Question

Area of right angle triangle is 24cm' and perimeter is 24cm. Find the lengths
of sides of triangle?​

Answer 1

Answer:

2p(a+b)=p2+4q

a+b=p/2+2q/p

b=p/2+2q/p−a

ab=2q

a(p/2+2q/p−a)=2q

p2a+4qa−2pa2=4pq

2pa2−(p2+4q)a+4pq=0

a=14p((p2+4q)±(p2+4q)2−32p2q−−−−−−−−−−−−−−−√)

The problem is symmetrical in sides a and b so we can just call a the + root and b the − root.

Plugging in the numbers, we have p=q=24. Let’s set p=q first.

(p2+4p)2−32p3=(p(p+4))2−32p3=p2[(p+4)2−32p]=p2[p2−24p+16]=242⋅42

a=14(24)(24(28)+24(4))=7+1=8

b=7−1=6

c2=62+82=100

c=10

Check: p=6+8+10=24q=12(6)(8)=24

Step-by-step explanation:

Hope you got it

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