that the ratio of the areas of two similar triangles is equal to the square of
the ratio of their corresponding sides.
Answers
Step-by-step explanation:
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 ΔABCΔABC and 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△PQR
ar△ABC
=(
PQ
AB
)
2
=(
QR
BC
)
2
=(
RP
CA
)
2
Since area of triangle =\dfrac { 1 }{ 2 } \times base\times altitude=
2
1
×base×altitude
To find the area of ΔABCΔABC and ΔPQRΔPQR draw the altitudes ADAD and PEPE from the vertex AA and PP of ΔABC ΔABC and ΔPQRΔPQR
Now, area of ΔABCΔABC =\dfrac { 1 }{ 2 } \times BC\times AD=
2
1
×BC×AD
area of ΔPQRΔPQR =\dfrac { 1 }{ 2 } \times QR\times PE=
2
1
×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△PQR
ar△ABC
=
2
1
×QR×PE
2
1
×BC×AD
\dfrac { ar\triangle ABC }{ ar\triangle PQR } =\dfrac { BC\times AD }{ QR\times PE }
ar△PQR
ar△ABC
=
QR×PE
BC×AD
Now in \triangle ABD△ABD and \triangle PQE△PQE it can be seen
∠ABC = ∠PQR∠ABC=∠PQR (Since ΔABC\sim ΔPQRΔABC∼ΔPQR )
∠ADB = ∠PEQ∠ADB=∠PEQ (Since both the angles are 90°90°)
From AAAA criterion of similarity ΔADB\sim ΔPEQΔADB∼ΔPEQ
\dfrac { AD }{ PE } =\dfrac { AB }{ PQ }
PE
AD
=
PQ
AB
Since it is known that ΔABC\sim ΔPQRΔABC∼ΔPQR
\dfrac { AB }{ PQ } =\dfrac { BC }{ QR } =\dfrac { CA }{ RP }
PQ
AB
=
QR
BC
=
RP
CA
Substituting this value in equation , we get
\dfrac { ar\triangle ABC }{ ar\triangle PQR } =\dfrac { AB }{ PQ } \times \dfrac { AD }{ PE }
ar△PQR
ar△ABC
=
PQ
AB
×
PE
AD
we can write
\dfrac { ar\triangle ABC }{ ar\triangle PQR } ={ \left( \dfrac { AB }{ PQ } \right) }^{ 2 }
ar△PQR
ar△ABC
=(
PQ
AB
)
2
Similarly we can prove
\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△PQR
ar△ABC
=(
PQ
AB
)
2
=(
QR
BC
)
2
=(
RP
CA
)
2
Step-by-step explanation:
I m in class 10 and I want to be ur friend.
brainliest please joi aai akhom ☺
