Question

prove that in a right triangle, the square on a hypotenuse is equal to the sum of squares on the other two sides​

Answer 1

$\orange{\underline{Proposition :}}$ Let in the triangle ABC, ∠B = 90°,the hypotenuse AC = b, AB = c and BC = a. It is required to prove that $\bf{AC^2=AB^2+BC^2}$.i.e. b²=c²+a².

$\orange{\underline{Construction :}}$ Produce BC up to D in such a way that CD = AB = c.Also, draw perpendicular DE at D on BC produced, so that DE = BC = a. Join C,E and A,E.

$\orange{\underline{Proof :}}$

• In △ABC and △CDE, AB =CD=c, BC = DE = a and included ∠ABC = included ∠CDE(each right angle)

Hence, △ABC ≅△CDE

∴AC = CE = b and ∠BAC=∠ECD.

• Again, since AB ⊥BD and ED⊥BD

∴AB || ED

Therefore, ABDE is a trapezium.

• Moreover ∠ACB+∠BAC= ∠ACB+∠ECD=1 right angle.

∴∠ACE = 1 right angle. ∴△ACE is a right angled triangle.

$\sf{Now\:the\:area\:of\:trapezium\:ABDE,}\tt{=(△\:region\:ABC+△\:region\:CDE+△\:region\:ACE)}$

$\:\:\:\:\:\:\tt{\leadsto\:\frac{1}{2}BD(AB+DE)=\frac{1}{2}ac+\frac{1}{2}ac+\frac{1}{2}b^2}$

$\:\:\:\:\:\:\tt{\leadsto\:\frac{1}{2}(BC+CD)(AB+DE)=\frac{1}{2}[2ac+b^2]}$

$\:\:\:\:\:\:\:\tt{\leadsto\:(a+c)(a+c)=2ac+b^2[by\:multiplying\:with\:2]}$

$\:\:\:\:\:\:\:\:\tt{\leadsto\:a^2+2ac+c^2=2ac+b^2}$

$\:\:\:\:\:\:\:\:\:\tt\orange{\underline{\boxed{\leadsto\:b^2=c^2+a^2}}}$

Answer 2

Step-by-step explanation:

Answer ⬇️

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.” Proof: Let ABC be a right triangle where ∠B = 90°. This proves the Pythagoras Theorem.

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