides is parallel to each other.
• The sum of adjacent angles on non-parallel sides is 180°.

Trapezoid Shapes in Real Life:

Properties of Rhombus

• All sides of the rhombus are equal.
• The opposite sides of a rhombus are parallel.
• Opposite angles of a rhombus are equal.
• In a rhombus, diagonals bisect each other at right angles.
• Diagonals bisect the angles of a rhombus.
• The sum of two adjacent angles is equal to 180 degrees.

Rhombus Shapes in Real Life:

Properties of Kite

• The pair of adjacent sides are equal.
• One pair of opposite angles is equal.
• Diagonals are perpendicular to each other
• The longer diagonal bisects the shorter diagonal.

Kite Shapes in Real Life:

How to calculate the Area of the Trapezoid, Rhombus, and Kite?

Area of Trapezoid

The bases of a trapezoid are the parallel lines and the height of a trapezoid is the perpendicular distance between the two bases.

The area of a trapezoid is one-half the product of the height and the sum of the lengths of the bases. (b₁ and b₂ are the length of the bases and h is the perpendicular distance between the two bases)

In a trapezoid, the average of the lengths of the bases is also the length of the midsegment. So, you can also find the area by multiplying the midsegment by the height.

Area of Rhombus

Method 1:

The area of a rhombus is one-half the product of the lengths of its diagonals.

Length of diagonals: d₁ and d₂

Method 2:

Rhombus is also a parallelogram.

If the length of base (b) and height (h) are given, then we can simply use the formula of the area of a parallelogram.

A = b × h

Area of Kite

The area of a kite is one-half the product of the lengths of its diagonals.

The lengths of the diagonals are d₁ and d₂.

Let’s Solve Some Examples!

Example: 1

The free-throw lane on an international basketball court is shaped like a trapezoid. Find the area of the free-throw lane.

Solution:

The height of the trapezoid is 5.8 m. The lengths of the bases are 3.6 m and 6 m.

b₁ = 3.6 m, b₂ = 6 m, h = 5.8 m

A =  1/2 × h × (b₁+b₂)

    =  1/2 × 5.8 × (3.6 + 6)

    = 27.84

The area of the free-throw lane is 27.84 square meters.

Example: 2

Rhombus PQRS represents one of the inlays on the guitar in the photo. Find the area of the inlay.

Solution:

1. QN = 9 mm, PN = 12mm

      The diagonals of a rhombus bisect each other so:

      QN = NS, PN = NR

      The lengths of the diagonals are:

      QS (d₁) = 18 mm, PR (d₂) = 24 mm

• Area of a rhombus:

      A =  1/2 × d₁ × d₂

          =  1/2 × 18 × 24 = 216

 The area of the inlay is 216 square mm.

Example: 3

One diagonal of a kite is twice as long as the other diagonal. The area of the kite is 72.25 square inches. What are the lengths of the diagonals?

Solution:

Let x be the length of one diagonal. Then 2x is the length of the other diagonal.

Area of kite:

A = 1/2 × d₁ × d₂

A = 1/2 × x × 2x

72.25 =  1/2 × 2 × x²

x² = 72.25

x = 8.5

The lengths of the diagonals are d₁ = 8.5 in and d₂ = 17 in.

Questions to Solve

Question: 1

The lengths of the bases of a trapezoid are 5.4 centimeters and 10.2 centimeters. The height is 8 centimeters. Draw and label a trapezoid that matches this description. Then find its area.

Solution:

The lengths of the bases are b₁=5.4 cm,

                                            b₂= 10.2 cm

Height (h) = 8 cm

Area = 1/2 × h × (b₁ + b₂) =  1/2 × 8 × (5.4 + 10.2) = 62.4 square cm.

Question: 2

Find the area of the below figure. Tell whether it is a rhombus or a kite.

Solution:

It is a kite because the longer diagonal is divided into two unequal parts.

Longer diagonal: d₁ = 4 + 5 = 9 unites

Shorter diagonal: d₂ = 2 + 2 = 4 units (the longer diagonal bisects the shorter diagonal in a kite)

A =  × d₁ × d₂ =  × 9 × 4 = 18 square units

Question: 3

The figure is a rhombus. Its side length is 13. The length of one of its diagonals is 24. Sketch the figure and determine its area.

Solution:

There is a right-angled triangle formed (ABC).

Sides of the triangle: AC =13, AB = 12, BC = x

Following the Pythagoras theorem,

12² + x² = 13²

x = 5 units

Lengths of the diagonals are: d₁ = 24, d₂ = 10

A = 1/2 × d₁ × d₂ = 1/2 × 24 × 10 = 120 square units

Exercise:

1. Find the area of a trapezoid with bases of 12 cm and 18 cm and a height of 10 cm.
2. Find the area of a trapezoid with bases 2 ft and 3 ft and height 1/3 ft.
3. The border of Tennessee resembles a trapezoid with bases of 342 mi and 438 mi and a height of 111 mi. Approximate the area of Tennessee by finding the area of this trapezoid.
4. The area of a kite is 120 square cm. The length of one diagonal is 20 cm. What is the length of the other diagonal?
5. The lengths of the sides of a rhombus and one of its diagonals are each 10 m. What is the area of the rhombus?
6. The area of a trapezoid is 100 square ft. The sum of the two bases is 25 ft. What is the height of the trapezoid?
7. Find the area of a trapezoid with bases of 3 cm and 19 cm and a height of 9 cm.
8. Find the area of a trapezoid in a coordinate plane with vertices at (1, 1), (1, 6),
   (5, 9), and (5, 1).
9. One base of a trapezoid is twice the other. The height is the average of the two bases. The area is 324 square cm. Find the height and the bases. (Hint: Let the smaller base be 2x.)
10. Find the area of the rhombus with diagonals 10 cm and 20 cm.