Question

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?

Answer 1

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, respectively

Such 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. $\frac{BD}{AB}=\frac{DE}{AC}$

Let AD = x ⇒ BD = 8 - x

By substituting the values,

$\frac{8-x}{8}=\frac{DE}{9}$

$\implies DE=\frac{9}{8}(8-x)$

Thus, the area of the rectangle ADEF would be,

$A(x) = AD\times DE$

$\implies A(x) = x(\frac{9}{8}(8-x))=\frac{9}{8}(8x-x^2)$

Differentiating with respect to x,

$A'(x) = \frac{9}{8}(8-2x)$

Again differentiating w.r.t. x,

$A''(x) = \frac{9}{8}(-2)=-\frac{9}{4}$

For maxima or minima,

A'(x) = 0

$\implies \frac{9}{8}(8-2x)=0$

$\implies 8-2x=0$

$\implies x = 4$

At x = 4, A''(x) = negative,

Hence, the area would be maximum if x = 4,

The maximum area of the rectangle ADEF,

$A(4) = \frac{9}{8}(8\times 4-(4)^2)=\frac{9}{8}(32-16)=\frac{9}{8}\times 16=9\times 2=18\text{ square cm}$