Question

Show that in a right triangle the square of the hypotenuse

Answer 1

$\huge\bold{\rm{\underline{Answer:}}}$

$\large\rm{\bigstar{\underline{Statement-}}}$

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

$\large\rm{\bigstar{\underline{Given-}}}$

ABC is a right triangle right triangled at A so that BC is its hypotenuse.

$\large\rm{\bigstar{\underline{To\:Prove-}}}$

$\sf{\implies {BC}^{2} = {AB}^{2} + {AC}^{2} }$

$\large\rm{\bigstar{\underline{Construction-}}}$

From A, draw AD perpendicular on BC.

$\large\rm{\bigstar{\underline{Proof-}}}$

$\large\sf{In\:\triangle DBA \:and\:\triangle ABC, }$

$\large\sf{\angle ABD=\angle ABC}$...(same angle)

$\large\sf{\angle ADB=\angle BAC}$...(each 90°)

$\large\sf{So, \triangle DBA \:\sim \triangle ABC}$....(By AA similarity criterion)

$\large\sf{\frac{AB}{BC}=\frac{BD}{AB}}$

$\large\sf{\implies {AB}^{2}=BD \times BC}$.......(i)

$\large\sf{ In\:\triangle DAC \:and\:\triangle ABC, }$

$\large\sf{ \angle ACD=\angle ACB }$...(same angle)

$\large\sf{ \angle ADC=\angle BAC }$...(each 90°)

$\large\sf{ So, \triangle DAC \:\sim \triangle ABC }$....(By AA similarity criterion)

$\large\sf{ \frac{AC}{BC}=\frac{DC}{AC} }$

$\large\sf{ \implies {AC}^{2}=DC \times BC}$.....(ii)

$\large\sf{On\:adding\:(i) \:and\:(ii) -}$

$\large\sf{{AB}^{2} + {AC}^{2}= BD \times\:BC + DC \times BC}$

$\large\sf{ {AB}^{2} + {AC}^{2}=(BD + DC) \times BC}$

$\large\sf{ {AB}^{2} + {AC}^{2}=BC \times BC }$

$\large\sf{\implies {AB}^{2} + {AC}^{2}={BC}^{2} }$

$\large\sf{Hence, {AB}^{2} + {AC}^{2}={BC}^{2} }$

$\large\sf{The\:\:above\:\:result\: \:\:is\:\:known }$

$\large\sf{as\:\:\:Pythagoras\:\:\:Theorem. }$

Answer 2

Answer:

Step-by-step explanation:

Given: A right triangle ABC right angled at B.

To Prove: AC2 = AB2 + BC2

Construction: Draw BD ⊥ AC

Proof:

In Δ ADB and Δ ABC,

∠ ADB = ∠ ABC (each 90°)

∠ BAD = ∠ CAB (common)

Δ ADB ~ Δ ABC (By AA similarity criterion)

Now, AD/AB = AB/AC (corresponding sides are proportional)

AB2 = AD × AC … (i)

Similarly, Δ BDC ~ Δ ABC

BC2 = CD × AC … (ii)

Adding (i) and (ii)

AB2 + BC2 = (AD × AC) + (CD × AC)

AB2 + BC2 = AC × (AD + CD)

AB2 + BC2 = AC2

Hence Proved.