# Kite Questions with Solutions

A kite is a quadrilateral with two pairs of adjacent sides equal in length. In the diagram below, sides \( AB \) and \( AD \) have equal lengths and sides \( CB \) and \( CD \) have equal lengths.

The diagonal axis \( AC \) and the second diagonal \( BD \) intersect at an angle of \( 90^{\circ} \)

## Kite Formulas

We define the length of segments \( AC \), \( BD \) and \( AO \) using small letters as follows: \( AC = e\), \( BD = f \) and \( AO = g \).

We now present the formulas that may be used to find the sides, the area, perimeter and angles of the kite.

Note that: \( OC = AC - AO = e - g\) and \( BO = OD = f/2 \)

Area of triangle \( AOB = (1/2)(BO)(AO) = (1/4) f g \)

Area of triangle \( BOC = (1/2)(BO)(CO) = (1/4) f (e-g) \)

Area of triangle \( ABC = (1/4) f g + (1/4) f (e-g) = (1/4) f g \)

Area \( A \) of Kite: \[ \displaystyle A = 2 \times \text{Area of tringle} ABC = \dfrac{f \cdot e}{2} \]

Use the Pythagorean theorem in the right triangle \( BOC \)

Sides \( a \) and \( b \): \[ \displaystyle a = b = \sqrt{ \left(\dfrac{f}{2}\right)^2 + (e-g)^2} \]

Use the Pythagorean theorem in the right triangle \( AOB \)

Sides \( c \) and \( d \): \[ \displaystyle d = c = \sqrt{ \left(\dfrac{f}{2}\right)^2 + g^2} \]

Perimeter: \( \displaystyle p = 2 a + 2 d \)

Use the right triangle \( AOB \) to write: \( \tan(\alpha / 2) = \dfrac{f/2}{g} \), which gives

Angle \( \alpha \): \[ \displaystyle \alpha = 2 \arctan (\dfrac{f}{2g}) \]

Use the right triangle \( BOC \) to write: \( \tan(\gamma / 2) = \dfrac{f/2}{e-g} \), which gives

Angle\( \gamma \) : \[ \displaystyle \gamma = 2 \arctan \left(\dfrac{f}{2(e-g)}\right) \]

The sum of all angles in triangle \( ABC \) is equal to \( 180^{\circ} \), hence

Angle \( \beta \): \[ \displaystyle \beta = 180 - \dfrac{\gamma}{2} - \dfrac{\alpha}{2} \]

A kite calculator is included and may be used to check answers to calculations.

## Questions with Solutions

Question 1

Calculate the sides \( a \) and \( d \), the area, the perimeter and the angles \( \alpha, \beta \) and \( \gamma \) of a kite with the diagonal axis of \( 0.8 \) meters, the second diagonal \( 0.40 \) meters and distance \( AO \) of \( 0.2 \) meters.

__Solution__

Given: \( e = 0.8 \), \( f = 0.4\) and \( g = 0.2\)

Area: \( \displaystyle A = \dfrac{f \cdot e}{2} = \dfrac{0.4 \cdot 0.8}{2} = 0.16 \) square meters

Sides: \( \displaystyle a = b = \sqrt{ \left(\dfrac{f}{2}\right)^2 + (e-g)^2} = \sqrt{ \left(\dfrac{0.4}{2}\right)^2 + (0.8-0.2)^2} \approx 0.63 \) meters

Sides: \( \displaystyle d = c = \sqrt{ \left(\dfrac{f}{2}\right)^2 + g^2} = \sqrt{ \left(\dfrac{0.4}{2}\right)^2 + 0.2^2} \approx 0.28 \) meters

Perimeter: \( \displaystyle p = 2 a + 2 d = 2 \cdot 0.68 + 2 \cdot 0.28 \approx 1.92 \) meters

Angle: \( \displaystyle \alpha = 2 \arctan (\dfrac{f}{2g}) = 2 \arctan (\dfrac{0.4}{2\cdot0.2}) = 90^{\circ}\)

Angle: \( \displaystyle \gamma = 2 \arctan (\dfrac{f}{2(e-g)}) = 2 \arctan (\dfrac{0.4}{2(0.8-0.2)}) = 36.87^{\circ}\)

Angle: \( \displaystyle \beta = 180 - \dfrac{\gamma}{2} - \dfrac{\alpha}{2} = 180 - \dfrac{36.87}{2} - \dfrac{90}{2} = 116.57^{\circ} \)

Question 2

Calculate the sides \( a \) and \( d \), the angles \( \beta \), \( \gamma \), the area, the perimeter of a kite with the diagonal axis of \( e = 1.5 \) meters, the second diagonal \( f = 0.50 \) meters and angle \( \alpha = 30^{\circ} \).

__Solution__

Use triangle \( AOB \) to write: \( \sin(\alpha/2) = \dfrac{f/2}{d} \)

The above gives: \( d = \dfrac{f/2}{\sin(\alpha/2)} = \dfrac{0.5/2}{\sin(15^{\circ}} = 0.97 \) meters

Use triangle \( AOB \) to write: \( \tan(\alpha/2) = \dfrac{f/2}{OA} \)

The above gives: \( g = OA = \dfrac{f/2}{\tan(\alpha/2)} = \dfrac{0.5/2}{\tan(15^{\circ}} = 0.933 \) meters

Use triangle \( BOC \) to write: \( \tan(\gamma/2) = \dfrac{f/2}{AC-AO} = \dfrac{0.5/2}{1.5 - 0.933} = 0.44091\)

The above gives: \( \gamma = 2 \arctan(0.44091) = 47.59^{\circ} \)

Side: \( a = \sqrt { (1.5 - 0.933)^2 + (0.5/2)^2} = 0.63 \) meters

Angle: \( \displaystyle \beta = 180 - \dfrac{\gamma}{2} - \dfrac{\alpha}{2} = 180 - \dfrac{47.59}{2} - \dfrac{30}{2} = 141.21^{\circ} \)

Perimeter: \( \displaystyle p = 2 a + 2 d = 2 \cdot 0.63 + 2 \cdot 0.97 \approx 3.2 \) meters

Area: \( \displaystyle A = \dfrac{f \cdot e}{2} = \dfrac{0.5 \cdot 1.5}{2} = 0.375 \) square meters

Question 3

Given the second diagonal \( f = 0.60 \) meters and angles \( \alpha = 30^{\circ} \) and \( \beta = 120^{\circ} \), calculate the lengths of the diagonal axis \( AC = e\) the second diagonal \( BD = f \) and angle \( \gamma \)

__Solution__

Use triangle \( AOB \) to write: \( \tan(\alpha/2) = \dfrac{f/2}{OA} \)

Which gives: \(OA = g = \dfrac{f/2}{\tan(\alpha/2)} = \dfrac{0.60/2}{\tan(15^{\circ})} = 1.12 \) meters

Use triangle \( ABC \) to write: \( \alpha / 2 + \gamma /2 + \beta = 180^{\circ} \)

Hence: \( \gamma = 2(180 - \alpha / 2 - \beta) = 2(180 - 15 - 120) = 90^{\circ} \)

Use triangle \( BOC \) to write: \( \tan(\gamma / 2) = \dfrac{BO}{OC} \)

Which gives: \( OC = e - g = \dfrac{BO}{\tan(\gamma / 2)} = \dfrac{0.6/2}{45^{\circ}} = 0.3 \)

Hence: \( e = 0.38 + g = 0.3 + 1.12 = 1.42 \) meters.

## More References and Links

Kite CalculatorGeometry Tutorials, Problems and Interactive Applets