RHOMBUS The diagonals of rhombus ABCD intersect at E. Given that

mZBAC = 53° and DE = 8, find the indicated measure.

B

А.

53°

mZAED

m2 DAC

m ZADC

AE

DB

8

E

с

AC

D

11. Name the four isosceles triangles in rhombus ABCD.

12. Name three properties of AC and BD.

Answer:

Question 11:

[tex]\angle DAC = 53^\circ[/tex]

[tex]\angle AED = 90^\circ[/tex]

[tex]\angle ADC = 74[/tex]

[tex]DB = 16[/tex]

[tex]AE = 6.03[/tex]

[tex]AC = 12.06[/tex]

Question 12:

[tex]\triangle ABD[/tex], [tex]\triangle BAC[/tex], [tex]\triangle CDA[/tex] and [tex]\triangle DAB[/tex]

Question 13:

AC and BD are perpendicular lines, and they are diagonals

Step-by-step explanation:

Question 11

Given

[tex]\angle BAC = 53^\circ[/tex]

[tex]DE = 8[/tex]

See attachment for Rhombus

Required

Determine the indicated sides

Solving (a): [tex]\angle DAC[/tex]

Diagonal CA divides [tex]\angle DAB[/tex] into 2 equal angles

i.e

[tex]\angle DAC = \angle BAC[/tex]

So:

[tex]\angle DAC = 53^\circ[/tex]

Solving (b): [tex]\angle AED[/tex]

The angles at E is 90 degrees because diagonals AC and BD meet at a perpendicular.

So:

[tex]\angle AED = 90^\circ[/tex]

Solving (c): [tex]\angle ADC[/tex]

First, we calculate [tex]\angle ADE[/tex], considering [tex]\triangle ADE[/tex]:

[tex]\angle ADE + \angle AED + \angle DAC = 180[/tex]

[tex]\angle ADE + 90 + 53 = 180[/tex]

[tex]\angle ADE + 143 = 180[/tex]

[tex]\angle ADE = -143 + 180[/tex]

[tex]\angle ADE = 37[/tex]

To calculate [tex]\angle ADC[/tex], we have:

[tex]\angle ADC = 2*\angle ADE[/tex]

[tex]\angle ADC = 2* 37[/tex]

[tex]\angle ADC = 74[/tex]

Solving (d): [tex]DB[/tex]

From the rhombus

[tex]DB = DE +EB[/tex]

Where

[tex]DE =EB[/tex]

So:

[tex]DB = 8 + 8[/tex]

[tex]DB = 16[/tex]

Solving (e): [tex]AE[/tex]

To do this we consider [tex]\triangle ADE[/tex]

Using the tan formula

[tex]tan(\angle ADE) = \frac{AE}{DE}[/tex]

[tex]\angle ADE = 37[/tex] and [tex]DE = 8[/tex]

So:

[tex]\tan(37) = \frac{AE}{8}[/tex]

[tex]AE = 8 * \tan(37)[/tex]

[tex]AE = 6.03[/tex]

Solving (f): [tex]AC[/tex]

This is calculated as:

[tex]AC = AE + EC[/tex]

Where

[tex]AE = EC[/tex]

[tex]AC = 6.03 +6.03[/tex]

[tex]AC = 12.06[/tex]

Question 12: Isosceles Triangle

In the rhombus, all 4 sides are equal;

So, the isosceles triangle are:

[tex]\triangle ABD[/tex], [tex]\triangle BAC[/tex], [tex]\triangle CDA[/tex] and [tex]\triangle DAB[/tex]

Question 13:

AC and BD are perpendicular lines, and they are diagonals

-2 1/4 = r - 4/5

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I’m super confused on what this means

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