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

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°.

to find the perimeter of the rectangle this is the expression:

x-9+x-9+3x-4+3x-4=138

if we simplify it:

8x-26=138

remember, 138 is what it all adds up to

the length is 11.5cm

the width is 57.5cm

the area is 661.25cm^2

the answers might not be correct use at own risk

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4