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.
i couldn't find in internet pls write the answer if you know only then
Attachments:
![](https://hi-static.z-dn.net/files/d3d/bdd3f540580cbd1363b35f4226554446.jpg)
Answers
Answered by
13
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
Attachments:
![](https://hi-static.z-dn.net/files/d20/a1424ef8db1e40a59fd8426c862e6581.jpg)
![](https://hi-static.z-dn.net/files/d71/4957c41c4dc231265ee229cc8e81eb3a.jpg)
Answered by
0
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
Similar questions
CBSE BOARD XII,
1 month ago
Geography,
1 month ago
Math,
1 month ago
Hindi,
3 months ago
Math,
8 months ago