Maharashtra Board Class 10 Maths Part II Chapter 2 Pythagoras Theorem PDF Download

Pythagoras Theorem

What We Will Learn

Pythagorean triplet

Theorem of geometric mean

Application of Pythagoras theorem

Similarity and right angled triangles

Pythagoras theorem

Apollonius theorem

Let's Recall

Pythagoras Theorem: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides.

In triangle PQR where angle PQR = 90°

\[l(PR)^2 = l(PQ)^2 + l(QR)^2\]

We will write this as,

\[PR^2 = PQ^2 + QR^2\]

The lengths PQ, QR and PR of triangle PQR can also be shown by letters r, p and q. With this convention, referring to figure 2.1, Pythagoras theorem can also be stated as

\[q^2 = p^2 + r^2\]

Pythagorean Triplet

In a triplet of natural numbers, if the square of the largest number is equal to the sum of the squares of the remaining two numbers then the triplet is called Pythagorean triplet.

For Example: In the triplet (11, 60, 61),

\[11^2 = 121, \quad 60^2 = 3600, \quad 61^2 = 3721 \quad \text{and} \quad 121 + 3600 = 3721\]

The square of the largest number is equal to the sum of the squares of the other two numbers.

Therefore, 11, 60, 61 is a Pythagorean triplet.

Verify that (3, 4, 5), (5, 12, 13), (8, 15, 17), (24, 25, 7) are Pythagorean triplets.

Numbers in Pythagorean triplet can be written in any order.

Teacher's Note

Pythagorean triplets are special number sets. For example, 3-4-5 is like the length-width-diagonal of a simple rectangle in real life.

Exam Trick

Remember: Always check the largest number first. If the square of the largest = sum of squares of other two, then it is a Pythagorean triplet.

Points to Remember

A Pythagorean triplet has three numbers.

The largest number is the hypotenuse.

The sum of squares of two smaller numbers = square of largest number.

Common triplets: (3,4,5), (5,12,13), (8,15,17), (11,60,61).

For More Information

Formula for Pythagorean triplet: If a, b, c are natural numbers and a > b, then \([(a^2 + b^2), (a^2 - b^2), (2ab)]\) is Pythagorean triplet.

\[(a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4 \quad \text{(I)}\]

\[(a^2 - b^2)^2 = a^4 - 2a^2b^2 + b^4 \quad \text{(II)}\]

\[(2ab)^2 = 4a^2b^2 \quad \text{(III)}\]

By (I), (II) and (III),

\[(a^2 + b^2)^2 = (a^2 - b^2)^2 + (2ab)^2\]

Therefore, \([(a^2 + b^2), (a^2 - b^2), (2ab)]\) is Pythagorean Triplet.

This formula can be used to get various Pythagorean triplets.

For example, if we take a = 5 and b = 3,

\[a^2 + b^2 = 34, \quad a^2 - b^2 = 16, \quad 2ab = 30\]

Check that (34, 16, 30) is a Pythagorean triplet.

Assign different values to a and b and obtain 5 Pythagorean triplets.

Last year we have studied the properties of right angled triangle with the angles 30° - 60° - 90° and 45° - 45° - 90°.

Property Of 30°-60°-90° Triangle

If acute angles of a right angled triangle are 30° and 60°, then the side opposite 30° angle is half of the hypotenuse and the side opposite to 60° angle is \(\frac{\sqrt{3}}{2}\) times the hypotenuse.

See figure 2.2. In triangle LMN, angle L = 30°, angle N = 60°, angle M = 90°

Side opposite 30° angle = MN = \(\frac{1}{2} \times LN\)

Side opposite 60° angle = LM = \(\frac{\sqrt{3}}{2} \times LN\)

If LN = 6 cm, we will find MN and LM.

\[MN = \frac{1}{2} \times LN \quad \text{and} \quad LM = \frac{\sqrt{3}}{2} \times LN\]

\[MN = \frac{1}{2} \times 6 = 3 \text{ cm} \quad \text{and} \quad LM = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3} \text{ cm}\]

Teacher's Note

A 30-60-90 triangle is very common. For example, if you cut an equilateral triangle in half, you get this special triangle.

Exam Trick

Remember: In 30-60-90 triangle, if hypotenuse is 2, then sides are 1 and √3. Just like Aadhaar divides things into two parts!

Points to Remember

The angle of 30° is opposite the shortest side.

The shortest side is half the hypotenuse.

The side opposite 60° is \(\frac{\sqrt{3}}{2}\) times the hypotenuse.

These are fixed ratios that always work.

Property Of 45°-45°-90° Triangle

If the acute angles of a right angled triangle are 45° and 45°, then each of the perpendicular sides is \(\frac{1}{\sqrt{2}}\) times the hypotenuse.

See Figure 2.3. In triangle XYZ,

\[XY = \frac{1}{\sqrt{2}} \times ZY\]

\[XZ = \frac{1}{\sqrt{2}} \times ZY\]

Therefore, \(XY = XZ = \frac{1}{\sqrt{2}} \times ZY\)

If ZY = \(3\sqrt{2}\) cm then we will find XY and ZX

\[XY = XZ = \frac{1}{\sqrt{2}} \times 3\sqrt{2}\]

\[XY = XZ = 3 \text{ cm}\]

In 7th standard we have studied theorem of Pythagoras using areas of four right angled triangles and a square. We can prove the theorem by an alternative method.

Activity

Take two congruent right angled triangles. Take another isosceles right angled triangle whose congruent sides are equal to the hypotenuse of the two congruent right angled triangles. Join these triangles to form a trapezium.

Area of the trapezium = \(\frac{1}{2} \times \text{(sum of the lengths of parallel sides)} \times \text{height}\)

Using this formula, equating the area of trapezium with the sum of areas of the three right angled triangles we can prove the theorem of Pythagoras.

Teacher's Note

The 45-45-90 triangle is an isosceles right triangle. It looks like a square cut diagonally in half.

Exam Trick

Remember: In 45-45-90 triangle, both equal sides are \(\frac{1}{\sqrt{2}}\) times the hypotenuse. Both angles are equal so both sides are equal!

Points to Remember

This triangle is isosceles (two equal sides).

Both acute angles are 45 degrees.

Both perpendicular sides are equal to each other.

Each side = \(\frac{1}{\sqrt{2}}\) times hypotenuse.

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