Please help me how do I solve and draw the problem ?

Side of the rhombus is 12 cm. Let d1 = 2x cm , so d2 = 2x+6 cm

Now, 12^2 = x^2 + (x+3)^2 = 2x^2 + 6x+9 = 144, or

2x^2 + 6x-135=0, or

x^2+3x-67.5 = 0

x1 = [-3 +(9+270)^0.5]/2

= [-3+279^0.5]/2

= [-3+16.70329309]/2 = 6.851646544 cm [x2 will be negative and so inadmissible]

So d1 = 2x6.851646544 = 13.70329309 cm and d2 = 19.70329309 cm.

So the area of the rhombus is 13.70329309x19.70329309/2 = 135 sq cm.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

answer

d. 10

explanation

the distance from point y to wx is the height of the triangle.

using pythagoras theorem,

group similar terms to get,

this implies that,

take positive square root of both sides to get,

this implies that,