Area of a triangle = (base)(height)/2

Let b1 = length of base of T1
So, area of T1 = (b1)(h1)/2

Let b2 = length of base of T2
So, area of T2 = (b2)(h2)/2

We get:
Quantity A: [(b1)(h1)/2]/h1
Quantity B: [(b2)(h2)/2]/h2

Simplify to get:
Quantity A: (b1)/2
Quantity B: (b2)/2

Multiply both quantities by 2 to get:
Quantity A: b1
Quantity B: b2

Since we aren't told anything about the lengths of the two bases, there's no way to determine which is bigger.

Answer: D
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Another approach is to look at different cases.

Area of a triangle = (base)(height)/2

case 1:
T1 has base of 4 and a height (h1) of 4
T2 has base of 4 and a height (h2) of 4

Area of T1 = (4)(4)/2 = 8
Area of T2 = (4)(4)/2 = 8
As you can see the two triangles have the same area.

We get:
Quantity A: (area of T1)/h1 = 8/4 = 2
Quantity B: (area of T2)/h2 = 8/4 = 2
In this case, the the two quantities are equal


case 2:
T1 has base of 4 and a height (h1) of 3
T2 has base of 3 and a height (h2) of 4

Area of T1 = (4)(3)/2 = 6
Area of T2 = (3)(4)/2 = 6
As you can see the two triangles have the same area.

We get:
Quantity A: (area of T1)/h1 = 6/3 = 2
Quantity B: (area of T2)/h2 = 6/4 = 1.5
In this case, the quantity A is greater

Answer: D
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Here : Let AREA of both triangles be 8 .
Then TI has base and height of one be 4 and 2 respectively . And T2 with base and height of other be 2 and 4 respectively.

For QA : only base remains which is 4
For QB : only base remains which is 2

Now A > B

But if heights and base value are interchanged then B > A

Hence D