Question

What is the largest size of a rectangle that can be inscribed in a semicircle of radius 1 unit so that two vertices lie on the diameter ?

Answer 1

Answer:

$\frac{4}{\sqrt{5} }$,$\frac{1}{\sqrt{5} }$

Step-by-step explanation:

See the diagram attached.

Let us assume that ABCD is the required rectangle with the maximum area, O is the center of the circle.

Let us assume, AB=p and BC=q

So, the area of ABCD= A= pq ........ (1)

Now, join O and C, so that, OC= radius of the circle =1 units.

So, from Δ OBC, OC²=OB²+BC²

⇒1²= $(\frac{p}{2}) ^{2}+q^{2}$

⇒$q^{2}=1-\frac{p^{2} }{4}$

⇒q= $\sqrt{(1-\frac{p^{2} }{4})}$....... (2)

Now, from equation (1), we get

A= $p.\sqrt{(1-\frac{p^{2} }{4})}$

Differentiating both sides with respect to p,we get

$\frac{dA}{dp}= \sqrt{(1-\frac{p^{2} }{4})}+\frac{\frac{p}{2}(-\frac{p}{8})}{2\sqrt{(1-\frac{p^{2} }{4})}} }=0$

{For area to be maximum, $\frac{dA}{dp}=0$}

⇒$\sqrt{(1-\frac{p^{2} }{4})}-\frac{\frac{p^{2} }{16} }{\sqrt{(1-\frac{p^{2} }{4})}}=0$

⇒$1-\frac{p^{2} }{4} -\frac{p^{2} }{16}=0$

⇒$\frac{5p^{2} }{16}=1$

⇒p= $\frac{4}{\sqrt{5} }$

Hence, from equation (2),

q= $\sqrt{1-\frac{16}{20} }$

⇒ q=$\frac{1}{\sqrt{5} }$.

Therefore, the dimension of the rectangle with maximum area will be [ $\frac{4}{\sqrt{5} }$,$\frac{1}{\sqrt{5} }$]

(Answer)