In the Figure FD\\AC, the area of AEF = 6sq.cm. AE = 3cm, BC = 3cm, CD = 5cm, < BCD is an obtuse angle. Find the length of BD.
Area of triangle AEF = 6sq. cm
area of triangle = \(\frac{1}{2}\)bh (Line DX makes right angles with the parallel lines)
6 = \(\frac{1}{2}\) x 3 x h
6 = 3h
3h = \(\frac{12}{3}\)
h = 4 = DX
From D, C x D, CX\(^2\) = 5\(^2\)
- 4\(^2\) = 25 - 16 = 9
Cx = 3. From angle B x D, Bx = 6(i.e. 3 + 3)
BD\(^2\) = 4\(^2\) + 6\(^2\)
= 16 + 36 = 52
BD = \(\sqrt{4 \times 13}\)
= 2\(\sqrt{13}\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}