Question

... cute girls and cute boys plz answer this question...


prove that the ratio of 2similar triangles are equal to the ratio of the square of their corresponding sides ​

Answer 1

Answer: its given in the NCERT but still ... Here it is

Answer 2

Answer:

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 ΔABCΔABCand ΔPQRΔPQR

According to the stated theorem

\dfrac { ar\triangle ABC }{ ar\triangle PQR } ={ \left( \dfrac { AB }{ PQ }  \right)  }^{ 2 }={ \left( \dfrac { BC }{ QR }  \right)  }^{ 2 }={ \left( \dfrac { CA }{ RP }  \right)  }^{ 2 }ar△PQRar△ABC=(PQAB ) 2=(QRBC ) 2=(RPCA ) 2

Since area of triangle =\dfrac { 1 }{ 2 } \times base\times altitude=21×base×altitude

To find the area of ΔABCΔABC and ΔPQRΔPQR draw the altitudes ADADand PEPE from the vertex AA and PP of ΔABCΔABC and ΔPQRΔPQR 

Now, area of ΔABCΔABC =\dfrac { 1 }{ 2 } \times BC\times AD=21×BC×AD

area of ΔPQRΔPQR =\dfrac { 1 }{ 2 } \times QR\times PE=21×QR×PE

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

\dfrac { ar\triangle ABC }{ ar\triangle PQR } =\dfrac { \dfrac { 1 }{ 2 } \times BC\times AD }{ \dfrac { 1 }{ 2 } \times QR\times PE }ar△PQRar△ABC=21×QR×PE21×BC×AD

\dfrac { ar\triangle ABC }{ ar\triangle PQR } =\dfrac { BC\times AD }{ QR\times PE }ar△P

hope u get ans....

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