Question

@ 1. The areas of two similar triangles area and ka.
What is the ratio of the corresponding side lengths of
the triangles?
(A) 1 : k2
(B) 1:k (C) 1: a? (D) 1:a plz tell the ans

Answer 1

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 ΔABC and ΔPQR

According to the stated theorem

ar△PQRar△ABC=(PQAB)2=(QRBC)2=(RPCA)2

Since area of triangle =21×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 =21×BC×AD

area of ΔPQR =21×QR×PE

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

ar△PQRar△ABC=21×QR×PE21×BC×AD

ar△PQRar△ABC

Answer 2

Given: ratio of area of two similar triangles $=1:k$.

To find: the ratio of the corresponding side lengths of  the triangles.

Solution:

Know that, 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.

Find the ratio of the sides.

$\[\frac{{sid{e_1}}}{{sid{e_2}}} = \sqrt {\frac{1}{k}} \]$

Therefore, the ratio of the sides is $\sqrt{1}:\sqrt{k}$.