Question

abc is a right angles triangle right angled at b. bc is perpendicular to ac, what is ac×dc​

Answer 1

Answer:

AB

2

=AC

2

−BC

2

=4−1=3

AB=

3

ar( Δ ABC) =

2

1

×AB×BC

=

2

1

×

3

×1×=

2

3

Also ar( Δ ABC) =

2

1

×BD×AC

=

2

1

×BD×2

⇒

2

3

=BD

Consider ΔBDF & ΔACB

∠BFD=∠ABC=90

o

∠DEC+∠DCB=90

o

=∠BAC+∠BCA

∠DBF+∠DCB=∠BAC+∠BCD

∠DBF+∠BACΔBDF∼ΔACB (AA axiom of similarity).

⇒

CB

DF

=

BD

AC=

AB

BF

⇒

1

DF

=

4

3

=

3

BF

⇒DF=

4

3

& BF=

4

3

ar(BEDF)=BF×DF=

4

3

×

4

3

=

16

3

3

solution