WILL GIVE BRAINLIEST!! Rhombus ABCD has perimeter 148, and one of its diagonals has length 24. How long is the other diagonal?

70

Step-by-step explanation:

The four sides of a rhombus all have equal length, so if the perimeter is $148$, then each side has length $\frac{148}{4} = 37$. Also, the diagonals of a rhombus cut each other in half and meet at right angles, so the diagonal of length $24$ is cut into two pieces of length $12$. We can show this information in a diagram:

[asy]

size(9cm);

pair a=(0,12); pair b=(-35,0); pair c=-a; pair d=-b;

dot(a); dot(b); dot(c); dot(d); dot((0,0));

draw(a--b--c--d--a);

draw(a--c,dashed);

draw(b--(0,0),dashed);

draw(d--(0,0),dashed);

label("$A$",a,N);

label("$B$",b,W);

label("$C$",c,S);

label("$D$",d,E);

label("37",(a+b)/2,NW);

label("37",(c+b)/2,SW);

label("37",(c+d)/2,SE);

label("37",(a+d)/2,NE);

label("12",a/2,W);

label("12",c/2,W);

label("$x$",2*b/5,S);

label("$x$",2*d/5,S);

draw(rightanglemark(a,(0,0),b,70));

[/asy]

Applying the Pythagorean Theorem to any of the four right triangles in our diagram, we have

$$12^2 + x^2 = 37^2.$$Solving this equation for positive $x$, we get $x=\sqrt{37^2-12^2} = \sqrt{1369-144} = \sqrt{1225} = 35$. The length of the long diagonal is $x+x = \boxed{70}$.