Question

theorem 6.9 of Chapter 6 class 10th​

Answer 1

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

A "right-angle triangle theorem" is nothing but a "Pythagoras theorem" that states the relationship between "hypotenuse, base and perpendicular of the triangle".

If $Hypotenuse^2 = Perpendicular^{2} + Base^{2$

then, ∠θ = 90°

Step-by-step explanation:

Theorem: "In a triangle, if the "square of one side" is equal to the "sum of the squares" of the "other two sides", then the "angle opposite" the first side is a "right angle".

To prove: ∠B = 90°

Proof: We have a Δ ABC in which $AC^2 = AB^{2} + BC^{2}$

We need to prove that ∠B=90°

To prove the above, we "construct a triangle PQR" which is "right-angled at "Q" such that:

PQ=AB and QR=BC

From triangle PQR, we have

$PR^{2} = PQ^{2} + QR^{2}$ (According to "Pythagoras theorem",as ∠Q=90°)

or, $PR^{2} = AB^{2} + BC^{2}$                            (By construction) …… (1)

We know that;

$AC^{2} = AB^{2} +BC^{2}$                                 (Which is given) …………(2)

So, AC=PR                                      [From equation (1) &(2)]

Now, in Δ ABC & Δ PQR,

AB=PQ                                      (By "construction")

BC=QR                                      (By "construction")

AC=PR                                        ["Proved above"]

So, Δ ABC≅Δ PQR                    (By "SSS" congruence")  (Triangles are "congruent" if all 3 sides in one triangle are "congruent" to the "corresponding sides" in the other)

Thus,  ∠B=∠Q                             ("CPCT")  ("Corresponding parts of congruent triangle")

However,  ∠Q=90°                     (By "construction")

Therefore, ∠B = 90°    

Hence, the theorem is proved.