Question

22. The area of a rhombus is 24 sq, cm and
one of its diagonals is 4cm.Its other
diagonal is​

Answer 1

Step-by-step explanation:

The area of a regular rhombus is [math]A = pq/2 [/math], where p and q are the diagonals of the shape.

given A = 24 and p = 4

[math]24 = 4q/2[/math]

[math]48 = 4q[/math]

[math]q = 12[/math]

so the other diagonal is 12 cm long.

now, by Pythagorean theorem, each side of the rhombus (annotate as s) is given by:

[math]s^2 = (p/2)^2 + (q/2)^2[/math]

[math]s^2 = (4/2)^2 + (12/2)^2[/math]

[math]s^2 = 4+36[/math]

[math]s^2 = 40[/math]

[math]s = 2 \sqrt{10}[/math]

since there are 4 sides in a rhombus, so the perimeter of it is:

[math]4*2 \sqrt{10}[/math]

[math]= 8 \sqrt{10}[/math]

Answer 2

Step-by-step explanation:

area of rhombus=1/2 d1d2

24=1/2×4×d2

48=4d2

D2=12cm , another diagonal=12cm if it is correct please follow me