Understanding the Problem

We are given an original triangle with base $b$ and altitude (height) $h$. The altitude is increased by a length $m$, making the new height $h+m$. We need to find the amount $x$ that must be subtracted from the base $b$ so that the area of the new triangle is exactly half the area of the original triangle.

Step-by-Step Solution

Step 1: Write the Formula for the Area of the Original Triangle

The area $A_{orig}$ of a triangle is given by: $A_{orig}=\frac{1}{2}\times b\times h$

Step 2: Write the Formula for the Area of the New Triangle

The new altitude is $h+m$. The new base is $b-x$ (since we are subtracting $x$ from the original base $b$). Therefore, the area $A_{new}$ of the new triangle is: $A_{new}=\frac{1}{2}\times (b-x)\times (h+m)$

Step 3: Set Up the Condition Given in the Problem

The area of the new triangle is one-half that of the original triangle. $A_{new}=\frac{1}{2}{A}_{orig}$

Substitute the area formulas from Step 1 and Step 2 into this equation: $\frac{1}{2}(b-x)(h+m)=\frac{1}{2}\times \left(\frac{1}{2}bh\right)$

Step 4: Simplify and Solve for $x$

First, we can multiply both sides of the equation by 2 to eliminate the fraction $\frac{1}{2}$ on the left side. $(b-x)(h+m)=\frac{1}{2}bh$

Next, expand the left side of the equation: $bh+bm-xh-xm=\frac{1}{2}bh$

Now, get all terms involving $x$ on one side and the other terms on the opposite side. Subtract $\frac{1}{2}bh$ from both sides and add $xh+xm$ to both sides. $bh+bm-\frac{1}{2}bh=xh+xm$

Simplify the left side. Note that $bh-\frac{1}{2}bh=\frac{1}{2}bh$. $\frac{1}{2}bh+bm=xh+xm$

Factor $x$ out of the right side of the equation. $\frac{1}{2}bh+bm=x(h+m)$

Finally, solve for $x$ by dividing both sides by $(h+m)$. $x=\frac{\frac{1}{2}bh+bm}{h+m}$

We can simplify this expression further. Factor $b$ out of the terms in the numerator. $x=\frac{b(\frac{1}{2}h+m)}{h+m}$

To write the numerator with a common denominator, note that $m=\frac{2m}{2}$. Therefore: $x=\frac{b(\frac{h}{2}+\frac{2m}{2})}{h+m}=\frac{b\left(\frac{h+2m}{2}\right)}{h+m}=\frac{b(h+2m)}{2(h+m)}$

Final Answer:

The amount $x$ that must be taken from the base is: $\frac{b(h+2m)}{2(h+m)}$

Comparing this with the provided options, the correct answer is the third option: $\frac{\mathrm{b}(2\mathrm{m}+\mathrm{h})}{2(\mathrm{h}+\mathrm{m})}$

Related Concepts and Formulas

Area of a Triangle: The fundamental formula used is $A=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$. This is crucial for solving any problem related to the area of triangles.

Algebraic Manipulation: This problem heavily relies on skills like setting up equations based on a given condition, expanding expressions, rearranging terms, and factoring to isolate the desired variable.

Inverse Relationship of Base and Height: For a triangle to have a certain area, its base and height are inversely proportional. If one increases, the other must decrease to keep the area constant. This problem explores a specific case of changing both dimensions to achieve a new, specific area.