Question

in the adjoining figure ABCD is a rhombus whose diagonal AC and BD intersect at O if ab is equal to 10 centimetre and diagonal BD is equal to 16 cm find the length of diagonal AC ??​

Answer 1

Answer:

according to question :-

DB = 16

so OB = 16/2 = 8

AB = 10

so by using Pythagoras theorem :-

AB ^2 = AO^2 + OB ^2

10^2 = AO^2 + 8^2

100-64 = AO^2

36 = AO ^2

√36 = AO

6 = AO

now AC = 2*AO = 2*6 = 12

So AC = 12 cm

hope you understand plz thank my answer

Answer 2

Answer:

bro diagonals of rhombus bisect each other at 90*

means BO=8cm

Step-by-step explanation:

IN TRIANGLE AOB using phythagorus theorem

AB^2= AO^2 + BO^2

AB^2 - BO^2 = AO^2

10^2 - 8^2 = AO^2

100-64 = AO^2

√36 = AO

AO = 6cm.

Therefore, AC = 2 × AO = 2 × 6 = 12 cm.