Question

the perimeter of a right angled triangle is 72 cm and its area is 216 cm square find the sum of the lengths of its perpendicular sides .

Answer 1

Hello mate
Here's ur answer


This gives you 3 equations in three unknowns.

Lets call the perpendicular sides aa and bb, the the remaining side cc.

From the perimeter value, our first equation is a+b+c=72a+b+c=72.
From  the area value, and knowing that the area of a right triangle is  12ab12ab, we have 12ab=21612ab=216.

The third equation comes from the fact that the pythagorean theorem holds for right triangles, so a2+b2=c2a2+b2=c2

So our equations are:
a+b+c=72a+b+c=72
12ab=21612ab=216
a2+b2=c2a2+b2=c2

We can rearrange the first equation into c=72−a−bc=72−a−b. Then we can plug this into the third equation:
a2+b2=(72−a−b)2a2+b2=(72−a−b)2
a2+b2=5184−144a+a2−144b+2ab+b2a2+b2=5184−144a+a2−144b+2ab+b2
0=5184−144a−144b+2ab0=5184−144a−144b+2ab
0=2592−72a−72b+ab0=2592−72a−72b+ab (divide out by 2)

Now we use the second equation to note that ab=432ab=432 and b=432ab=432a:
0=2592−72a−31104a+4320=2592−72a−31104a+432
0=36−a−432a+60=36−a−432a+6 (divide out by 72)
0=42a−a2−4320=42a−a2−432

So now we just need the roots of the equation 0=a2−42a+4320=a2−42a+432, which are a=18a=18and a=24a=24.

We might get a little nervous at this point, because there are two possible aa. However, this is to be expected, because we have 2 perpendicular sides.

We are looking for the sum of the two perpendicular sides, a+ba+b. Recalling that b=432ab=432a, this means we are looking for a+432aa+432a.

For a=18a=18, 18+43218=18+24=4218+43218=18+24=42.
For a=24a=24, 24+43224=24+18=4224+43224=24+18=42.

Therefore, either choice of a yields the same sum of perpendicular sides, and that sum is 4242.


Hope it may helps you
@neha7755