Question

1. ∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from
point S is 9. Find the ratio A(∆MNT):
A (∆QRS)​

Answer 1

Given:

∆MNT ~ ∆QRS.

Length of altitude drawn from point T is 5 and length of altitude drawn from  point S is 9

To find:

The ratio A(∆MNT): A (∆QRS)​

Solution:

Let's assume,

"TO" → represents the altitude of  ΔMNT

"SP" → represents the altitude of  ΔQRS

It is given,

TO = 5

SP = 9

We know that,

The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Since  ∆MNT ~ ∆QRS, so according to the above theorem we get

$\frac{Area (\triangle MNT)}{Area (\triangle QRS )} = \frac{(TO)^2}{(SP)^2}$

now, substituting the values of TO = 5 & SP = 9, we get

$\implies \frac{Area (\triangle MNT)}{Area (\triangle QRS )} = \frac{(5)^2}{(9)^2}$

$\implies \bold{\frac{Area (\triangle MNT)}{Area (\triangle QRS )} = \frac{25}{81} }$

Thus, $\boxed{\bold{\frac{Area (\triangle MNT)}{Area (\triangle QRS )} = \underline{\frac{25}{81} }}}$.

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Answer 2

Answer:

∆MNT- ∆QRS [Given] ∴ ∠M ≅ ∠Q (i) [Corresponding angles of similar triangles] In ∆MLT and ∆QPS, ∠M ≅ ∠Q [From (i)] ∠MLT ≅ ∠QPS [Each angle is of measure 90°] ∴ ∆MLT ~ ∆QPS [AA test of similarity]