Question

If the area of a rhombus is 60 sq.cm and one of its diagonal is 15 cm then​

Answer 1

The area of a rhombus is give by 1/2 times the product of the diagonals. Therefore:

[math]60 cm^2 = \frac{1}{2} (d_{1} \cdot d_{2})[/math]

where [math]d_{1}[/math] and [math]d_{2}[/math] are the diagonals. Letting [math]d_{1} = 15 cm [/math] and solving for [math]d_{2}[/math] gives [math]d_{2} = 8cm.[/math]

The two diagonals bisect each other and are perpendicular to each other. Therefore the four triangles created by the diagonals in the rhombus are right triangles. Using the Pythagorean theorem on any of these triangles yields that the sides s of the rhombus are given by

[math](\frac{15}{2}[/math])[math]^2[/math]+ [math](\frac{8}{2})^2[/math] = s[math]^2[/math]

or s = [math]\frac{17}{2}cm.[/math]

The perimeter is therefore 4 X [math]\frac{17}{2}[/math] = 34 cm.

Answer 2

Answer:

8cm

Step-by-step explanation:

area of rhombus=1/2×d1×d2

so there D1 = 15

let D2 be value of x

so 1/2×15×x=60

15×x = 60×2

15x = 120

X= 120/15

X= 8cm