To find the area of a kite using diagonals you use the following equation $\displaystyle \frac{a*b}{2}=Area\: of a\: kite$ 

That diagonals ($\displaystyle a$ and $\displaystyle b$)are the lines created by connecting the two sides opposite of each other.

Plug in the diagonals for $\displaystyle a$ and $\displaystyle b$ to get $\displaystyle \frac{5*7}{2}=Area$

Then multiply and divide to get the area. $\displaystyle \frac{35}{2}=17.5$

The answer is $\displaystyle 17.5$

The formula for the area of a kite is:

$\displaystyle A = \frac{1}{2}(d_{1}\cdot d_{2})$

Where $\displaystyle d_{1}$ is the length of one diagonal and $\displaystyle d_{2}$ is the length of the other diagonal

Plugging in our values, we get:

$\displaystyle A = \frac{1}{2}(d_{1}\cdot d_{2})$

$\displaystyle A = \frac{1}{2}(6m\cdot 13m) = 39m^2$

The formula for the area of a kite is:

$\displaystyle A = \frac{1}{2} (d_1)(d_2)$

where $\displaystyle d_1$ is the length of one diagonal and $\displaystyle d_2$ is the length of another diagonal.

Use the formulas for a $\displaystyle 45-45-90$ triangle and a $\displaystyle 30-60-90$ triangle to find the lengths of the diagonals. The formula for a $\displaystyle 45-45-90$ triangle is $\displaystyle a-a-a \sqrt{2}$ and the formula for a $\displaystyle 30-60-90$ triangle is $\displaystyle a-a \sqrt{3}-2a$.

Our $\displaystyle 45-45-90$ triangle is: $\displaystyle 3 \sqrt{2}m-3 \sqrt{2}m-6m$

Our $\displaystyle 30-60-90$ triangle is: $\displaystyle 3 \sqrt{2}m-3 \sqrt{6}m-6\sqrt{2}m$

Plugging in our values, we get:

$\displaystyle A = \frac{1}{2} (d_1)(d_2)$

$\displaystyle A = \frac{1}{2} (6\sqrt{2}m) (3\sqrt{6}m + 3\sqrt{2}m)$

$\displaystyle A = \frac{1}{2} (18\sqrt{12}+18\sqrt{4}) = 18\sqrt{3}+18m^2$

In order to find the length of the two shorter edges, use a Pythagorean triple:

$\displaystyle 3-4-5$

$\displaystyle 3m-4m-5m$

In order to find the length of the two longer edges, use the Pythagorean theorem:

$\displaystyle A^2+B^2=C^2$

$\displaystyle (9)^2+(3)^2=C^2$

$\displaystyle C^2=90$

$\displaystyle C=3\sqrt{10}m$

The formula of the perimeter of a kite is:

$\displaystyle P = 2(side_{short})+2(side_{long})$

Plugging in our values, we get:

$\displaystyle P = 2(5m)+2(3\sqrt{10}m) = 10+6\sqrt{10}m$

The formula for the perimeter of a kite is:

$\displaystyle P = 2(s_{long}) + 2(s_{short})$

Where $\displaystyle s_{long}$ is the length of the longer side and $\displaystyle s_{short}$ is the length of the shorter side

Use the formulas for a $\displaystyle 45-45-90$ triangle and a $\displaystyle 30-60-90$ triangle to find the lengths of the longer sides. The formula for a $\displaystyle 45-45-90$ triangle is $\displaystyle a-a-a \sqrt{2}$ and the formula for a $\displaystyle 30-60-90$ triangle is $\displaystyle a-a \sqrt{3}-2a$.

Our $\displaystyle 45-45-90$ triangle is: $\displaystyle 3 \sqrt{2}m-3 \sqrt{2}m-6m$

Our $\displaystyle 30-60-90$ triangle is: $\displaystyle 3 \sqrt{2}m-3 \sqrt{6}m-6\sqrt{2}m$

Plugging in our values, we get:

$\displaystyle P = 2(6\sqrt{2}m) + 2(6m)$

$\displaystyle P = 12\sqrt{2}m + 12m$

When polygons are similar, the sides will have the same ratio to one another. Set up the appropriate proportions.

$\displaystyle \small \frac{2}{8}=\frac{x}{14}$

Cross multiply.

$\displaystyle \small 28=8x$

$\displaystyle \small x=3.5$

The formula for the area of a trapezoid is:

$\displaystyle A = \frac{1}{2}(b_{1}+b_{2})(h)$

Where $\displaystyle b_1$ is the length of one base, $\displaystyle b_2$ is the length of the other base, and $\displaystyle h$ is the height.

To find the height of the trapezoid, use a Pythagorean triple:

$\displaystyle 5-12-13$

$\displaystyle 5m-12m-13m$

Plugging in our values, we get:

$\displaystyle A = \frac{1}{2}(b_{1}+b_{2})(h)$

$\displaystyle A = \frac{1}{2}(16m+26m)(12m)=252m^2$

Use the formula for $\displaystyle 30-60-90$ triangles in order to find the length of the bottom base and the height.

The formula is:

$\displaystyle a-a\sqrt{3}-2a$

Where $\displaystyle a$ is the length of the side opposite the $\displaystyle \measuredangle 30$.

Beginning with the $\displaystyle 12m$ side, if we were to create a $\displaystyle 30-60-90$ triangle, the length of the base is $\displaystyle 6\sqrt{3}m$, and the height is $\displaystyle 6m$.

Creating another $\displaystyle 30-60-90$ triangle on the left, we find the height is $\displaystyle 6m$, the length of the base is $\displaystyle 2\sqrt{3}m$, and the side is $\displaystyle 4\sqrt{3}m$.

The formula for the area of a trapezoid is:

$\displaystyle A = \frac{1}{2}(b_{1}+b_{2})(h)$

Where $\displaystyle b_1$ is the length of one base, $\displaystyle b_2$ is the length of the other base, and $\displaystyle h$ is the height.

Plugging in our values, we get:

$\displaystyle A = \frac{1}{2}(b_{1}+b_{2})(h)$

$\displaystyle A = \frac{1}{2}(16m+16m+6\sqrt{3}m+2\sqrt{3}m)(6m)$

$\displaystyle A = \frac{1}{2}(32m+8\sqrt{3}m)(6m)$

$\displaystyle A = (32m+8\sqrt{3}m)(3m)=96+24\sqrt{3}m^2$

The formula for the area of a trapezoid is:

$\displaystyle A=\frac{1}{2}(b_{1}+b_{2})(h)$,

where $\displaystyle b_1$ is the length of one base, $\displaystyle b_2$ is the length of another base, and $\displaystyle h$ is the length of the height.

Plugging in our values, we get:

$\displaystyle A=\frac{1}{2}(7m+5m)(4m)=24m^2$

The formula for the area of a trapezoid is:

$\displaystyle A=\frac{1}{2}(b_{1}+b_{2})(h)$,

where $\displaystyle b_1$ is the length of one base, $\displaystyle b_2$ is the length of another base, and $\displaystyle h$ is the length of the height.

Use the Pythagorean Theorem to find the height of the trapezoid:

$\displaystyle A^2+B^2=C^2$

$\displaystyle 6m^2+B^2=20m^2$

$\displaystyle B=2\sqrt{91}m$

Plugging in our values, we get:

$\displaystyle A=\frac{1}{2}(18m+30m)(2\sqrt{91}m)=48\sqrt{91}m^2$