Answer:
a) AC = [tex]\sqrt{13}[/tex]
b) Area = [tex]\frac{\sqrt 3}{4}[/tex] × [tex]a^{2}[/tex]
Step-by-step explanation:
a) From question,
AC = BC, CD⊥AB
Now in ΔCAD and ΔCBD
AC=BC, ∠A = ∠B and AD=BD (because in isosceles triangle perpendicular bisects the side).
then, from SAS potulates
ΔCAD≅ΔCBD
So,
AD = [tex]\frac{AB}{2}[/tex] = [tex]\frac{4}{2}[/tex] = 2 in
From Pythagorean theorem in ΔADC
[tex]AC^{2}[/tex] = [tex]AD^{2}[/tex] + [tex]CD^{2}[/tex]
[tex]AC^{2}[/tex] = [tex]2^{2}[/tex] + [tex]3^{2}[/tex]
[tex]AC^{2}[/tex] = 4 + 9 = 13
AC = [tex]\sqrt{13}[/tex]
b) In given ΔABC,
AB = BC = AC = a, means ΔABC is a equilateral triangle.
So, area of equilateral triangle is
Area = [tex]\frac{\sqrt 3}{4}[/tex] × [tex]side^{2}[/tex]
side = a
then,
Area = [tex]\frac{\sqrt 3}{4}[/tex] × [tex]a^{2}[/tex]
