Question

one of the diagonals of a rhombus is 12cm and area
is 96 cm². The perimeter of the rhombus is​

Answer 1

Answer:

40 cm

Step-by-step explanation:

Area = 96 $cm^{2}$

Diagonal 1 ($d_{1}$) = 12 cm

Area of a rhombus = 0.5 * diagonal 1 * diagonal * 2

                         96 = 0.5 * 12 * $d_{2}$

                         $d_{2}$ = $\frac{96}{0.5*12} = \frac{96}{6} = 16$

now we know that the diagonals of a rhombus are perpendicular bisectors of each other and also all sides of rhombus are equal

therefore when we draw both the diagonals of a rhombus we are left with 4 right angled triangles. Also each triangle has its arms equal to half of the corresponding diagonals and the side of the rhombus becomes the hypotenuse of triangle.

Therefore by applying pythagoras theorem

$side^{2} = (\frac{d_{1}}{2})^{2} + (\frac{d_{2}}{2})^{2}\\\\side^{2} = (\frac{12}{2})^{2} + (\frac{16}{2})^{2}\\\\side^{2} = 6^{2} +8^{2}\\side^{2} = 36 + 64\\side^{2} = 100\\side = \sqrt{100} = 10$

therefore the side = 10cm

Now perimeter = 4*side = 4*10 = 40cm