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PSAT Math : How to find the perimeter of a rhombus

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Example Question #1 : Rhombuses

If the area of a rhombus is 24 and one diagonal length is 6, find the perimeter of the rhombus.

Possible Answers:

Correct answer:

Explanation:

The area of a rhombus is found by

A = 1/2(d₁)(d₂)

where d₁ and d₂ are the lengths of the diagonals. Substituting for the given values yields

24 = 1/2(d₁)(6)

24 = 3(d₁)

8 = d₁

Now, use the facts that diagonals are perpendicular in a rhombus, diagonals bisect each other in a rhombus, and the Pythagorean Theorem to determine that the two diagonals form 4 right triangles with leg lengths of 3 and 4. Since 3² + 4² = 5², each side length is 5, so the perimeter is 5(4) = 20.

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Example Question #2 : How To Find The Perimeter Of A Rhombus

Rhombus

Note: Figure NOT drawn to scale.

Calculate the perimeter of Quadrilateral ABCD in the above diagram if:

AC = 48

BD = 20

AB=BC=CD=DA

Possible Answers:

Insufficient information is given to answer the question.

136

104

120

Correct answer:

104

Explanation:

AB=BC=CD=DA , so Quadrilateral ABCD is a rhombus. Its diagonals are therefore perpendicular to each other, and the four triangles they form are right triangles. Therefore, the Pythagorean theorem can be used to determine the common sidelength of Quadrilateral ABCD .

We focus on $\Delta AXB$ . The diagonals are also each other's bisector, so

$AX = \frac{1}{2} (AC) = \frac{1}{2} \cdot 48= 24$

$XB = \frac{1}{2} (BD) = \frac{1}{2} \cdot 20=10$

By the Pythagorean Theorem,

$AB = \sqrt{(AX)^{2}+ (XB)^{2}} = \sqrt{24^{2}+10^{2}} = \sqrt{576+100} = \sqrt{676} = 26$

26 is the common length of the four sides of Quadrilateral ABCD , so its perimeter is $4 \times 26 = 104$ .

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PSAT Math : How to find the perimeter of a rhombus

Study concepts, example questions & explanations for PSAT Math

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Example Question #1 : Rhombuses

Example Question #2 : How To Find The Perimeter Of A Rhombus

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