Using the Pythagorean Theorem: Using Pythagoras' theorem

The rectangle ABCD is shown below.

$BD=25,BC=7$

Calculate the area of the rectangle.

We will use the Pythagorean theorem in order to find the side DC:

$(BC)^2+(DC)^2=(DB)^2$

We begin by inserting the existing data into the theorem:

$7^2+(DC)^2=25^2$

$49+DC^2=625$

$DC^2=625-49=576$

Finally we extract the root:

$DC=\sqrt{576}=24$

168

The trapezoid ABCD and the rectangle ABGE are shown in the figure below.

Given in cm:

AB = 5

BC = 5

GC = 3

Calculate the area of the rectangle ABGE.

Let's calculate side BG using the Pythagorean theorem:

$BG^2+GC^2=BC^2$

We'll substitute the known data:

$BG^2+3^2=5^2$

$BG^2+9=25$

$BG^2=16$

$BG=\sqrt{16}=4$

Now we can calculate the area of rectangle ABGE since we have the length and width:

$5\times4=20$

20

The trapezoid DECB forms part of triangle ABC.

AB = 6 cm
AC = 10 cm

Calculate the area of the trapezoid DECB, given that DE divides both AB and AC in half.

DE crosses AB and AC, that is to say:

$AD=DB=\frac{1}{2}AB=\frac{1}{2}\times6=3$

$AE=EC=\frac{1}{2}AC=\frac{1}{2}\times10=5$

Now let's look at triangle ADE, two sides of which we have already calculated.

Now we can find the third side DE using the Pythagorean theorem:

$AD^2+DE^2=AE^2$

We substitute our values into the formula:

$3^2+DE^2=5^2$

$9+DE^2=25$

$DE^2=25-9$

$DE^2=16$

We extract the root:

$DE=\sqrt{16}=4$

Now let's look at triangle ABC, two sides of which we have already calculated.

Now we can find the third side (BC) using the Pythagorean theorem:

$AB^2+BC^2=AC^2$

We substitute our values into the formula:

$6^2+BC^2=10^2$

$36+BC^2=100$

$BC^2=100-36$

$BC^2=64$

We extract the root:

$BC=\sqrt{64}=8$

Now we have all the data needed to calculate the area of the trapezoid DECB using the formula:

(base + base) multiplied by the height divided by 2:

Keep in mind that the height in the trapezoid is DB.

$S=\frac{(4+8)}{2}\times3$

$S=\frac{12\times3}{2}=\frac{36}{2}=18$

18

ABCD is a parallelogram.

CE is its height.

CB = 5
AE = 7
EB = 2

What is the area of the parallelogram?

To find the area,

first, the height of the parallelogram must be found.

To conclude, let's take a look at triangle EBC.

Since we know it is a right triangle (since it is the height of the parallelogram)

the Pythagorean theorem can be used:

$a^2+b^2=c^2$

In this case: $EB^2+EC^2=BC^2$

We place the given information: $2^2+EC^2=5^2$

We isolate the variable:$EC^2=5^2+2^2$

We solve:$EC^2=25-4=21$

$EC=\sqrt{21}$

Now all that remains is to calculate the area.

It is important to remember that for this, the length of each side must be used.
That is, AE+EB=2+7=9

$\sqrt{21}\times9=41.24$

41.24

Using the rhombus in the drawing:

Calculate the area?

Remember there are two options to calculate the area of a rhombus:

1: The diagonal multiplied by the diagonal divided by 2.

2: The base multiplied by the height.

In the question, we are only given the data for one of the diagonals and one of the sides, which means we cannot use either of the above formulas.

We need to find more data. Let's begin by finding the second diagonal:

Remember that the diagonals of a rhombus are perpendicular to one another, which means that they form a 90-degree angle.

Therefore, all the triangles in a rhombus are right-angled.

Now we can focus on the triangle where the side and the height are given, and we will calculate the third side using the Pythagorean theorem:

$a²+b²=c²$Insert the given data:

$3^2+x^2=5^2$$9+x^2=25$$x^2=25-9=16$$x=\sqrt{16}=4$

Now that we have found the second half of the diagonal, we can calculate the area of the rhombus by multiplying the two diagonals together.

Since the diagonals in a rhombus are perpendicular and cross each other, they are equal. Hence, our diagonals are equal:

$3+3=6$$4+4=8$Therefore, the area of the rhombus is:

$\frac{6\times8}{2}=\frac{48}{2}=24$

24

The area of triangle ABC is equal to 2X+16 cm².

Work out the value of X.

The area of triangle ABC is equal to:

$\frac{AD\times BC}{2}=2x+16$

As we are given the area of the triangle, we can insert this data into BC in the formula:

$\frac{AD\times(BD+DC)}{2}=2x+16$

$\frac{AD\times(x+5+3)}{2}=2x+16$

$\frac{AD\times(x+8)}{2}=2x+16$

We then multiply by 2 to eliminate the denominator:

$AD\times(x+8)=4x+32$

Divide by: $(x+8)$

$AD=\frac{4x+32}{(x+8)}$

We rewrite the numerator of the fraction:

$AD=\frac{4(x+8)}{(x+8)}$

We simplify to X + 8 and obtain the following:

$AD=4$

We now focus on triangle ADC and by use of the Pythagorean theorem we should find X:

$AD^2+DC^2=AC^2$

Inserting the existing data:

$4^2+(x+5)^2=(\sqrt{65})^2$

$16+(x+5)^2\text{ }=65/-16$

$(x+5)^2=49/\sqrt{}$

$x+5=\sqrt{49}$

$x+5=7$

$x=7-5=2$

2 cm

Given the rectangle ABCD

It is known that:

AB=4

AD=3

What is the length of the diagonal BD?

We will use the Pythagorean theorem in order to find BD:

$BD^2=AD^2+AB^2$

Let's input the known data:

$BD^2=3^2+4^2$

$BD^2=9+16$

$BD^2=25$

We'll take the square root:

$BD=\sqrt{25}=5$

$5$

Look at the following rectangle:

Calculate the perimeter of the rectangle ABCD.

Let's focus on triangle BCD in order to find side DC

We'll use the Pythagorean theorem and input the known data:

$BC^2+DC^2=BD^2$

$6^2+DC^2=10^2$

$DC^2=100-36=64$

Let's take the square root:

$DC=8$

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

$DC=AB=8$

$BC=AD=6$

Now we can calculate the perimeter of the rectangle by adding all sides together:

$8+6+8+6=16+12=28$

28

ABC is an isosceles triangle.

AD is the height of triangle ABC.

AF = 5

AB = 17
AG = 3

AD = 8

What is the perimeter of the trapezoid EFBC?

To find the perimeter of the trapezoid, all its sides must be added:

We will focus on finding the bases.

To find GF we use the Pythagorean theorem: $A^2+B^2=C^2$in the triangle AFG

We replace

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We perform the same process with the side DB of the triangle ABD:

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

We start by finding FB:

$FB=AB-AF=17-5=12$

Now we reveal EF and CB:

$GF=GE=4$

$DB=DC=15$

This is because in an isosceles triangle, the height divides the base into two equal parts so:

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

All that's left is to calculate:

$30+8+12\times2=30+8+24=62$

62

The parallelogram ABCD contains the rectangle AEFC inside it, which has a perimeter of 24.

AE = 8

BC = 5

What is the area of the parallelogram?

In the first step, we must find the length of EC, which we will identify with an X.

We know that the perimeter of a rectangle is the sum of all its sides (AE+EC+CF+FA),

Since in a rectangle the opposite sides are equal, the formula can also be written like this: 2AE=2EC.

We replace the known data:

$2\times8+2X=24$

$16+2X=24$

We isolate X:

$2X=8$

and divide by 2:

$X=4$

Now we can use the Pythagorean theorem to find EB.

(Pythagoras: $A^2+B^2=C^2$)

$EB^2+4^2=5^2$

$EB^2+16=25$

We isolate the variable

$EB^2=9$

We take the square root of the equation.

$EB=3$

The area of a parallelogram is the height multiplied by the side to which the height descends, that is$AB\times EC$.

$AB=\text{ AE}+EB$

$AB=8+3=11$

And therefore we will apply the area formula:

$11\times4=44$

44

Given that the triangle ABC is isosceles,
and inside it we draw EF parallel to CB:

AF=5 AB=17
AG=3 AD=8
AD the height in the triangle

What is the area of the trapezoid EFBC?

To find the area of the trapezoid, you must remember its formula:$\frac{(base+base)}{2}+\text{altura}$We will focus on finding the bases.

To find GF we use the Pythagorean theorem: $A^2+B^2=C^2$ In triangle AFG

We replace:

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We will do the same process with side DB in triangle ABD:

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

From here there are two ways to finish the exercise:

1. Calculate the area of the trapezoid GFBD, prove that it is equal to the trapezoid EGDC and add them up.

2. Use the data we have revealed so far to find the parts of the trapezoid EFBC and solve.

Let's start by finding the height of GD:

$GD=AD-AG=8-3=5$

Now we reveal that EF and CB:

$GF=GE=4$

$DB=DC=15$

This is because in an isosceles triangle, the height divides the base into two equal parts then:

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

We replace the data in the trapezoid formula:

$\frac{8+30}{2}\times5=\frac{38}{2}\times5=19\times5=95$

95

Shown below is the rectangle ABCD.

Given in cm:

AK = 5

DK = 4

The area of the rectangle is 24 cm².

Calculate the side AB.

Let's look at triangle ADK to calculate side AD:

$AD^2+DK^2=AK^2$

Let's input the given data:

$AD^2+4^2=5^2$

$AD^2+16=25$

We'll move 16 to the other side and change the appropriate sign:

$AD^2=25-16$

$AD^2=9$

We'll take the square root and get:

$AD=3$

Since AD is a side of rectangle ABCD, we can calculate side AB as follows:

$S=AB\times AD$

Let's input the given data:

$24=3\times AB$

We'll divide both sides by 3:

$AB=8$

8

What is the area of the triangle in the drawing?

There are two ways to solve the exercise:

It is possible to drop a height from one of the vertices, as we know

In an equilateral triangle, the height intersects the base,

This creates a right triangle whose two sides are 6 and 3,

Using the Pythagorean theorem$A^2+B^2=C^2$

We can find the length of the missing side.

$3^2+X^2=6^2$

We convert the formula

$6^2-3^2=X^2$

$36-9=27$

Therefore, the height of the triangle is equal to:$\sqrt{27}$

From here we calculate with the usual formula for the area of a triangle.

$\frac{6\times\sqrt{27}}{2}=15.588$

Option B for the solution is through the formula for the area of an equilateral triangle:

$S=\frac{\sqrt{3}\times X^2}{4}$

Where X is one of the sides.

$\frac{\sqrt{3}\times6^2}{4}=\frac{62.353}{4}=15.588$

15.588

Shown below is a rectangle and an isosceles right triangle.

What is the area of the rectangle?

To find the missing side, we use the Pythagorean theorem in the upper triangle.

Since the triangle is isosceles, we know that the length of both sides is 7.

Therefore, we apply Pythagoras$A^2+B^2=C^2$$7^2+7^2=49+49=98$

Therefore, the area of the missing side is:$\sqrt{98}$

The area of a rectangle is the multiplication of the sides, therefore:

$\sqrt{98}\times10=98.99\approx99$

$\approx99$

Calculate AE given that triangle ABC is isosceles.

$8\frac{1}{3}$

The area of a concave deltoid is 9 cm².
What is the value of X?

$$

1 cm

ABCD is a square with a side length of 8 cm.

EB = 10

What is the area of the parallelogram EBFC?

112 cm²

Look at the isosceles trapezoid ABCD below.

DF = 2 cm
AD =$\sqrt{20}$ cm

Calculate the area of the trapezoid given that the quadrilateral ABEF is a square.

24

Given ABCD deltoid AB=AC DC=BD

The diagonals of the deltoid intersect at the point O

Given in cm AO=12 OD=4

The area of the deltoid is equal to 48 cm².

Calculate the side CD

5 cm

The trapezoid ABCD is drawn inside a rectangle.

DC = 12 cm
BK = 3 cm
Height of the trapezoid (H) = 4

Calculate the area of the trapezoid.

36