Triangle Practice Problems: Types, Properties & Geometry

First, let's recall the formula for the area of a rhombus:

(Diagonal 1 * Diagonal 2) divided by 2

Now we will substitute the known data into the formula, giving us the answer:

(12*16)/2
192/2=
96

First, let's remind ourselves of the formula for the area of a trapezoid:

$A=\frac{\left(Base\text{ }+\text{ Base}\right)\text{ h}}{2}$

We substitute the given values into the formula:

(2.5+4)*6 =
6.5*6=
39/2 = 
19.5

First, we need to remind ourselves of how to work out the area of a trapezoid:

$\frac{(Base+Base)\cdot h}{2}=Area$

Now let's substitute the given data into the formula:

(10+6)*5 =
2

Let's start with the upper part of the equation:

16*5 = 80

80/2 = 40

First, we recall the formula for the area of a kite: multiply the lengths of the diagonals by each other and divide the product by 2.

We substitute the known data into the formula:

 $\frac{8\cdot DB}{2}=32$

We reduce the 8 and the 2:

$4DB=32$

Divide by 4

$DB=8$

Given that in a rectangle every pair of opposite sides are equal to each other, we can state that:

$AB=CD=2$

$AD=BC=7$

Now we can add all the sides together and find the perimeter:

$2+7+2+7=4+14=18$