Respuesta :

saItair

Answer :

  • 100 units

Explanation :

In a llgm, the diagonals bisect each other i.e. divide each other in two equal parts.

thus,

  • HD = DF
  • 6x - 16 = 3x + 17
  • 6x - 3x = 17 + 16
  • 3x = 33
  • x = 33/3
  • x = 11

plugging in the value of x in the expression for HD & DF,

  • HD = 6x - 16 = 6*11 - 16 = 66 - 16 = 50
  • DF = 3x +17 = 3*11 + 17 = 33 + 17 = 50

since both DF and HD together make HF, hence,

  • HF = HD + DF
  • HF = 50 + 50
  • HF = 100

therefore, the measure of HF is 100 units.

msm555

Answer:

[tex]\sf HF = 100 [/tex]

Step-by-step explanation:

To find the length of the diagonal [tex]\sf HF [/tex] in parallelogram [tex]\sf EFGH [/tex] , we need to apply the properties of parallelograms. In particular, we'll use the fact that diagonals of a parallelogram bisect each other.

Let's denote [tex]\sf HD [/tex] as [tex]\sf a [/tex] and [tex]\sf DF [/tex] as [tex]\sf b [/tex] . According to the properties of parallelograms, [tex]\sf HD = DF [/tex] .

Given:

[tex]\sf HD = 6x - 16 [/tex]

[tex]\sf DF = 3x + 17 [/tex]

Since [tex]\sf HD = DF [/tex] , we can set up an equation:

[tex]\sf 6x - 16 = 3x + 17 [/tex]

Now, solve for [tex]\sf x [/tex] :

[tex]\sf 6x - 3x = 17 + 16 [/tex]

[tex]\sf 3x = 33 [/tex]

[tex]\sf x =\dfrac{33}{3}[/tex]

[tex]\sf x = 11 [/tex]

Now that we have found the value of [tex]\sf x [/tex] , we can substitute it back into one of the expressions to find the length of [tex]\sf HF [/tex] .

Let's use [tex]\sf HD = 6x - 16 [/tex] :

[tex]\sf HD = 6(11) - 16 [/tex]

[tex]\sf HD = 66 - 16 [/tex]

[tex]\sf HD = 50 [/tex]

Since [tex]\sf HF = 2 \times HD [/tex] (due to the property of parallelograms), we have:

[tex]\sf HF = 2 \times 50 = 100 [/tex]

So, the length of diagonal [tex]\sf HF [/tex] in parallelogram [tex]\sf EFGH [/tex] is 100 units.

Otras preguntas