Pythagoras Theorem Questions (with Answers)

Pythagoras Theorem

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

Right Triangle with Pythagoras Statement

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.

Jump to the Questions & Answers ➤

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.

Right triangle with unknown 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: a² + b² = c²

3² + b² = 5²

9 + b² = 25

b² = 25 − 9 = 16 = 4²

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.

Right triangle with unknown hypotenuse

Pythagoras Formula: a² + b² = c²

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:

20² + 21² = c²

400 + 441 = c²

841 = c²

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 (= c²)
Find the sum of squares of the other two sides (= a² + b²)
If a² + b² ≠ c² it is a not right triangle
If a² + b² = c² 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)

c² = 15² = 225

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

a² + b² = 8² + 11² = 64 + 121 = 185

185 ≠ 225

a² + b² ≠ c²

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

c² = 17² = 289

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

a² + b² = 8² + 15² = 64 + 225 = 289

So a² + b² = c²

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 sidesThese questions are the direct application of the theorem (formula) and are easiest to solve.
Find sides, given a direct relationship between any two sidesTo solve these questions:
1. Express the relation between the two sides in an equation
2. Substitute one side by the other using the first equation in the Pythagoras Formula
Find sides, given an indirect relationship between any two sidesThese 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 $= \frac{1}{2} \times (ba se \times a lt i t u d e)$ . 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 cm². Find its sides.

Right triangle ABC with hypotenuse AC = 6 cm

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.

Right triangle ABC with medians to side AB and BC

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.

Right triangle ABC with altitude BD to the hypotenuse AC

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 a² + b² = c²:

(2√5)² + (4√5)² = c²

4(5) + 16(5) = c²

c² = 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 $= \frac{1}{2} \times (ba se \times a lt i t u d e)$ . So:

$\frac{1}{2} \times (ab) = 9$

ab = 18 (Equation 1)

Using the Pythagoras Formula:

a² + b² = 6² = 36

We add and subtract 2ab to complete the square :

a² + b² − 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:

b² = 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 a² + b² = c², we get:

$(410)^{2} + b^{2} = 1 3^{2}$

(16 × 10) + b² = 169

160 + b² = 169

b² = 169 − 160 = 9

b = 3 cm

4 Answer

Given:

AD = $601$

CE = $261$

Let AE = EB = x

BD = DC = y

Using triangle ABD:

AB² + BD² = AD²

$(2 x)^{2} + y^{2} = (601)^{2}$

$4 x^{2} + y^{2} = 601$ (Equation 1)

Using triangle EBC :

EB² + BC² = EC²

$x^{2} + (2 y)^{2} = (261)^{2}$

$x^{2} + 4 y^{2} = (261)^{2}$

$x^{2} + 4 y^{2} = 244$ (Equation 2)

Adding Equation 1 and 2:

4x² + y² + x² + 4y² = 601 + 244

5x² + 5y² = 845

$x^{2} + y^{2} = 169$ (Equation 3)

Let us solve for the hypotenuse using the triangle ABC:

AB² + BC² = AC²

(2x)² + (2y)² = AC²

4x² + 4y² = AC²

4(x² + y²) = AC²

Substituting the value of $x^{2} + y^{2}$ from Equation 3:

4(169) = AC²

$A C = 4 (169)$

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:

a² + b² = c²

a² + (4√3)² = 8²

a² + 16(3) = 64

a² + 48 = 64

a² = 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:

a² + b² = c²

(b − 31)² + b² = (b + 1)²

b² − 62b + 31² + b² = b² + 2b + 1

b² − 62b + 961 = 2b + 1

b² − 64b + 960 = 0

b² − 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:

a² + b² = c²

Using Equation 1 to substitute the value of a

(20 − c)² + (10)² = c²

400 − 40c + c² + 100 = c²

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 c² = 12² = 144

The Pythagoras Formula: a² + b² = c²

We can assume any side to be a or b.

Let a = 5 cm, b = 9 cm.

a² + b² = 5² + 9² = 25 + 81 = 106

106 ≠ 144

So a² + b² ≠ c²

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:

a² + b² = c²

b² + b² = (7√2)²

2b² = 49(2)

b² = 49

b = 7

Each side is 7 cm.

10 Answer

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

Given $B D = 910 D C = 2710$

Applying Pythagoras Theorem to triangle BCD:

BD² + DC² = BC²

$(910)^{2} + (2710)^{2} = B C^{2}$

(9² × 10) + (27² × 10) = b²

810 + 7290 = b²

b² = 8100

b = 90

Applying Pythagoras Theorem to triangle ABD:

BD² + AD² = AB²

$a^{2} = (910)^{2} + x^{2}$ (Equation 1)

From the figure:

AC = AD + BD

$c = x + 2710$ (Equation 2)

Applying Pythagoras Theorem to triangle ABC:

a² + b² = c²

Using b = 90 and value of a² from Equation 1 and c from Equation 2:

$(910)^{2} + x^{2} + 9 0^{2} = (x + 2710)^{2}$

$9^{2} (10) + x^{2} + 9 0^{2} = x^{2} + 5410 x + 2 7^{2} (10)$

$810 + 8100 = 5410 x + 7290$

$5410 x = 8910 - 7290 = 1620$

$x = \frac{30}{10}$

$x = 310$

Putting value of x in Equation 1:

$a^{2} = (910)^{2} + (310)^{2}$

a² = 810 + 90 = 900

a = 30

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

a² + b² = c²

30² + 90² = c²

900 + 8100 = c²

9000 = c²

$c = 3010$

AB = 30cm, BC = 90cm, $A C = 3010 c m$

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