+0

# Geometry

0
4
2
+867

An isosceles triangle has two sides of length $7$ and an area of $14.$ What is the product of all possible values of its perimeter?

Jun 4, 2024

#1
+729
-1

Let the base of the triangle be $b$ and the height be $h.$ Since the area of the triangle is 14, we have

$\frac{bh}{2} = 14 \Rightarrow bh = 28.$

By the Pythagorean Theorem, we have

$b^2 = 7^2 - \left(\frac{b}{2}\right)^2 = 49 - \frac{b^2}{4}.$

Multiplying both sides by 4, we get

$4b^2 = 196 - b^2 \Rightarrow 5b^2 = 196 \Rightarrow b = \frac{4\sqrt{5}}{5}.$

The perimeter of the triangle is $2b+7 = 2\left(\frac{4\sqrt{5}}{5}\right) + 7 = \frac{8\sqrt{5}}{5} + 7.$ The product of all possible values of the perimeter is

$\left(\frac{8\sqrt{5}}{5} + 7\right)\left(\frac{-8\sqrt{5}}{5} + 7\right) = 49 - \frac{320}{25} = \boxed{\frac{825}{25} = 33}.$

Jun 4, 2024
#2
+129745
+1

bh =  28

h =  28/b

So.....by the Pythagorean Theorem

(b/2)^2  + (28/b)^2   =7^2

b^2/4 + 784 / b^2  = 49

b^4 /4 + 784  = 49b^2

b^4/4 - 49/b^2  + 784  = 0

The  only  positive value for b ≈ .25

Only possible perimeter ≈    7 + 7 + .25  ≈  14.25

Jun 4, 2024