Question

what is the lenght of altitude of an equilateral triangle of side 8cm[use pythagoran theorem]

Answer 1

Answer:

$\sqrt{48}$ cm

Step-by-step explanation:

We call the equilateral triangle of side 8cm is ΔABC with AM is its altitude

In ΔABM and ΔACM both square at M, we have:

AB = AC (ΔABC is an equilateral triangle)

AM is the common edge

⇒ΔABM = ΔACM (hypotenuse - Right angle's edge)

⇒BM = CM = $\frac{BC}{2}$ = $\frac{8cm}{2\\}$ = 4cm(Corresponding edges)

In ΔABM square at M, we have:

$AM^{2} + MB^{2} = AB^{2}$ (Pythagoras theorem)

⇒$4^{2} + AM^{2} = 8^{2}$

⇒$AM^{2}$ = $8^{2} - 4^{2}$ = 64 - 16 = 48

⇒AM = $\sqrt{48} cm$

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