Question

parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm. point F is on RS exactly 5 cm from S. let T be the intersection of PF and QS. draw a diagram and find the lengths of TS and TQ.

Answer 1

Solution-

Given that,

In the parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm.

Then considering ΔPQT and ΔSTF,

1-    ∠FTS ≅ ∠PTQ            ( ∵ These two are vertical angles)

2-   ∠TFS ≅ ∠TPQ            ( ∵ These two are alternate interior angles)

3-   ∠TSF ≅ ∠TQP            ( ∵ These two are also alternate interior angles)

If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.

∴ ΔFTS ∼ ΔPTQ, so corresponding side lengths are in proportion.

$\Rightarrow \frac{PQ}{FS} =\frac{TQ}{TS} =\frac{TP}{TF}$

As QS = TQ + TS = 10 (given)

If TS is x, then TQ will be 10-x. Then putting these values in the equation

$\Rightarrow \frac{PQ}{FS} =\frac{TQ}{TS}$

$\Rightarrow \frac{8}{5} =\frac{10-x}{x}$

$\Rightarrow x=3.85$

∴ So TS = 3.85 cm and TQ is 10-3.85 = 6.15 cm