The question asksTo solve this for the problem, area of the we largest rectangle that can need to find be the area of ins the largest rectanglecribed in ins a rightcribed in triangle a with right legs triangle of where two lengths 4 cm and sides5 of the rectangle lie along cm the, legs where two of the sides triangle of. the The rectangle given lie right along triangle the legs has of legs the of lengths triangle (.

4### \ Solution,:

1 .text **{Areacm of}\ a) Right and Triangle (**5: \ , The \ areatext of{ thecm given}).

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triangle### is Step: -by -Step \ Solution[ :

1 .text {AreaEquation of} the = hypoten \usefrac{:1}{2
} \ The hypcdototenuse \ oftext the{base right} triangle \ connectscd points (ot0 \,text {4height)\}) = and \ \frac(({51,}{ 20})\cdot 4 \cdot on5 a = Cartesian plane . Its equation10 is obtained \ using the, slope-inter \textcept{ formcm:}
^ 2
[ \ y =]

2 -.\frac Largest{ Ins4cribed}{5 Rectangle}x:

• The4 largest. rectangle \ that can be]

2 ins. Rectangle dimensionscribed in a: right
triangle occurs Let when the the rectangle rectangle have's one diagonal lies along vertex at the the hypoten originuse .(( The0 maximum, area0 of such)\ a rectangle) is and exactly the ** opposite vertexhalf at the ((x, area of the y triangle)**.

), where Thus, $x the area \le ofq the largest rectangle5$ and is: ( \y[ \ leq \4text{).Area The of area rectangle of} the = rectangle \ isfrac given{ by1:}{
2 } \ $$cd ot A \ =text x{ \Areacd ofot triangle y}. = \ frac{ \1]

}{23}. ** \Subcdstituteot $y10$ from the = hyp 5 ,oten \textuse{ equationcm**:}
^ 2 From the equation$$

of### the Final hyp Answeroten: useThe, area substitute of ( they largest = rectangle - is: frac{[ 4}{boxed5{}5x \ +, \4text{)cm into} the^ area2 formula} :
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\Let[ me know A if = you x want \ a detailedleft(-\frac{4 explanation}{ of5 why} thex area is + maximized in this way4!

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Related \ Questions]

: 1 Simpl.ify How do: you find \ the[ hyp Aoten =use - of a\frac triangle{ with legs of 4 cm4}{5 and}x^ 2 +5 cm 4? x2. . Why \ is]

the4. Maximize the area:
To find the maximum area, differentiate $A$ with respect to $x$, and set (\frac{d largestA rectangle}{'s areadx half} the area of = the triangle0? \3.): What
\ happens[ to the area of the rectangle \ if the triangle'sfrac dimensions are doubled{? d4A.}{ Candx a rectangle of any size larger} than = this fit inside - the triangle? 5. How can calculusfrac be used to derive the{ maximum8 area of the rectangle?

Tip:

When}{ maximizing5} areas orx volumes + ,4 symmetry. and proportional \ity] often lead Set to this the equal optimal to solution zero!: $-\frac{8}{5}x + 4 = 0 \implies x = \frac{20}{8} = 2.5.$

5. Find $y$ when $x = 2.5$:
   Substitute $x = 2.5$ into the hypotenuse equation: $y = -\frac{4}{5}(2.5) + 4 = -2 + 4 = 2.$

6. Calculate the area:
   The area of the rectangle is: $A = x \cdot y = 2.5 \cdot 2 = 5 \, \text{cm}^2.$

Final Answer:

The area of the largest rectangle is: $\boxed{5 \, \text{cm}^2}.$

Do you want a detailed explanation of any step?
Here are five related questions for deeper understanding:

1. How is the equation of the hypotenuse derived?
2. Why do we substitute $y$ from the hypotenuse equation into the area formula?
3. What is the role of calculus in maximizing the area?
4. How does the derivative determine the maximum area?
5. How can we verify the answer using alternative methods?

Tip: Always check the geometry of the problem visually to ensure accurate setup of equations.