Examples with solutions for Using the Pythagorean Theorem: Using Pythagoras' theorem

Exercise #1

ABCD is a parallelogram.

CE is its height.

CB = 5
AE = 7
EB = 2

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What is the area of the parallelogram?

Video Solution

Step-by-Step Solution

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:

a2+b2=c2 a^2+b^2=c^2

In this case: EB2+EC2=BC2 EB^2+EC^2=BC^2

We place the given information: 22+EC2=52 2^2+EC^2=5^2

We isolate the variable:EC2=52+22 EC^2=5^2+2^2

We solve:EC2=254=21 EC^2=25-4=21

EC=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

21×9=41.24 \sqrt{21}\times9=41.24

Answer

41.24

Exercise #2

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.

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Video Solution

Step-by-Step Solution

DE crosses AB and AC, that is to say:

AD=DB=12AB=12×6=3 AD=DB=\frac{1}{2}AB=\frac{1}{2}\times6=3

AE=EC=12AC=12×10=5 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:

AD2+DE2=AE2 AD^2+DE^2=AE^2

We substitute our values into the formula:

32+DE2=52 3^2+DE^2=5^2

9+DE2=25 9+DE^2=25

DE2=259 DE^2=25-9

DE2=16 DE^2=16

We extract the root:

DE=16=4 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:

AB2+BC2=AC2 AB^2+BC^2=AC^2

We substitute our values into the formula:

62+BC2=102 6^2+BC^2=10^2

36+BC2=100 36+BC^2=100

BC2=10036 BC^2=100-36

BC2=64 BC^2=64

We extract the root:

BC=64=8 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=(4+8)2×3 S=\frac{(4+8)}{2}\times3

S=12×32=362=18 S=\frac{12\times3}{2}=\frac{36}{2}=18

Answer

18

Exercise #3

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.

S=24S=24S=24555444AAABBBCCCDDDKKK

Video Solution

Step-by-Step Solution

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

AD2+DK2=AK2 AD^2+DK^2=AK^2

Now let's substitute in our values:

AD2+42=52 AD^2+4^2=5^2

AD2+16=25 AD^2+16=25

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

AD2=2516 AD^2=25-16

AD2=9 AD^2=9

Next, we'll take the square root and get:

AD=3 AD=3

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

S=AB×AD S=AB\times AD

Let's substitute in our values:

24=3×AB 24=3\times AB

Finally, we'll divide both sides by 3:

AB=8 AB=8

Answer

8

Exercise #4

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

EB = 10

What is the area of the parallelogram EBFC?

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Video Solution

Answer

112 cm²

Exercise #5

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.

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Video Solution

Answer

36

Exercise #6

Look at the isosceles trapezoid ABCD below.

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

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

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Video Solution

Answer

24

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