What is the perimeter of an isosceles triangle that has a perpendicular height of 5.32 cm and an area of 21.2268 square cm showing all work with answers?

Answer 1

21.2268 = 0.5 x base x 5.32

base = (21.2268)/(0.5 x 5.32) = (21.2268/2.66) = 7.98

Half the base = 3.99

Now one of the equal sides of the isosceles triangle (using pythagoras theorem) is

sqrt(3.99^2 + 5.32^2) = sqrt(15.92 + 28.30) = sqrt(44.22) = 6.65

Therefore perimeter of triangle is = base + 2x6.65 = 7.98 + 13.3 = 21.28 cm

Answer 2

There are two possible configurations. The base of the triangle could be the "third" side of the isosceles triangle or it could be one of the equal sides. For the following answer it is assumed that it is the third side. This assumption is made on the basis that the fact that you ask this question indicates that you are still working at a fairly basic level of geometry and would not be expected to handle the longer process complicated

Answer: The perimeter is 21.28 cm.

The area, A, is 21.2268 sq cm and the height, H, is 5.32 cm.

Therefore the base, B, is 2*A/H = 2*21.2268/5.32 = 7.98 cm.

Now, suppose the slant sides are S cm each. The slant side, the height (altitude) and half the base make a right angled triangle whose hypotenuse is the slant side.

So, by Pythagoras, S^2 = H^2 + (B/2)^2 = 5.32^2 + (7.98/2)^2 = 28.3024 + 15.9201 = 44.2225

and so S = 6.65 cm.

Then perimeter = 2S + B = 21.28 cm.