Question

Show that in a right triangle the square of the hypotenuse is equal to the sum of the square of the other two sides 4marks

Answer 1

pythagoras theorem:

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

construct a right triange right angled at B.

construction:construct a perpendicular BD on side AC.

given:angleB=90

to prove:AB2+BC2=AC2

proof: in triangle ABD and tri ABC,

angle A=angle A

angle ADB=angleBDC=90

therefore,triADB is similar to triABC.

which implies AD/AB=AB/AC (sides are in proportion)

which implies AD*AC=AB2-------1

IIIly, tri BDC is similar to tri ABC

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

which implies AC*DC=BC2-----2

add 1 and 2

AD.AC=AB2

AD.AC+AC.DC=AB2+BC2

=AC(AD+DC)=AB2+BC2

=AC2=AB2+BC2

Hence proved.

Answer 2

Answer:

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

∠ A = ∠ A    [ Common angle ]

∠ ADB = ∠ ABC   [ Both are 90° ]

∴  Δ  ADB  Similar to Δ ABC   [ By AA similarity ]

So , AD / AB = AB / AC   [ Sides are proportional ]

= > AB² = AD . AC  ... ( i )

Now in Δ BDC and Δ ABC

∠ C = ∠ C    [ Common angle ]

∠ BDC = ∠ ABC   [ Both are 90° ]

∴  Δ  BDC Similar to Δ ABC   [ By AA similarity ]

So , CD / BC = BC / AC

= > BC² = CD . AC   ... ( ii )

Now adding both equation :

AB² + BC² = CD . AC +  AD . AC

AB² + BC² = AC ( CD + AD )

AB² + BC² = AC² .

AC² = AB² + BC² .

Hence proved .