Please solve the 15th problem Please do not provide any link Q 15 AD - Maths - Area and Perimeter of Plane Figure

Question

Please solve the 15th problem

Please do not provide any link
Q 15 AD is altitude of an isosceles triangle ABC in which AB = AC =
30 cm and BC = 36 cm A point O is marked on AD in such a way that
∠ B O C = 90 ° Find the area of the quadrilateral ABOC

Surabhi Grover · Accepted Answer

Dear student
Consider the figure:

<img alt="" src=
"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/ck_59e8b77686d09%20jpg"
style="width: 250px; height: 189px;">
Since ABC is an isosceles triangle and AD is an altitudeBD=DC=18 cm⇒AD2=AC2-DC2⇒AD2=900-324=576⇒AD=24 cmLet OD be x cmConsider ∆BOD and ∆CODDC=BD= 18cm∠D=90°OD is common⇒∆BOD ≅∆COD (By SAS criteria)∠1=∠2∠3=∠4 (By CPCT)Also ∠1+∠2=90°∠2 +∠3=90°S0∠1+∠3=90°Also ∠2+∠4=90⇒∠1=∠2=∠3=∠4=45°⇒OD=DC (SIdes opposite to equal angles are equal)⇒x=18cmArea (∆ABC)-Area(∆BOC)=Required ARea=12×24×36-12×18×36=12×36×6=108cm2 
Regards

Please solve the 15th problem. Please do not provide any link. Q.15. AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that ∠ B O C = 90 ° . Find the area of the quadrilateral ABOC.

Please solve the 15th problem.

Please do not provide any link.
Q.15. AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that $∠ B O C = 90 °$ . Find the area of the quadrilateral ABOC.