Question

Solve the problems below. Please answer with completely simplified exact value(s) or expression(s).

Answer 1

Answer:

a) AC = $\sqrt{13}$

b) Area = $\frac{\sqrt 3}{4}$ × $a^{2}$

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 = $\frac{AB}{2}$ = $\frac{4}{2}$ = 2 in

From Pythagorean theorem in ΔADC

$AC^{2}$ = $AD^{2}$ + $CD^{2}$

$AC^{2}$ = $2^{2}$ + $3^{2}$

$AC^{2}$ = 4 + 9 = 13

AC = $\sqrt{13}$

b) In given ΔABC,

               AB = BC = AC = a,  means ΔABC is a equilateral triangle.

So, area of equilateral triangle is

                              Area = $\frac{\sqrt 3}{4}$ ×  $side^{2}$

                                side = a

then,

       Area = $\frac{\sqrt 3}{4}$ × $a^{2}$