# Find the area of the kite in cm2. Round to 2 decimal places. the area?

Mar 12, 2018

The total area of the kite is $41.52$ $c {m}^{2}$.

We use the sine rule, Pythagoras theorem and our big brains to solve this as shown below.

#### Explanation:

If we draw in the lines AC and BD the kite becomes 4 right-angled triangles. Call the point where those lines cross E.

The lower left triangle is AED. We know that side AD is the same length as side CD, which is $11$ $c m$, and we know that the angle which is half of angle D will be ${17}^{o}$. We know that the angle in the centre is ${90}^{o}$.

We want to find the lengths of the lines AE and DE: once we know those it will be easy to calculate the area of triangle AED and triangle CED has the same area.

Use the sine rule:

$\frac{11}{\sin 90} = \frac{A E}{\sin 17}$

$A E = \frac{11 \left(\sin 17\right)}{\sin 90} = 3.22$ $c m$

Now we can use Pythagoras' theorem to find the length of DE:

${11}^{2} = D {E}^{2} + {3.22}^{2}$

$D E = \sqrt{{11}^{2} - {3.2}^{2}} = 10.52$ $c m$

Now the area of this lower left triangle, and the area of the lower right triangle, will be $A = \frac{b h}{2} = \frac{3.22 \times 10.52}{2} = 16.94$, so the combined area of the lower section of the kite will be $33.88$ $c {m}^{2}$.

Now for the upper left triangle, AEB. We already know the lengths of two sides, so we can just use Pythagoras:

${4}^{2} = B {E}^{2} + {3.22}^{2}$

$B E = \sqrt{{4}^{2} - {3.22}^{2}} = 2.37$ $c m$.

The area of this triangle is $A = \frac{b h}{2} = \frac{2.37 \times 3.22}{2} = 3.82$ $c {m}^{2}$. There are two triangles of this size, so the total area of the upper section will be $7.64$ $c {m}^{2}$.

Combining the areas for the lower and upper sections, the total area of the kite will be $33.88 + 7.64 = 41.52$ $c {m}^{2}$