In D PQR; PQ = 8 , QR = 5 , PR = 3 . Is D PQR a right angled triangle ? If yes, which angle is of 90° ?
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 2.6 | Page 44
In D PQR; PQ = 8 , QR = 5 , PR = 3 . Is D PQR a right angled triangle ?
If yes, which angle is of 90° ?
Longest side of ∆PQR = PQ = √8
∴ PQ2 = (√8)2 = 8
Now, the sum of the squares of the remaining sides is
QR2 + PR2 = (√5)2 + (√3)2 = 5 + 3 = 8
∴ PQ2 = QR2 + PR2
∴ The square of the longest side is equal to the sum of the squares of the remaining two sides.
by Converse of Pythagoras theorem,
∴ ∆PQR is a right-angled triangle.
Now, PQ is the hypotenuse.
∴ ∠PRQ = 90° ...(Angle opposite to hypotenuse)
∴ ∆PQR is a right-angled triangle in which ∠PRQ is 90°.
Explanation:-
We can use the Pythagorean theorem to determine if D PQR is a right-angled triangle. According to the Pythagorean theorem, in a right-angled triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.
Let’s label the sides of triangle D PQR as follows: PQ = 8, QR = 5, and PR = 3. Then, we can check if the Pythagorean theorem is satisfied for this triangle:
PR^2 + QR^2 = 3^2 + 5^2 = 9 + 25 = 34 PQ^2 = 8^2 = 64
Since PQ^2 is not equal to PR^2 + QR^2, we can conclude that the triangle D PQR is not a right-angled triangle.
Therefore, we cannot determine which angle would be 90° if D PQR were a right-angled triangle.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 2.6 | Page 44