Question

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio
of their corresponding medians.

Answer 1

Step-by-step explanation:

ANSWER

Given:- Let △ABC∼△PQR

AD and PS are corresponding medians.

To prove:-

ar(△PQR)

ar(△ABC)

=(

PS

AD

)

2

Proof:- In △ABC

∵AD is median.

∴BD=CD=

2

1

BC

Similarly, in △PQR

PS is median.

∴QS=RS=

2

1

QR

Now,

△ABC∼△PQR(Given)

∠B=∠Q.....(1)(Corresponding angles of similar triangles are equal)

PQ

AB

=

QR

BC

(Corresponding sides of similar triangles are proportional)

⇒

PQ

AB

=

2QS

2BD

(∵AD and PS are medians)

PQ

AB

=

QS

BD

.....(2)

Now, in △ABD and △PQS,

∠B=∠Q(From (1))

PQ

AB

=

QS

BD

(From (2))

∴△ABD∼△PQS(By SAS Property)

Therefore,

PQ

AB

=

PS

AD

.....(3)(∵Corresponding sides of similar triangles are proportonal)

Now,

∵△ABC∼△PQR

As we know that ratio of area of similar triangles is always equal to the square of ratio of their corresponding side.

Therefore,

ar(△PQR)

ar(△ABC)

=(

PQ

AB

)

2

ar(△PQR)

ar(△ABC)

=(

PS

AD

)

2

(From (3))

Hence proved.