# The ratio of the areas of two similar triangles is equal to the square of the ratios of their corresponding *​

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If two triangles are similar, then the ratio of the area of both triangles is proportional to square of the ratio of their corresponding sides.

To prove this theorem, consider two similar triangles ΔABC and ΔPQR

According to the stated theorem

ar△PQR

ar△ABC

=(

PQ

AB

)

2

=(

QR

BC

)

2

=(

RP

CA

)

2

Since area of triangle =

2

1

×base×altitude

To find the area of ΔABC and ΔPQR draw the altitudes AD and PE from the vertex A and P of ΔABC and ΔPQR

Now, area of ΔABC =

2

1

area of ΔPQR =

2

1

×QR×PE

The ratio of the areas of both the triangles can now be given as:

ar△PQR

ar△ABC

=

2

1

×QR×PE

2

1

ar△PQR

ar△ABC

=

QR×PE

Now in △ABD and △PQE it can be seen

∠ABC=∠PQR (Since ΔABC∼ΔPQR )

∠ADB=∠PEQ (Since both the angles are 90°)

From AA criterion of similarity ΔADB∼ΔPEQ

PE

=

PQ

AB

Since it is known that ΔABC∼ΔPQR

PQ

AB

=

QR

BC

=

RP

CA

Substituting this value in equation , we get

ar△PQR

ar△ABC

=

PQ

AB

×

PE

we can write

ar△PQR

ar△ABC

=(

PQ

AB

)

2

Similarly we can prove

ar△PQR

ar△ABC

=(

PQ

AB

)

2

=(

QR

BC

)

2

=(

RP

CA

)

2

157
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