## Pythagoras Theorem

**In a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.**

- Length of the hypotenuse is
*c* - The
**hypotenuse**is the**longest side** - Lengths of the other sides are
*a, b*

### Right Triangle Questions – using the theorem

The Theorem helps us in:

**Finding Sides:**If two sides are known, we can find the third side.**Determining if a triangle is right-angled:**If the sides of a triangle are known and satisfy the Pythagoras Formula, it is a right-angled triangle.

There is a proof of this theorem by a US president. Its simplicity makes it is **easy** enough for the **grade 8** kids to understand.

## Finding the missing sides (side lengths) of a Right Triangle

The theorem gives a relation among the three sides of a right-angled triangle. We can find one side if we know the other two sides. How?

**Example:** We are given (see figure) below the two sides of the right triangle. Find the third side.

Given: a = 3, c = 5

Which side is the hypotenuse?

AC = c = 5

**✩ Always identify the hypotenuse first**

Unknown side = BC = b?

Putting the values in the Pythagoras Formula: a^{2} + b^{2} = c^{2}

3^{2} + b^{2} = 5^{2}

9 + b^{2} = 25

b^{2} = 25 − 9 = 16 = 4^{2}

b = 4

### Finding the Hypotenuse of a Triangle

Using the Pythagoras formula, finding hypotenuse is no different from any other side.

**Example:** Sides of a right triangle are 20 cm and 21 cm, find its hypotenuse.

Pythagoras Formula: a^{2} + b^{2} = c^{2}

**c is the length of the hypotenuse****a, b are the lengths of the other two sides**(you can assume any length as a or b).

Let AB = a = 20, BC = b = 21

Putting values in the formula:

20^{2} + 21^{2} = c^{2}

400 + 441 = c^{2}

841 = c^{2}

c = 29 cm

## Finding Right Triangle

Given the sides, we can determine if a triangle is right-angled by applying the Pythagoras Formula. How?

- Assume the longest side to be hypotenuse Length = c. Find its square (= c
^{2}) - Find the sum of squares of the other two sides (= a
^{2}+ b^{2}) - If a
^{2}+ b^{2}≠ c^{2}it is a**not**right triangle - If a
^{2}+ b^{2}= c^{2}it is a right triangle

**Example:** A triangle has sides 8 cm, 11 cm, and 15 cm. Determine if it is a right triangle

Longest side = 15 cm. Let us assume it to be hypotenuse = c (as we know that it is always the longest)

c^{2} = 15^{2} = 225

Other sides a = 8 cm b = 11 cm. (You can **assume any side length** to be **a** or **b**).

a^{2} + b^{2} = 8^{2} + 11^{2} = 64 + 121 = 185

185 ≠ 225

a^{2} + b^{2} ≠ c^{2}

So this is not a right-angled triangle.

**Example:** The sides of a triangle are 8 cm, 17 cm, and 15 cm. Find if it is a right triangle.

Longest side = 17 cm. Let us assume it to be hypotenuse = c

c^{2} = 17^{2} = 289

Other sides a = 8 cm, b = 15 cm.

a^{2} + b^{2} = 8^{2} + 15^{2} = 64 + 225 = 289

So a^{2} + b^{2} = c^{2}

It is a right-angled triangle.

## Pythagoras Questions Types

You will encounter the following types of questions related to this theorem:

**Find a side, given two sides**These questions are the direct application of the theorem (formula) and are easiest to solve.**Find sides, given a direct relationship between any two sides**To solve these questions:- Express the relation between the two sides in an equation
- Substitute one side by the other using the first equation in the Pythagoras Formula

**Find sides, given an indirect relationship between any two sides**These questions may involve geometrical construction or other concepts of geometry/algebra. Some examples of this type of question are:*Given the Perimeter and one side, find other sides*– Perimeter is the sum of the three sides. Since one side is known, we subtract it from the perimeter to get a relationship between the other two sides.*Given Area and one side find other sides*– Area $=21 ×(base×altitude)$. Base and altitude can be the sides with the right angle**OR**the hypotenuse and the altitude.

## Pythagoras Questions

The questions chosen have minimal use of other concepts, yet, some of these are hard Pythagoras questions (See Ques 4 and Ques 10).

### 1 Question

ABC is a right triangle. AC is its hypotenuse. Length of side AB is 2√5. Side BC is twice of side AB. Find the length of AC.

#### Solution

Can you express BC in terms of AB and apply the Pythagoras Theorem?

### 2 Question

The hypotenuse of a right triangle is 6 cm. Its area is 9 cm^{2}. Find its sides.

#### Solution

Can you form two equations – one using area and the other using the Pythagoras formula?

### 3 Question

One side of a right triangle is $410 $ cm. Find the length of its other side if the hypotenuse is 13 cm.

#### Solution

Can you directly apply the Pythagoras Theorem?

### 4 Question

In a right triangle ABC, length of the medians to the sides AB and BC are $261 $ and $601 $ respectively. Find the length of its hypotenuse.

#### Solution

Given:

AD = $601 $

CE = $261 $

Let AE = EB = x

BD = DC = y

Can you use the Pythagorean Theorem to form an equation between x, y, and AD?

### 5 Question

In a right triangle, the longest side is 8 cm. One of the remaining sides is 4√3 cm long. Find the length of the other side.

#### Solution

Can you apply the Pythagoras Theorem directly?

### 6 Question

The first side of a right triangle is shorter than the second side by 1 cm. It is longer than the third side by 31 cm. Find the sides of the triangle.

#### Solution

Can you form equations between the first side and the other two sides? Which side is the hypotenuse?

### 7 Question

The perimeter of a right triangle is equal to 30 cm. The length of one of its sides is 10 cm. Find its hypotenuse.

#### Solution

Can you find the relation between the unknown side and the perimeter in terms of the hypotenuse?

### 8 Question

The sides of a triangle are 5 cm, 9 cm, and 12 cm. Is it a right-angled triangle?

#### Solution

Can you identify the possible hypotenuse? Also, test if the sides satisfy the Pythagoras Formula.

### 9 Question

In a right triangle, two sides are equal. The longest side is 7√2 cm, find the remaining sides.

#### Solution

Can you apply the Pythagoras Theorem directly?

### 10 Question

In the following right triangle altitude BD = $910 $ cm and DC = $2710 $ cm. Find the sides of the triangle.

#### Solution

Can you apply the Pythagoras Theorem to the triangle BCD?

## Answers to Pythagoras Questions

### 1 Answer

Let AB = a, BC = b and AC = c.

AB = a = 2√5

BC is twice of AB, b = 2a = 4√5

AC = Hypotenuse = c

Applying the Pythagoras Theorem a^{2} + b^{2} = c^{2}:

(2√5)^{2} + (4√5)^{2} = c^{2}

4(5) + 16(5) = c^{2}

c^{2} = 20 + 80 = 100

c = 10

AC = 10

### 2 Answer

Let AB = a, BC = b

In triangle ABC, base = b and altitude = a

Area of Triangle $=21 ×(base×altitude)$. So:

$21 ×(ab)=9$

ab = 18 (Equation 1)

Using the Pythagoras Formula:

a^{2} + b^{2} = 6^{2} = 36

We add and subtract 2ab to complete the square :

a^{2} + b^{2} − 2ab + 2ab = 36

(a − b)^{2} + 2ab = 36

(a − b)^{2} + 36 = 36 Using ab = 18 from Equation 1

(a − b)^{2} = 0

a = b

Substituting a by b in Equation 1:

b^{2} = 18

b = 3√2

Side AB = BC = 3√2 cm

### 3 Answer

Let the length of sides be a, b and c, such that:

$a=410 $ cm

b = unknown

c = 13 cm

From the Pythagoras formula a^{2} + b^{2} = c^{2}, we get:

$(410 )_{2}+b_{2}=13_{2}$

(16 × 10) + b^{2} = 169

160 + b^{2} = 169

b^{2} = 169 − 160 = 9

b = 3 cm

### 4 Answer

Given:

AD = $601 $

CE = $261 $

Let AE = EB = x

BD = DC = y

Using triangle ABD:

AB^{2} + BD^{2} = AD^{2}

$(2x)_{2}+y_{2}=(601 )_{2}$

$4x_{2}+y_{2}=601$ (Equation 1)

Using triangle EBC :

EB^{2} + BC^{2} = EC^{2}

$x_{2}+(2y)_{2}=(261 )_{2}$

$x_{2}+4y_{2}=(261 )_{2}$

$x_{2}+4y_{2}=244$ (Equation 2)

Adding Equation 1 and 2:

4x^{2} + y^{2} + x^{2} + 4y^{2} = 601 + 244

5x^{2} + 5y^{2} = 845

$x_{2}+y_{2}=169$ (Equation 3)

Let us solve for the hypotenuse using the triangle ABC:

AB^{2} + BC^{2} = AC^{2}

(2x)^{2} + (2y)^{2} = AC^{2}

4x^{2} + 4y^{2} = AC^{2}

4(x^{2} + y^{2}) = AC^{2}

Substituting the value of $x_{2}+y_{2}$ from Equation 3:

4(169) = AC^{2}

$AC=4(169) $

AC = 2 × 13 = 26

### 5 Answer

Let the lengths of sides be a, b and c (hypotenuse).

Hypotenuse is the longest side. So c = 8.

Let b = 4√3.

From the Pythagoras Theorem:

a^{2} + b^{2} = c^{2}

a^{2} + (4√3)^{2} = 8^{2}

a^{2} + 16(3) = 64

a^{2} + 48 = 64

a^{2} = 16

a = 4

The third side is 4 cm.

### 6 Answer

The second side is the longest. It is the hypotenuse. Let its length be c.

Let the length of first side be b and third side a.

c = b + 1

a = b − 31

Applying the Pythagoras formula:

a^{2} + b^{2} = c^{2}

(b − 31)^{2} + b^{2} = (b + 1)^{2}

b^{2} − 62b + 31^{2} + b^{2} = b^{2} + 2b + 1

b^{2} − 62b + 961 = 2b + 1

b^{2} − 64b + 960 = 0

b^{2} − 24b − 40b + 960 = 0

b(b − 24) − 40(b − 24) = 0

(b − 40)(b − 24) = 0

b = 40 Or b = 24

For b = 24, we get a = 24 − 31 = − 7. Length of a side cannot be negative, so we reject b = 24.

For b = 40, we get a = 40 − 31 = 9 and c = 40 + 1 = 41

The sides of triangle are 9 cm, 40 cm and 41 cm.

### 7 Answer

Let AB = a, BC = b and AC = c.

Side BC = b = 10 cm

Perimeter = Sum of the sides

= a + b + c = 30 (Given)

a + 10 + c = 30

a + c = 20

a = 20 − c ( Equation 1 )

Applying the Pythagoras Theorem to find the hypotenuse:

a^{2} + b^{2} = c^{2}

Using Equation 1 to substitute the value of a

(20 − c)^{2} + (10)^{2} = c^{2}

400 − 40c + c^{2} + 100 = c^{2}

500 − 40c = 0

40c = 500

c = 12.5

The length of hypotenuse = 12.5 cm

### 8 Answer

Longest side = 12 cm. Let us assume it to be the hypotenuse = c

So c^{2} = 12^{2} = 144

The Pythagoras Formula: a^{2} + b^{2} = c^{2}

We can assume any side to be a or b.

Let a = 5 cm, b = 9 cm.

a^{2} + b^{2} = 5^{2} + 9^{2} = 25 + 81 = 106

106 ≠ 144

So a^{2} + b^{2} ≠ c^{2}

This is a not a right angled triangle.

### 9 Answer

Let the length of the sides be a, b, and c (hypotenuse).

In a right triangle hypotenuse is the longest side. So c = 7√2

Other sides are equal. So a = b.

Applying the Pythagoras theorem:

a^{2} + b^{2} = c^{2}

b^{2} + b^{2} = (7√2)^{2}

2b^{2} = 49(2)

b^{2} = 49

b = 7

Each side is 7 cm.

### 10 Answer

Let AB = a, BC = b, AC = c and AD = x

Given $BD=910 DC=2710 $

Applying Pythagoras Theorem to triangle BCD:

BD^{2} + DC^{2} = BC^{2}

$(910 )_{2}+(2710 )_{2}=BC_{2}$

(9^{2} × 10) + (27^{2} × 10) = b^{2}

810 + 7290 = b^{2}

b^{2} = 8100

b = 90

Applying Pythagoras Theorem to triangle ABD:

BD^{2} + AD^{2} = AB^{2}

$a_{2}=(910 )_{2}+x_{2}$ (Equation 1)

From the figure:

AC = AD + BD

$c=x+2710 $ (Equation 2)

Applying Pythagoras Theorem to triangle ABC:

a^{2} + b^{2} = c^{2}

Using b = 90 and value of a^{2} from Equation 1 and c from Equation 2:

$(910 )_{2}+x_{2}+90_{2}=(x+2710 )_{2}$

$9_{2}(10)+x_{2}+90_{2}=x_{2}+5410 x+27_{2}(10)$

$810+8100=5410 x+7290$

$5410 x=8910−7290=1620$

$x=10 30 $

$x=310 $

Putting value of x in Equation 1:

$a_{2}=(910 )_{2}+(310 )_{2}$

a^{2} = 810 + 90 = 900

a = 30

Using value of a and b in Pythagoras Formula for triangle ABC:

a^{2} + b^{2} = c^{2}

30^{2} + 90^{2} = c^{2}

900 + 8100 = c^{2}

9000 = c^{2}

$c=3010 $

AB = 30cm, BC = 90cm, $AC=3010 cm$