Respuesta :

konrad509

[tex] \dfrac{1}{2}\cdot(x+1)\cdot 2=x\cdot 1\\\\
x+1=x\\\\
0=1 [/tex]

Therefore, there aren't any.

altavistard

Let's assume that there is a value (or values) of x for which the areas of these two figures aer the same:

Triangle Rectangle

A = (1/2)(2)(x + 1) = (x)(1)

Simplifying,

x + 1 = x

This becomes: 1 = 0

This is never true, regardless of what x is chosen.