Examkin

Soru

Zorluk: ZorPerimeter and Area of Plane Shapes

A rhombus has an area of 120 cm2120\text{ cm}^2. If the length of one of its diagonals exceeds the length of the other diagonal by 14 cm14\text{ cm}, what is the perimeter of the rhombus in centimeters?

Cevap: 52 cm

Cevap

The perimeter of the rhombus is 52 cm.
The area of a rhombus is given by A = \frac{1}{2} d_1 d_2. Setting \frac{1}{2} d_1(d_1 + 14) = 120$ yields the quadratic equation d_1^2 + 14d_1 - 240 = 0, which factors to (d_1 - 10)(d_1 + 24) = 0. Taking the positive solution d_1 = 10\text{ cm} gives d_2 = 24\text{ cm}. The diagonals intersect at right angles, dividing the rhombus into four congruent right triangles with legs of 5 cm and 12 cm. The hypotenuse (side length s) is \sqrt{5^2 + 12^2} = 13\text{ cm}. Therefore, the perimeter is 4 \times 13\text{ cm} = 52\text{ cm}.

Adım Adım Çözüm

1
Set up the area formula for a rhombus in terms of its diagonals
d1⋅d2=240d_1 \cdot d_2 = 240
The area of a rhombus is given by A = \frac{1}{2} d_1 d_2, so \frac{1}{2} d_1 d_2 = 120.
2
Substitute d_2 = d_1 + 14 into the area equation
d_1^2 + 14d_1 - 240 = 0
The difference between the diagonal lengths is 14 cm.
3
Solve the quadratic equation for d_1
d_1 = 10\text{ cm} \text{ and } d_2 = 24\text{ cm}
Factoring gives (d_1 - 10)(d_1 + 24) = 0. Discarding the negative root yields d_1 = 10 cm.
4
Calculate the side length s using the right-angled triangle formed by the perpendicular bisecting diagonals
s = 13\text{ cm}
s = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2} = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\text{ cm}.
5
Multiply side length by 4 to get the total perimeter
P = 52\text{ cm}
All four sides of a rhombus are equal, so Perimeter = 4 \times s = 4 \times 13 = 52 cm.

Anahtar Kavram

Perimeter and Area of a Rhombus using Diagonals and Pythagorean Theorem
Bu soruyu puanla

Konular

Perimeter and Area of Plane Shapes

Bu konudan daha fazla soru