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2 years ago

XY bisects DF at point E. If DE=2y and EF=8y-3, find the value of DF

Mathematics

1 answer:

nirvana33 [79]2 years ago

7 0

Answer:

Step-by-step explanation:

If XY bisects DF at point E, then DE = EF.

Given DE = 2y and EF = 8y-3

2y = 8y-3

subtract 8y from both sides

2y-8y = 8y-8y-3

-6y = -3

y = 3/6

y = 1/2

y = 0.5

DF = DE+EF

DF= 2y+8y-3

DF = 10y - 3

DF = 10(0.5)-3

DF = 5-3

DF = 2

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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

Doss [256]

<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.

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

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

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3 years ago

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Minchanka [31]

Answer:

$- \frac{16}{ 25 } \div ( - \frac{4}{5} ) = \frac{16}{25} \times \frac{5}{4} = \frac{4}{5} \:or \: 0.8$

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2 years ago

Read 2 more answers

How to graph y=-4|x+2|-5

Elina [12.6K]

$y= -4|x+2| -5 \\ \\x= -6 \Rightarrow y=-4|-6+2|- 5= -4|-4|- 5= -4*4 -5 =\\ \\=-16-5=-21\\ \\ \\x= -4\Rightarrow y=-4|-4+2|- 5= -4|-2|- 5= -4*2 -5 =\\ \\ =-8-5=-13\\ \\ \\x= -2 \Rightarrow y=-4|-2+2|- 5= -4|0|- 5= -4*0 -5 =-5\\ \\ \\x= 0 \Rightarrow y=-4|0+2|- 5= -4|2|- 5= -4*2 -5 =-8-5=-13\\ \\ \\x= 2 \Rightarrow y=-4|2+2|- 5= -4| 4|- 5= -4*4 -5 =-16-5=-21$