Answer:
see explanation
Step-by-step explanation:
ΔABC is isosceles and AM is the perpendicular bisector to BC, hence
BM = MC = 4 cm
Using Pythagoras' identity on ΔABM, then
AB² = AM² + BM², that is
7² = AM² + 4²
49 = AM² + 16 ( subtract 16 from both sides )
33 = AM² ( take the square root of both sides )
AM = [tex]\sqrt33}[/tex]
The area (A) of ΔABC is found using
A = [tex]\frac{1}{2}[/tex] bh ( b is the base BC and h the height AM )
A = [tex]\frac{1}{2}[/tex] × 8 × [tex]\sqrt{33}[/tex] = 4[tex]\sqrt{33}[/tex] cm² ≈ 23 cm²
