Chapter 11.8, Problem 11PSC

a.

To determine

To state: that what happens to a quadrilateral $PQRS$ when $\overline{PQ}$ gets smaller and smaller.

Answer to Problem 11PSC

The quadrilateral becomes a triangle when $\overline{PQ}$ gets smaller and smaller.

Explanation of Solution

Given Information:

Sides of the cyclic quadrilateral $PQRS$ are

  $PQ=d,QR=c,RS=b\text{and}SP=a$

Side $\overline{PQ}$ gets smaller and smaller.

The quadrilateral $PQRS$ is a cyclic quadrilateral with sides

  $PQ=d,QR=c,RS=b\text{and}SP=a$

Now, $\overline{PQ}$ gets smaller and smaller

This means, P is coming closer and closer to Q.

As $\overline{PQ}$ gets smaller and smaller, the value $d$ will decrease.

The smallest value of $d$ will be zero because it is the length of the side.

After certain point, $d$ will attain its minimum value which is zero.

When $d$ is zero, the points P and Q will coincide.

So, the quadrilateral will transform into a polygon with three sides only.

Thus, it transforms into a triangle.

Hence, the quadrilateral becomes a triangle when $\overline{PQ}$ gets smaller and smaller.

b.

To determine

To state: What happens to Brahmagupta’s formula when $P$ and $Q$ become the same point.

Answer to Problem 11PSC

Brahmagupta’s formula becomes Heron’s formula when $P$ and $Q$ become the same point.

Explanation of Solution

Given Information:

Sides of the cyclic quadrilateral are:

  $PQ=d,QR=c,RS=b\text{and}SP=a$

At some point of time, $P$ and $Q$ become the same point.

Formula used:

Brahmagupta's formula for area of quadrilateral is

  $\left(A\right)=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}$

Here, $s=\frac{a+b+c+d}{2}$ is the semi-perimeter of the quadrilateral

  $a,b,c\text{and}d$ are the sides of an inscribed quadrilateral

Calculation:

Sides of the cyclic quadrilateral are:

  $PQ=d,QR=c,RS=b\text{and}SP=a$

Brahmagupta's formula for area of quadrilateral is

  $\left(A\right)=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}$

  $"s"$ is the semi-perimeter of the quadrilateral and $s=\frac{a+b+c+d}{2}$

When $P$ and $Q$ become the same point, then

Distance between $P$ and $Q$ will be zero.

  $\therefore PQ=d=0$

  $\begin{array}{l}\therefore A=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}\\ \therefore A=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-0\right)}\\ \therefore A=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)s}\\ \therefore A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\end{array}$

This is Heron’s formula for the area of a triangle.

So, when $P$ and $Q$ become the same point, Brahmagupta’s formula becomes Heron’s formula.