kensiehp
Answered

The lengths of the diagonals of a rhombus are 4x and 6x. What expressions give the perimeter and area of the rhombus? PLEASE HELP

The lengths of the diagonals of a rhombus are 4x and 6x. What expressions give the perimeter and area of the rhombus? PLEASE HELP class=

Answer :

Answer:

Area = 12x²

Perimeter = 4x√13

Step-by-step explanation:

Formula to be used for the area of a rhombus,

Area = [tex]\frac{1}{2}(\text{Diagonal 1)}(\text{Diagonal 2})[/tex]

        = [tex]\frac{1}{2}(4x)(6x)[/tex]

        = [tex]12x^2[/tex]

Since, diagonals of a rhombus bisect each other at 90°.

OD = [tex]\frac{1}{2}(BD)[/tex]

      = [tex]\frac{1}{2}(4x)[/tex]

      = 2x

AO = [tex]\frac{1}{2}(AC)[/tex]

      = [tex]\frac{1}{2}(6x)[/tex]

      = 3x

By using Pythagoras theorem in ΔAOD,

AD² = AO² + DO²

AD² = (2x)² + (3x)²

AD² = 13x²

AD = x√13

Perimeter of a rhombus = 4(Side)

                                        = 4(x√13)

                                        = 4x√13

${teks-lihat-gambar} eudora
akposevictor

Perimeter of the rhombus = 4x√13 units.

Area of the rhombus = 12x² units².

Area and Perimeter of a Rhombus

  • Area of a rhombus is half the product of the lengths of the diagonals, which is, (d1 × d2)/2.
  • Perimeter = 4(a), where a is length of the side of the rhombus.

Thus:

  • d1 = 4x
  • d2 = 6x

Area = (4x × 6x)/2 = 24x²/2

Area = 12x² units².

To find the perimeter, first, find the length of a side (a), using the Pythagorean Theorem.

Thus:

Diagonals bisect each other, therefore,

a = √[(½4x)² + (½6x)²]

a = √[4x² + 9x²]

a = √13x² = x√13

Therefore:

Perimeter of the rhombus = 4(x√13) = 4x√13 units.

Learn more about rhombus on:

https://brainly.com/question/3050890

Other Questions