Question

The sides of a right triangle are in an AP. The area and the perimeter of the triangle are numerically equal. Find its perimeter.

• A. 24 units
• B. 34 units
• C. 44 units
• D. Can’t say, the common difference should be given

Answer 1

The correct option is A

24 units

Let us take the sides of the first triangle to be (a - d), a and (a+d), with a and d both positive. We assume this so that we have positive values for the sides.
Perimeter, P = (a - d) + a + (a + d) = 3a.
Since the triangle is right angled, (a + d) is the hypotenuse as it is the largest side.
Hence, by Pythagoras theorem,
$(a+d{)}^2=a^2+(a-d{)}^2$
Simplifying the equation, we get $2ad=a^2-2ad$
$\Rightarrow a^2-4ad=0$
$\Rightarrow a(a-4d)=0$ so, a = 0 and a = 4d (we cannot take a = 0 as side of a triangle cannot be zero).
Since (a + d) is the hypotenuse, the other sides a and (a - d) form the legs of the right angled triangle, as shown:



So, area, $A=\frac{1}{2}\times a\times (a-d)=\frac{1}{2}\times 4d\times (4d-d)=6d^2$
Since it is given that area = perimeter, we have, $P=3a=6d^2$
a = 4d and hence, we have, $12d=6d^2$
$\Rightarrow 6d^2-12d=0\Rightarrow d(6d-12)$ = 0 so, d = 0 and d = 2 (we cannot take d = 0, because it will make all sides equal)
Hence, d = 2 and a = 4d = 8.
Therefore, the sides are a-d, a, a+d which is 6, 8 and 10

$\therefore$Perimeter = 6+8+10 = 24 units.