Given, hypotenuse = 10 units.
Let Base be b units.
And perpendicular be p units.
Then by pythagoras theorem,
perpendicular2+base2=hypotenuse2
p2+b2=102⇒p2+b2=100
⇒p=100−b2 ...(i)
Now, area of right triangle =21×base×height=21×p×b
A=21pb ...(ii)
Putting value from (i) into (ii),
A=21b100−b2
On differentiating w.r.t b,
dbdA=21[(b)′100−b2+b(100−b2)′]=21[(1)100−b2+b×2100−b21×(0−2b)]=21[100−b2−100−b2b2] ...(iii)
Now, put dbdA=0
21[100−b2−100−b2b2]=0⇒100−b2−100−b2b2=0⇒100−b2=100−b2b2⇒100−b2×100−b2=b2⇒(100−b2)=b2⇒100=2b2⇒b2=50⇒b=50=52 ....(iv)
Consider (iii)
dbdA=21[100−b2(100−b2)−b2]=21[100−b2100−2b2]
⇒dbdA=100−b250−b2
Again differentiating w.r.t b,
db2d2A=100−b2(100−b2)(0−2b)−(50−b2)(2100−b21)(0−2b)
=100−b2−2b100−b2+(100−b2b(50−b2))
=(100−b2)3/2−2b(100−b2)+b(50−b2)
=(100−b2)3/2−200b+2b2+50b−b3=(100−b2)3/2−150b+2b2−b3
Putting value from (iv) here,
db2d2A=(100−50)3/2−150(52)+2(50)−(50)3/2
=(50)3/2−750(2)+100−(50)3/2<0
So, area is maximum.
Now, put value of b from (iv) in (i),
p=100−50=50=52
Answer: Other two sides are 52,52 units.
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