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Geometry Practice Test 1

Question : 8 of 40

Marks: +1, -0

Solution: 👈: Video Solution

Area of triangle ADE = (Area of DEBC)

⇒ Area of DEBC =

\frac{1}{2}

area of ΔABC

So,

\frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle ADE}

\frac{2}{1}

The ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding sides.

So,

\left(\frac{AB}{AD}\right)^2

\frac{2}{1}

\frac{AB}{AD}

\frac{\sqrt{2}}{1}

\frac{AD+AB}{AD}

\frac{\sqrt{2}}{1}

1 +

\frac{BD}{AD}

\frac{\sqrt{2}}{1}

\frac{BD}{AD}

\frac{\sqrt{2}}{1} - 1

\frac{BD}{AD}

\frac{\sqrt{2}-1}{1}

\frac{AD}{BD}

\frac{1}{\sqrt{2}-1}

So, the required ratio = 1 :

\sqrt{2}

- 1

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