Let ABC be a right triangle with legs AB = 8 cm and AC = 9 cm. What is the largest possible area of a rectangle ADEF if the points D, E and F belong to the sides AB, BC and CA, respectively?

Accepted Solution

Answer:The largest possible area would be 18 square cm.Step-by-step explanation:Given,ABC is a right triangle,Having legs,AB = 8 cm, AC = 9 cm,Also, points D, E and F belong to the sides AB, BC and CA, respectivelySuch that we obtain a rectangle ADEF,Since, Δ BDE is similar to Δ BAC,( by AA similarity postulate, because ∠BDE ≅ ∠BAC, both are right angles and ∠DBE ≅ ∠ABC, both are same angles )∵ Corresponding sides of similar triangle are proportional,I.e. [tex]\frac{BD}{AB}=\frac{DE}{AC}[/tex]Let AD = x ⇒ BD = 8 - xBy substituting the values,[tex]\frac{8-x}{8}=\frac{DE}{9}[/tex][tex]\implies DE=\frac{9}{8}(8-x)[/tex]Thus, the area of the rectangle ADEF would be,[tex]A(x) = AD\times DE[/tex][tex]\implies A(x) = x(\frac{9}{8}(8-x))=\frac{9}{8}(8x-x^2)[/tex]Differentiating with respect to x,[tex]A'(x) = \frac{9}{8}(8-2x)[/tex]Again differentiating w.r.t. x,[tex]A''(x) = \frac{9}{8}(-2)=-\frac{9}{4}[/tex]For maxima or minima,A'(x) = 0[tex]\implies \frac{9}{8}(8-2x)=0[/tex][tex]\implies 8-2x=0[/tex][tex]\implies x = 4[/tex]At x = 4, A''(x) = negative,Hence, the area would be maximum if x = 4,The maximum area of the rectangle ADEF,[tex]A(4) = \frac{9}{8}(8\times 4-(4)^2)=\frac{9}{8}(32-16)=\frac{9}{8}\times 16=9\times 2=18\text{ square cm}[/tex]