Question

11. In the given figure, O is the centre of the circle and
AB is a straight line, then ACDO is always a

(1) Equilateral triangle
(2) Isosceles triangle
(3) Right triangle
(4) Isosceles right triangle​

Answer 1

Answer:

Given that:- △ABC is an isosceles triangle and ∠ABC=90°

∴AB=BC

△ABE∼△ACD(∵All equilateral triangles are similar)

To find:-

ar(△ACD)

ar(△ABE)

=?

Solution:-

In △ABC,

Using pythagoras theorem,

AC

2

=AB

2

+BC

2

AC

2

=AB

2

+AB

2

[∵AB=AC]

AC

2

=2AB

2

.....(i)

Now In △ABE and △ACD

△ABE∼△ACD(Given),

As we know that ratio of area of similar triangles is equal to the ratio of squares of their corresponding sides.

∴

ar(△ACD)

ar(△ABE)

=

AC

2

AB

2

⇒

ar(△ACD)

ar(△ABE)

=

2AB

2

AB

2

[From(i)]

⇒

ar(△ACD)

ar(△ABE)

=

2

1

⇒ar(△ABE):ar(△ACD)=1:2

Step-by-step explanation:

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