Question

In a right triangle ABC, in which ZC = 90° and CD LAB. If BC = a, CA = b.
AB = c and CD = p.
A
1
II
+
P
1
1
#
+
b-D.
b
a
-la-la-la-la
-15 -1's -15-15
1511-15
1
1
с
+
P
10
p р
b
I
1
1
1
19
a
B
+
(iv)​

Answer 1

Answer:

Let CD⊥AB. Then, CD=p

∴ Area of ΔABC=21(Base×Height)

=21(AB×CD)=21cp

Also,

Area of ΔABC=21(BC×AC)=21ab

∴21cp=21ab

⇒cp=ab

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

∴AB2=BC2+AC2

⇒c2=a2+b2

⇒(pab)2=a2+b2[∵cp=ab⇒c=pab]

⇒p2a2b

Answer 2

Answer:

Please ask your question in a proper way