\frac{1}{2} \times (b + 4) \times h'
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- \frac{1}{2} \times b \times \sqrt{100 - \frac{b^2}{4}} = 48
- Solving for \(b\) involves trial or algebraic manipulation, but instead, let's consider the changes. The new base is \(b + 4\). The new height \(h'\) is:
- h' = \sqrt{100 - \frac{(b+4)^2}{4}}
- \frac{1}{2} \times (b + 4) \times h' - 48
- Assuming \(b = 8\) (from solving the quadratic equation derived from area), then:
- h = \sqrt{100 - 16} = \sqrt{84} \approx 9.165
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