Question

20.
ABC is a right triangle right angled at 'C'. If BC = a, CA = b.
AB=c and 'p' be the length of perpendicular from 'c' to
'AB', then.
A.1/p²=1/a²+1/b²

B.1/p²=1/b²-1/a²

C.1/p²=1/a²-1/b²

D.1/p²=1/a²+1/b²


​

Answer 1

Step-by-step explanation:

Correct option is

A

True

Let CD⊥AB. Then, CD=p

∴ Area of ΔABC=

2

1

(Base×Height)

=

2

1

(AB×CD)=

2

1

cp

Also,

Area of ΔABC=

2

1

(BC×AC)=

2

1

ab

∴

2

1

cp=

2

1

ab

⇒cp=ab

(ii) Since ΔABC is a right triangle, right angled at C.

∴AB

2

=BC

2

+AC

2

⇒c

2

=a

2

+b

2

⇒(

p

ab

)

2

=a

2

+b

2

[∵cp=ab⇒c=

p

ab

]

⇒

p

2

a

2

b

2

=a

2

+b

2

⇒

p

2

1

=

b

2

1

+

a

2

1

⇒

p

2

1

=

a

2

1

+

b

2

1