Solve 3x^2+19x +6 please help
1 answer:
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<u>Solution-</u>
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)
<em>If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.</em>
∴ 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
