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.
22.jpg


We have,

ABC is an equilateral triangle

Let one side AB be 2X
Since 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,
AB2 = AD+ BD2

AD2=AB2 - BD2
AD2 = (2x)2-(x)2
AD2 =3x2

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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