Question

The area of rhombus is 1080 sq. m and one of its diagonals is 72m. Find perimeter of rhombus

Answer 1

Answer:

DB & AC are diagonals of Rhombus ABCD

Diagonal DB= 72 m

Area = 1080 m^2

Area of the rhombus = 1/2( product of two diagonals)

1/2*(72) * d2 = 1080

36d2 = 1080

d2= 1080/36

d2 = 30 m

Diagonal AC =30 m

Diagonals of a rhombus are perpendicular bisectors of each other.

Seg DO = OB = 36 cm (72/2 = 36)

Seg AO = OC = 15 cm ( 30/2 = 15)

∆ODC is a right angled Triangle

DO^2 + OC^2= DC^2 ( Pythagoras Theorem)

36^2 + 15^2 = DC^2

1296 + 225 = DC^2

1521=DC^2

DC= + or - 39

Discarding negative sign .

Side of rhombus (DC) = 39 m

Four sides of rhombus are equal in measures.

Perimeter of rhombus = 4*side

= 4*39

= 156 m

perimeter of the rhombus = 156m

Ans : 156 m

Step-by-step explanation:

Answer 2

ANSWER: d2=30

STEP - BY - STEP EXPLANATION:
$area \: of \: rombus = \frac{1}{2 = } \times d1 \times d2 \\ area = 1080 \: {m}^{2} \\ d1 = 72 \: m \\ \\ a.t.q. \\ \frac{1}{2} \times 72 \times d2 = 1080 \\ d2 = \frac{1080}{36} \\ d2 = 30$