Question

In a quadrilateral ABCD, two diagonals AC and BD intersect at O. If ∠DAB = ∠BCD =  60°, DO = 4 cm and AO = 5 cm, then find the value of AC + BC.

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Answer 1

to find out

AC+BC

solution

we know when ∠A = ∠C = 60°

so than there will be ∠B = ∠D = 120° as per symmetry

and if opposite angles of a quadrilateral are equal, it will become a parallelogram  

so here

AC = 2 OA = 10cm (the diagonals of //gm bisect each other )

and

we take here triangle ABC

so here sin(120) = BC/AC 

put here value and we get BC

sin(120) = {BC}{10}

BC = 8.66

so that AC + BC will be

AC + BC = 10 + 8.66

AC + BC = 18.66 cm

Answer 2

Answer:

18 cm

Step-by-step explanation:

we know when ∠A = ∠C = 60

and if *opposite angles of a quadrilateral are equal, it will become a parallelogram*  

so here

AC = 2 OA = 2*5 = 10cm (the diagonals of //gm bisect each other )

BD = 2 DO = 2*4 = 8 cm (the diagonals of //gm bisect each other )

AB + CD = 18 cm