Examples with solutions for Area of a Triangle: Using Pythagoras' theorem

Exercise #1

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.

555555333AAABBBCCCDDDEEEGGG

Video Solution

Step-by-Step Solution

Let's calculate side BG using the Pythagorean theorem:

BG2+GC2=BC2 BG^2+GC^2=BC^2

We'll substitute the known data:

BG2+32=52 BG^2+3^2=5^2

BG2+9=25 BG^2+9=25

BG2=16 BG^2=16

BG=16=4 BG=\sqrt{16}=4

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

5×4=20 5\times4=20

Answer

20

Exercise #2

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

Work out the value of X.

333X+5X+5X+5BBBAAACCCDDD

Video Solution

Step-by-Step Solution

The area of triangle ABC is equal to:

AD×BC2=2x+16 \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:

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

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

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

We then multiply by 2 to eliminate the denominator:

AD×(x+8)=4x+32 AD\times(x+8)=4x+32

Divide by: (x+8) (x+8)

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

We rewrite the numerator of the fraction:

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

We simplify to X + 8 and obtain the following:

AD=4 AD=4

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

AD2+DC2=AC2 AD^2+DC^2=AC^2

Inserting the existing data:

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

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

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

x+5=49 x+5=\sqrt{49}

x+5=7 x+5=7

x=75=2 x=7-5=2

Answer

2 cm

Exercise #3

What is the area of the triangle in the drawing?

666666666AAABBBCCC

Video Solution

Step-by-Step Solution

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 theoremA2+B2=C2 A^2+B^2=C^2

We can find the length of the missing side.

32+X2=62 3^2+X^2=6^2

We convert the formula

6232=X2 6^2-3^2=X^2

369=27 36-9=27

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

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

6×272=15.588 \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=3×X24 S=\frac{\sqrt{3}\times X^2}{4}

Where X is one of the sides.

3×624=62.3534=15.588 \frac{\sqrt{3}\times6^2}{4}=\frac{62.353}{4}=15.588

Answer

15.588

Exercise #4

What is the area of the triangle in the figure?

999444999

Video Solution

Answer

277 2\sqrt{77} cm²

Exercise #5

ABCD is a right-angled trapezoid.

Given in cm:

AD = 10

DC = 8

Calculate the area of triangle ACD.

888101010AAABBBDDDCCC

Video Solution

Answer

24

Exercise #6

Given the following shape, calculate the area

αααααα777777222AAABBBCCCDDDEEE

Video Solution

Answer

65 6\sqrt{5} cm².

Exercise #7

What is the area of the triangle ADE?

888121212333AAABBBCCCDDDEEE

Video Solution

Answer

945 \frac{9}{4}\sqrt{5} cm²..

Exercise #8

Look at the triangle in the figure.

AD is used to form a semicircle with a radius of 2.5 cm.

Calculate the area of the triangle ABC.

999AAACCCBBBDDD12

Video Solution

Answer

514+30 5\sqrt{14}+30 cm².

Exercise #9

What is the area of the triangle ABC?

888333jjj555AAACCCBBBDDDEEE

Video Solution

Answer

434 4\sqrt{34} cm².

Exercise #10

Express the area of the triangle ABC in terms of X.

2X2X2XAAABBBCCCDDD8X+1

Video Solution

Answer

X+923X22X1 \frac{X+9}{2}\sqrt{3X^2-2X-1}