Area of a Rectangle - Examples, Exercises and Solutions

Remember that the formula for the area of a rectangle is width times height

We are given that the width of the rectangle is 6

and that the length of the rectangle is 4

 Therefore we calculate:

6*4=24

We begin by multiplying side AB by side BC

We then substitute the given data and we obtain the following:

$4.5\times2=9$

Hence the area of rectangle ABCD equals 9

Let's begin by multiplying side AB by side BC

If we insert the known data into the above equation we should obtain the following:

$10\times2.5=25$

Thus the area of rectangle ABCD equals 25.

The triangle we are looking at is the large triangle - ABC

The triangle is formed by three sides AB, BC, and CA.

Now let's remember what we need for the calculation of a triangular area:

(side x the height that descends from the side)/2

Therefore, the first thing we must find is a suitable height and side.

We are given the side AC, but there is no descending height, so it is not useful to us.

The side AB is not given,

And so we are left with the side BC, which is given.

From the side BC descends the height AD (the two form a 90-degree angle).

It can be argued that BC is also a height, but if we delve deeper it seems that CD can be a height in the triangle ADC,

and BD is a height in the triangle ADB (both are the sides of a right triangle, therefore they are the height and the side).

As we do not know if the triangle is isosceles or not, it is also not possible to know if CD=DB, or what their ratio is, and this theory fails.

Let's remember again the formula for triangular area and replace the data we have in the formula:

(side* the height that descends from the side)/2

Now we replace the existing data in this formula:

$\frac{CB\times AD}{2}$

$\frac{11.6\times3}{2}$

$\frac{34.8}{2}=17.4$

This question is a bit confusing. We need start by identifying which parts of the data are relevant to us.

Remember the formula for the area of a triangle:

The height is a straight line that comes out of an angle and forms a right angle with the opposite side.

In the drawing we have a height of 6.

It goes down to the opposite side whose length is 5.

And therefore, these are the data points that we will use.

We replace in the formula:

$\frac{6\times5}{2}=\frac{30}{2}=15$

First, we will identify the data points we need to be able to find the area of the triangle.

the formula for the area of the triangle: height*opposite side / 2

Since it is a right triangle, we know that the straight sides are actually also the heights between each other, that is, the side that measures 5 and the side that measures 7.

We multiply the legs and divide by 2

$\frac{5\times7}{2}=\frac{35}{2}=17.5$

Formula for the area of a trapezoid:

$\frac{(base+base)}{2}\times altura$

We substitute the data into the formula and solve:

$\frac{9+12}{2}\times5=\frac{21}{2}\times5=\frac{105}{2}=52.5$

Let's begin by reminding ourselves of the formula for the area of a kite

$\frac{Diagonal1\times Diagonal2}{2}$

Both these values are given to us in the figure thus we can insert them directly into the formula:

(4*7)/2

28/2

14

To solve the exercise, we first need to know the formula for calculating the area of a kite:

It's also important to know that a concave kite, like the one in the question, has one of its diagonals outside the shape, but it's still its diagonal.

Let's now substitute the data from the question into the formula:

(6*5)/2=
30/2=
15

To solve the exercise, we first need to remember how to calculate the area of a rhombus:

(diagonal * diagonal) divided by 2

Let's plug in the data we have from the question

10*6=60

60/2=30

And that's the solution!

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, 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

$DB=8$

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



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, 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

The formula for the area of a trapezoid is:

We are given the following dimensions:

• Base $AB = 5$ cm
• Base $DC = 9$ cm
• Height $h = 7$ cm

Substituting these values into the formula, we have:

First, add the lengths of the bases:

Now substitute back into the formula:

Calculate the multiplication:

Then multiply by the height:

Thus, the area of the trapezoid is 49 cm$^2$.