Examples with solutions for Types of Triangles: Using Pythagoras' theorem

Exercise #1

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

171717888AAABBBCCCDDDEEEFFFGGG53 AF=5 AB=17
AG=3 AD=8
AD the height in the triangle

What is the area of the trapezoid EFBC?

Video Solution

Step-by-Step Solution

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

To find GF we use the Pythagorean theorem: A2+B2=C2 A^2+B^2=C^2  In triangle AFG

We replace:

32+GF2=52 3^2+GF^2=5^2

We isolate GF and solve:

9+GF2=25 9+GF^2=25

GF2=259=16 GF^2=25-9=16

GF=4 GF=4

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

82+DB2=172 8^2+DB^2=17^2

64+DB2=289 64+DB^2=289

DB2=28964=225 DB^2=289-64=225

DB=15 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=ADAG=83=5 GD=AD-AG=8-3=5

Now we reveal that EF and CB:

GF=GE=4 GF=GE=4

DB=DC=15 DB=DC=15

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

EF=GF×2=4×2=8 EF=GF\times2=4\times2=8

CB=DB×2=15×2=30 CB=DB\times2=15\times2=30

We replace the data in the trapezoid formula:

8+302×5=382×5=19×5=95 \frac{8+30}{2}\times5=\frac{38}{2}\times5=19\times5=95

Answer

95

Exercise #2

ABC is an isosceles triangle.

AD is the height of triangle ABC.

555333171717888AAABBBCCCDDDEEEFFFGGG

AF = 5

AB = 17
AG = 3

AD = 8

What is the perimeter of the trapezoid EFBC?

Video Solution

Step-by-Step Solution

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: A2+B2=C2 A^2+B^2=C^2 in the triangle AFG

We replace

32+GF2=52 3^2+GF^2=5^2

We isolate GF and solve:

9+GF2=25 9+GF^2=25

GF2=259=16 GF^2=25-9=16

GF=4 GF=4

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

82+DB2=172 8^2+DB^2=17^2

64+DB2=289 64+DB^2=289

DB2=28964=225 DB^2=289-64=225

DB=15 DB=15

We start by finding FB:

FB=ABAF=175=12 FB=AB-AF=17-5=12

Now we reveal EF and CB:

GF=GE=4 GF=GE=4

DB=DC=15 DB=DC=15

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

EF=GF×2=4×2=8 EF=GF\times2=4\times2=8

CB=DB×2=15×2=30 CB=DB\times2=15\times2=30

All that's left is to calculate:

30+8+12×2=30+8+24=62 30+8+12\times2=30+8+24=62

Answer

62

Exercise #3

Shown below is a rectangle and an isosceles right triangle.

777101010

What is the area of the rectangle?

Video Solution

Step-by-Step Solution

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 PythagorasA2+B2=C2 A^2+B^2=C^2 72+72=49+49=98 7^2+7^2=49+49=98

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

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

98×10=98.9999 \sqrt{98}\times10=98.99\approx99

Answer

99 \approx99