Q. 224.3( 23 Votes )

# AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area () : Area () = 3 : 4.

Answer :

**Given:**AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed

**To prove:**Area () : Area () = 3 : 4.

**Proof:**

Construct the figure according to the conditions given.

We have,

ABC is an equilateral triangle

Let one side AB be 2XSince in equilateral triangle all the sides are of equal length.

⇒ AB=BC=AC= 2X

∵ AD⊥BC

Since perpendicular bisects the given side into two equal parts,

then BD=DC=x

Now, In ADB

By Pythagoras theorem,

AB

^{2}= AD

^{2 }+ BD

^{2}

AD^{2}=AB^{2} - BD^{2}AD^{2} = (2x)^{2}-(x)^{2}AD^{2} =3x^{2}

AD= cm

ABC and ADE both are equilateral triangles

Since, all the angles of the equilateral triangle are of 60^{°}.

∴ABCADE [By AA similarity]

By the theorem which states that the areas of two similar triangles are in the ratio

of the squares of the any two corresponding sides.

Hence,Area () : Area () = 3 : 4

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