Question

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

Answer 1

Answer:

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

Step-by-step 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

$tan(\angle ADE) = \frac{AE}{DE}$

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

So:

$\tan(37) = \frac{AE}{8}$

$AE = 8 * \tan(37)$

$AE = 6.03$

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