Pythagoras Theorem | Class 7 | Practice Set 48

Pythagoras Theorem | Class 7 | Practice Set 48

Pythagoras Theorem Class 7 Practice Set 48, simple solutions of all the examples of this practice set are embedded in the pdf file attached below.

Practice-set-48

In this practice set all the examples are based on the Pythagoras theorem. The Pythagoras theorem is only applicable for the right angle triangles.

Also read : Perimeter and Area | Class 7 | Practice Set 47

The statement of the Pythagoras theorem is that the square of the hypotenuse is equal to the sum of squares of its remaining sides.

The formula for the Pythagoras theorem is,

(Hypotenuse)^2 = (Base)^2 + (Height)^2

We can also write the above mentioned formula as,

Hypotenuse)^2 = (one side)^2 + (other side)^2

Also read : Perimeter and Area | Class 7 | Practice Set 46

In these type of examples first we have to identify the hypotenuse. The hypotenuse is that side which is opposite to the 90 degree angle of the right angled triangle.

Once we are sure about the hypotenuse then we can write the formula and can put the given values in it and by doing simple mathematical calculations we can find the required side of that particular right angle triangle.

The examples solved in the above pdf file are,

Practice Set 48

1. the figures below, find the value of ‘x’.
2. In the right-angled ∆PQR, ∠P = 90°. If l(PQ) = 24 cm and l(PR) = 10 cm, find the length of segment QR.
3. In the right-angled ∆LMN, ∠M = 90°. If l(LM) = 12 cm and l(LN) = 20 cm, find the length of segment MN.
4. The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder ?

This is how the examples are solved using the Pythagoras theorem.