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.

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.

As QS = TQ + TS = 10 (given)

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

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

1. 8

2. 9

3. 20

4. 3

5. 17

6. 2.2

7. 11.25

8. 1

9. 14

10. 2

step-by-step explanation:

thats the first half