The diagonals of rhombus ABCD intersect at E. Given that m∠CAD=20° and CE=4, find the m∠CDA.

The Diagonals Of Rhombus ABCD Intersect At E. Given That MCAD=20 And CE=4, Find The MCDA.

Answers

Answer 1

Answer:

< CDA = 140 degrees

Step-by-step explanation:

For a rhombus, the sum of the interior angles is 360 degrees and also the opposite angles are equal i.e <ABC = <CDA and <BAD = <BCD

Given

<CAD = 20 degrees

<BCE = 20 degrees (based on geometry)

<BAD = <CAD + <BCE

<BAD = 20 + 20

<BAD = 40 degrees

Since  <BAD =  <BCD, then  <BCD = 40 degrees

Get <CDA

<CDA + <ABC + <BAD +   <BCD  = 360

<CDA + <CDA + <BAD +   <BCD  = 360

2 < CDA + 40 + 40 = 360

2< CDA + 80 = 360

2< CDA = 360-80

2< CDA = 280

< CDA = 280/2

< CDA = 140 degrees


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The diagonals of rhombus ABCD intersect at E. Given that m∠CAD=20° and CE=4, find the m∠CDA.

Answers

Given :

The diagonals of rhombus ABCD intersect at E.

∠CAD = 20°.

To Find :

The angle ∠CDA.

Solution :

We know, diagonals of a rhombus bisects each other perpendicularly.

So, ∠DEA = 90°.

In triangle ΔEAD :

∠EAD + ∠AED + ∠EDA = 180°

20° + 90° + ∠EDA = 180°

∠EDA = 70°

Now, we know diagonal of rhombus also bisect the angle between two sides .

So, ∠CDA = 2∠EDA

∠CDA = 2×70°

∠CDA =140°

Therefore, ∠CDA is 140°.

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[tex]\frac{6x+4}{3} =\frac{5x-5}{8} \\\\24(\frac{6x+4}{3} =\frac{5x-5}{8})\\\\48x+32=15x-15\\\\48x+32-32=15x-15-32\\\\48x-15x=-47+15x-15x\\\\33x=-47\\\\x=-\frac{47}{33}x[/tex]

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