AD is altitude of an isosceles triangle ABC in which AB=BC=30cm and BC=36cm.A point O is marked on AD in such a way that <BOC=90°.Find the area of the quadrilateral ABOC.
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AD is altitude of an isosceles triangle ABC in which AB=BC=30cm and BC=36cm.A point O is marked on AD in
Answer:
Correct option is
A
108 cm
2
Given, in an isosceles △ABC,AB=AC=30 cm BC=36 cm
AD is altitude on BC
O is an ,midpoint on AD such that ∠BOC=90
Since, AD⊥BC and AB=AC
Therefore, D is the midpoint of BC, also BD=DC
=
2
36
= 18 cm
In △ADB, AB
2
=BD
2
+AD
2
Length AD=
AB
2
−BD
2
=
30
2
−18
2
=
900−324
=
576
= 24 cm
Therefore, area of △ABC=
2
1
BC×AD
=
2
1
36×24
= 432cm
2
In △OBC,OB=OC
Let OB=OC=a,BC=36cm
Then, BC
2
=a
2
+a
2
⇒BC
2
=2a
2
⇒36
2
=2a
2
⇒a
2
=
2
1296
= 648
Area of △OBC=
2
1
×a×a
=
2
1
×a
2
=
2
1
×648
= 324cm
2
Hence, the area of quadrilateral ABOC = Area of △ABC − Area of △OBC
= (432−324)cm
= 108cm