the perimeter of a right angle triangle is 24 and its area is 24 cm square find the sides of the triangle
Answers
Answer:
Step-by-step explanation:
a2+b2=c2
c=p−a−b
c2=p2+a2+b2+2(ab−ap−bp)
2p(a+b)−p2=2ab=4q
2p(a+b)=p2+4q
EDIT: [simpler]
2p(p−c)=p2+4q
2p2−2pc=p2+4q
−2pc=−p2+4q
c=p/2−2q/p
Now we plug in p=q=24 and get c=12−2=10
We know a+b=p−c=14 and ab=2q=48 so a and b satisfy
x2−14x+48=0
x=7±49−48−−−−−−√=7±1
The problem is symmetrical in sides a and b so we can just call a the + root and b the − root.
a=8
b=6
Check: p=6+8+10=24q=12(6)(8)=24✓
ORIGINAL 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✓