How would I find the length of RH?

This can be done with the principles of similar triangles. Let x = RH = RB, and find RV in terms of x. This gives you the height and base of triangle BRV both in terms of x. Since you already have the numerical height and base of triangle AHV, you just need to show that the two triangles are similar, which lets you take ratios to get an equation that you can solve for x.

EDIT: I did assume RB is perpendicular to the base, as another commentor pointed out. If this isn't given then you would indeed need more information.
If one makes the assumption that RB is parallel to HA, then this becomes a “similar triangles” exercise.

For two triangles with the same internal angles, the sides all scale the same. I.e. the ratio of HA to RB will be the same as the ratio of HV to RV.

That is, RB/HA = RV/HV -> RB/18 = (24-RB)/24

And you can then solve for RB (= RH).
1. Triangle ahv similar to triangle Brv

2. Ah:BR=HV:Rv

It’s a very typical geometry practice+algebra, which grade is OP? 7-8? Maybe?
Count the squares, it’s 4 /s

1. BR:RV = 18:24 = 3:4 because AHV is similar to BRV.
2. BR = HR = 3x, RV = 4x (we don’t know x yet, just the ratio).
3. HV = 24, HV = 7x (HR to RV = HV).
4. x = 24/7 or 3 3/7.
5. HR = 3x = 10 2/7 or 10.29
With only that info, you cannot. You need something more, for example, the angle between HV and RB.
24/18 = (24-x)/x
x=10.285
By using criteria of similarities.
Crazy coincidence, AH is 18 cm or 7 squares long. RH is 4 squares long. So 4(18)/7 would be 10.28, which is what the actual answer is.
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