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

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 Calculator
Geometry Tutorials, Problems and Interactive Applets

Popular Pages

More Info