Question

Ad is altitude of an isosceles triangle abc in which ab=ac=30 cm and bc=36cm. A point o is marked on ad in such a way that angle boc=90 degree.Find the area of quadrilateral aboc

Answer 1

Refer the attachment for figure

In ∆ABD and ∆ADC,


AB = AC (given)
AD = AD (common)
Angle ADB = angle ADC = 90° (Since AD is altitude)

=> ∆ABD is congruent to ∆ADC (RHS)

=> BD = DC (CPCT)

Now, in ∆BOD and ∆DOC

BD = DC (shown above)
angle BDO = angle ODC = 90°
DO = DO (common)


=> ∆BOD is congruent to ∆DOC (SAS)

=> BO = CO (CPCT)

Now in ∆BOC

BO = CO (shown above)

and angle BOC = 90° (given)

So by Pythagoras theorem,

BO² + CO² = BC²

Given BC = 36
and BO = CO

=> CO² + CO² = (36)²

=> 2CO² = 1296

=> CO² = 1296/2

=> CO² = 648

=> CO = √648


Now BO = CO

=> BO = CO = √648


Now area of ∆ABC

= 1/2 × base × height


We have to find height


In ∆ ABD

BD = 1/2 BC

=> BD = 18

So by Pythagoras Theorem,

AD = √(30² - 18²)

=> AD = √(900 - 324)

=> AD = √576

=> AD = 24


so height = 24

=> area of ∆ABC = 1/2 × BC × AD

= 1/2 × 36 × 24

= 18 × 24

= 432 cm²


Now area of ∆ BOC

= 1/2 × BO ×CO

= 1/2 × √648 × √648

= 1/2 × 648

= 324


Area of Quadrilateral ABOC =

area of ∆ABC - area of ∆BOC

=> 432 - 324

= 108 cm²


Your answer :- 108

Answer 2

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