Please solve the 15th problem.
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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.

Dear student.
Consider the figure:


Since ABCis an isosceles triangle and AD is an altitudeBD=DC=18 cmAD2=AC2-DC2AD2=900-324=576AD=24 cmLet OD be x cmConsider BODand CODDC=BD= 18cmD=90°OD is commonBODCOD (By SAScriteria)1=23=4 (By CPCT)Also 1+2=90°2 +3=90°S01+3=90°Also 2+4=901=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

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