Question

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 1

Question 11:

`\angle DAC = 53^\circ`

`\angle AED = 90^\circ`

`\angle ADC = 74`

`DB = 16`

`AE = 6.03`

`AC = 12.06`

Question 12:

`\triangle ABD`,
`\triangle BAC`,
`\triangle CDA` and
`\triangle DAB`

Question 13:

AC and BD are perpendicular lines, and they are diagonals

Explanation:

Question 11

Given

`\angle BAC = 53^\circ`

`DE = 8`

See attachment for Rhombus

Required

Determine the indicated sides

Solving (a):
`\angle DAC`

Diagonal CA divides
`\angle DAB` into 2 equal angles

i.e

`\angle DAC = \angle BAC`

So:

`\angle DAC = 53^\circ`

Solving (b):
`\angle AED`

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

So:

`\angle AED = 90^\circ`

Solving (c):
`\angle ADC`

First, we calculate
`\angle ADE`, considering
`\triangle ADE`:

`\angle ADE + \angle AED + \angle DAC = 180`

`\angle ADE + 90 + 53 = 180`

`\angle ADE + 143 = 180`

`\angle ADE = -143 + 180`

`\angle ADE = 37`

To calculate
`\angle ADC`, we have:

`\angle ADC = 2*\angle ADE`

`\angle ADC = 2* 37`

`\angle ADC = 74`

Solving (d):
`DB`

From the rhombus

`DB = DE +EB`

Where

`DE =EB`

So:

`DB = 8 + 8`

`DB = 16`

Solving (e):
`AE`

To do this we consider
`\triangle ADE`

Using the tan formula

`tan(\angle ADE) = (AE)/(DE)`

`\angle ADE = 37` and
`DE = 8`

So:

`\tan(37) = (AE)/(8)`

`AE = 8 * \tan(37)`

`AE = 6.03`

Solving (f):
`AC`

This is calculated as:

`AC = AE + EC`

Where

`AE = EC`

`AC = 6.03 +6.03`

`AC = 12.06`

Question 12: Isosceles Triangle

In the rhombus, all 4 sides are equal;

So, the isosceles triangle are:

`\triangle ABD`,
`\triangle BAC`,
`\triangle CDA` and
`\triangle DAB`

Question 13:

AC and BD are perpendicular lines, and they are diagonals