Question

Prove that in a right triangle the square of the
hypotenuse is equal to the sum of the squares of the other
two sides prove that
PR2=PQ2+QR2-2QM.QR

Answer 1

Hi friend,

construct a right triangle right angled at B.

construction:construct a perpendicular BD on side AC.

GIVEN:angleB=90

TO PROVE:AB^2+BC^2=AC^2

PROOF: In triangle ABD and triangle ABC,

angle A=angle A

angle ADB=angle BDC=90°

therefore, triangle ADB is similar to triangle ABC.

=>AD/AB=AB/AC (sides are in proportion)

=>AD×AC=AB^2-------1

Similarly, triangle BDC is similar to triangle ABC

BC/DC=AB/BC (sides are in proportion)

which implies AC×DC=BC^2-----2

adding 1 and 2

AD.AC=AB^2

AD.AC+AC.DC=AB^2+BC^2

=AC(AD+DC)=AB^2+BC^2

=AC^2=AB^2+BC²

Hence proved.


HOPE THIS HELPS YOU:-))

Answer 2

Given : A right triangle ABC right angled at B.

To prove : AC^2 = AB^2 + BC^2

Construction : Draw BD ⊥ AC

Proof : In ΔADB and Δ ABC

∠ADB = ∠ABC (each 90°)

∠BAD =∠CAB (common)

ΔADB ~ ΔABC (By AA similarity criterion)

AD  ÷ AB  =  AB  ÷ AC (corresponding sides are proportional)

⇒ AB^2 = AD × AC …(i)

Similarly ΔBDC ~ ΔABC

⇒ BC^2 = CD × AC …(ii)

Adding (1) and (2)

AB^2 + BC^2 = AD × AC + CD × AC

⇒ AB^2 + BC^2 = AC × (AD + CD)

⇒ AB^2 + BC^2 = AC^2