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alisha [4.7K]
4 years ago
8

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

Mathematics
1 answer:
jenyasd209 [6]4 years ago
6 0

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}              

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