Triangle perimeter SSAT Q: if area of rt triangle is 9, one leg is 3, find perimeter

[tathagata7]

Hello,

I have the below question on my sample SSAT test and I cannot figure it out.

What is the perimeter, in ft, of a right triangle which has an area of 9 ft² and one leg has a length of 3 ft?

Answer:

\(\displaystyle 9\, +\, 3\, \sqrt{\strut 5\,}\)

I understand that:

a + b + c = perimeter

height (base) /2 = area

but I do not know how they derive that answer

 

[Steven G]

Hello,

I have the below question on my sample SSAT test and I cannot figure it out.

What is the perimeter, in ft, of a right triangle which has an area of 9 ft² and one leg has a length of 3 ft?

Answer:

I understand that:

a + b + c = perimeter

height (base) /2 = area

but I do not know how they derive that answer

Well you know that height (base) /2 = area and you know two of the three (height, base and area). So figure out the third. Then.....

 

[pka]

What is the perimeter, in ft, of a right triangle which has an area of 9 ft² and one leg has a length of 3 ft?

If \(\displaystyle a~\&~b\) are the lengths of the legs of the right triangle, then its area is \(\displaystyle \frac{1}{2}ab\).
So \(\displaystyle \frac{3}{2}a=9\) or \(\displaystyle a=6\). The perimeter is \(\displaystyle 6+3+\sqrt{6^2+3^2}\).

 

[tathagata7]

Well you know that height (base) /2 = area and you know two of the three (height, base and area). So figure out the third. Then.....

Okay, is this correct?

if hb/2 = 9 with b=3; 3h/2 = 9; h = 18/3; h = 6

a² + b² = c²

6² + 3^{2 =} c²

36 + 9 = c²

45 = c²

9*5 = c²

3 sq rt 5


resulting in 3+6+3 sq rt 5

 

[tathagata7]

If \(\displaystyle a~\&~b\) are the lengths of the legs of the right triangle, then its area is \(\displaystyle \frac{1}{2}ab\).
So \(\displaystyle \frac{3}{2}a=9\) or \(\displaystyle a=6\). The perimeter is \(\displaystyle 6+3+\sqrt{6^2+3^2}\).

Thank you - I think I worked it out. . . .