In this chapter we will learn the concept of Pythagoras theorem with solved examples.
What is Pythagoras Theorem ?
It gives a formula to find side length of right angles triangle.
According to the formula, the square of hypotenuse length is equal to sum of square of other sides of right triangle.
Given above is right angled triangle with ∠ABC = 90 degree and AC as hypotenuse.
According to Pythagoras theorem;
\mathtt{hypotenuse^{2} =\ side^{2} +\ side^{2}}
\mathtt{AC^{2} =\ AB^{2} +\ BC^{2}}
The above concept and formula is very important and is used in various math topics. Make sure you remember this formula for your examination.
I hope you understood the concept, let us solve some problems for better understanding.
Pythagoras theorem formula – Solved Examples
(01) Given below is the right angle triangle ABC. Find the length of side AC.
Solution
Since ABC is a right angled triangle, we can apply Pythagoras theorem formula.
\mathtt{AC^{2} =\ AB^{2} +\ BC^{2}}
Putting the values, we get;
\mathtt{AC^{2} =\ 4^{2} +\ 3^{2}}\\\ \\ \mathtt{AC^{2} =\ 16\ +\ 9}\\\ \\ \mathtt{AC^{2} =\ 25}\\\ \\ \mathtt{AC\ =\ \sqrt{25}}\\\ \\ \mathtt{AC\ =\ 5}
Hence, length of side AC = 5 cm.
(02) Given below is rectangle ABCD with side length = 5 cm and diagonal length = 13 cm. Find the length of breadth AB.
Solution
Since the rectangle ABCD is a rectangle, the measure of all angles is 90 degree.
So, ∠ABC = 90 degree.
Applying Pythagoras theorem formula in triangle ABC.
\mathtt{AC^{2} =\ AB^{2} +\ BC^{2}}
Putting the values in formula.
\mathtt{13^{2} =\ AB^{2} +\ 5^{2}}\\\ \\ \mathtt{169\ =\ AB^{2} \ +\ 25}\\\ \\ \mathtt{AB^{2} =\ 169\ -\ 25}\\\ \\ \mathtt{AB^{2} =\ 144}\\\ \\ \mathtt{AB\ =\ \sqrt{144}}\\\ \\ \mathtt{AB\ =\ 12\ cm}
Hence, side of rectangle AB measure 12 cm.
(03) In the below triangle ABC, BC = 12 cm, AB = 10 cm and AD is the perpendicular bisector of side BC. Find the length of side BC.
Solution
Since AD is perpendicular bisector it will divide side BC into two equal parts.
So, BD = DC = 12 / 2 = 6 cm
Also, ∠ADB = 90 degree.
Applying Pythagoras formula in triangle ADB.
\mathtt{AB^{2} =\ BD^{2} +\ AD^{2}}\\\ \\ \mathtt{10^{2} =\ 6^{2} +\ AD^{2}}\\\ \\ \mathtt{100\ =\ 36\ +\ AD^{2}}\\\ \\ \mathtt{AD^{2} =\ 100\ -\ 36}\\\ \\ \mathtt{AD^{2} =\ 64}\\\ \ \mathtt{AD\ =\ \sqrt{64}}\\\ \\mathtt{AD\ =\ 8\ cm}
Hence, length of side AD = 8 cm.
(04) Observe the below image and find length of side CD.
Solution
From vertex D, draw perpendicular line to side BC as shown below.
Triangle DOC is a right triangle.
Using Pythagoras theorem.
\mathtt{DC^{2} =\ OD^{2} +\ OC^{2}}\\\ \\ \mathtt{DC^{2} =\ 12^{2} +\ 9^{2}}\\\ \\ \mathtt{DC^{2} \ =\ 144\ +\ 81}\\\ \\ \mathtt{DC^{2} =\ 225}\\\ \\ \mathtt{DC\ =\ \sqrt{225}}\\\ \\ \mathtt{DC\ =\ 15\ cm}
Hence, length of side DC = 15 cm.
(05) A boy traveled 6 km towards north then right and move 8 km towards east. What is the shortest distance between initial and final position.
Solution
The shortest distance between any two point is a straight line.
Observe the below image.
Here point A is the starting position and B is the final location.
The line AB is the shortest distance between initial and final location.
Note that AOB form a right angle triangle.
Using Pythagoras theorem.
\mathtt{AB^{2} =\ OA^{2} +\ OB^{2}}\\\ \\ \mathtt{AB^{2} =\ 6^{2} +\ 8^{2}}\\\ \\ \mathtt{AB^{2} \ =\ 36\ +\ 64}\\\ \\ \mathtt{AB^{2} =\ 100}\\\ \\ \mathtt{AB\ =\ \sqrt{100}}\\\ \\ \mathtt{AB\ =\ 10\ cm}
Hence, length of side AB is 10 cm.