Question

51. AD is altitude of an isosceles triangle ABC in
which AB = AC = 30 cm and BC = 36 cm. A
point is marked on AD in such a way that
BOC = 90°. Find the area of quadrilateral
ABOC .
Area of quadrilateral ABOC
= Art (A ABC) - Ar: (A BOC)
1
I
= = XBCXAD X OB XOC
It can easily be shown that A BOD = A COD
= OB = 0C = rcm (let) and
OB-OC-= BC as BOC = 90°​

Answer 1

Answer:

ok wel come to the day yyyyi,❤️❤️