Pythagorean Theorem

Pythagorean Theorem is also known as ‘Pythagoras theorem’ and is related to the sides of a right angled triangle.

Statement of ‘Pythagoras theorem’:

In a right triangle the area of the square on the hypotenuse is equal to the sum of the areas of the squares of its remaining two sides.

(Length of the hypotenuse)2 = (one side)2 + (2nd side)2

In the given figure, ∆PQR is right angled at Q; PR is the hypotenuse and PQ, QR are

the remaining two sides, then

(PR)2 = PQ2 + QR2

(h)2 = p2 + b2

[Here h → hypotenuse, p → perpendicular, b → base]

Pythagorean Theorem

Verification of Pythagoras theorem by the method of dissection:

Verification of Pythagoras Theorem

In the adjoining figure, ∆ PQR is a right angled triangle where QR is its hypotenuse and PR > PQ.

Square on QR is QRBA, square on PQ is PQST and the square on PR is PRUV.

The point of intersection of the diagonal of the square PRUV is O.

The straight line through the point O parallel to the QR intersects PV and RU at the point J and K respectively.

Again the straight line through the point O perpendicular to JK intersects PR and VU at the point L and  respectively.



As a result, the square PRUV is divided into four parts which is marked as 1, 2, 3, 4 and the square PQST is marked 5.

You can draw the same figure on a thick paper and cut it accordingly and now cut out the squares respectively from this figure. Cut the squares PRUV along JK and LM dividing it in four parts. Now, place the parts 1, 2, 3, 4and 5 properly on the square QRBA.

Note:

(i) these parts together exactly fit the square. Thus, we find that QR2 = PQ2 + PR2

(ii) Square drawn on side PQ, which means the area of a square of side PQ is denoted by PQ2.

1. Find the value of x using Pythagorean theorem:

Value of x using Pythagoras Theorem










Solution:

Identify the sides and the hypotenuse of the right angle triangle.

The one sides length = 8 m and the other side length = 15.

'X' is the length of hypotenuse because it is opposite side of the right angle.

Substitute the values into the Pythagorean formula (here 'x' is the hypotenuse) 

(h)2 = p2 + b2

[Here h → hypotenuse, p → perpendicular, b → base]

x2 = 82 + 152

Solve to find the known value of ‘x’

x2 = 64 + 225

x2 = 289

x = √289

x = 17

Therefore, value of x (hypotenuse) = 17 m


2. Use the formula of Pythagorean theorem to determine the length of ‘a’.

Formula of Pythagorean Theorem










Solution:

Identify the perpendicular, base and the hypotenuse of the right angle triangle.

Length of perpendicular = 24 cm and the length of base = a.

Length of hypotenuse = 25 cm. Since hypotenuse is the opposite side of the right angle.

Substitute the values into the Pythagorean formula (here 'a' is the base) 

(h)2 = p2 + b2

[Here h → hypotenuse, p → perpendicular, b → base]

252 = 242 + a2

Solve to find the known value of ‘a’

625 = 576 + a2

625 – 576 = 576 – 576 + a2

49 = a2

a2 = 49

a = √49

a = 7

Therefore, length of ‘a’ (base) = 7 cm

 

3. Solve to find the missing value of the triangle using the formula of Pythagorean Theorem:

Find the Missing Value of the Triangle










Solution:

Identify the perpendicular, base and the hypotenuse of the right angle triangle.

Perpendicular = k and Base = 7.5

Hypotenuse = 8.5, since hypotenuse is the opposite side of the right angle.

Substitute the values into the Pythagorean formula (here 'k' is the perpendicular) 

(h)2 = p2 + b2

[Here h → hypotenuse, p → perpendicular, b → base]

8.52 = k2 + 7.52

Solve to find the known value of ‘k ’

72.25 = k2 + 56.25

72.25 – 56.25 = k2 + 56.25 – 56.25

16 = k2

k2 = 16

k = √16

k = 4

Therefore, missing value of the triangle ‘k’ (perpendicular) = 4

Congruent Shapes

Congruent Line-segments

Congruent Angles

Congruent Triangles

Conditions for the Congruence of Triangles

Side Side Side Congruence

Side Angle Side Congruence

Angle Side Angle Congruence

Angle Angle Side Congruence

Right Angle Hypotenuse Side congruence

Pythagorean Theorem

Proof of Pythagorean Theorem

Converse of Pythagorean Theorem





7th Grade Math Problems

8th Grade Math Practice

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