Chapter 11, Problem 28RP

a.

To determine

To find: The ratio of theareas of region I to that of region II.

Answer to Problem 28RP

The ratio of the area of region I to that of region II is 16:81.

Explanation of Solution

Given Information:

Sides of the quadrilateral are

  $a=8,b=9,c=18\text{and}d=10$

Formula used:

Ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Calculation:

Sides of the quadrilateral $ABCD$ are

  $AB=8,BC=9,CD=18\text{and}DA=10$

Let the diagonals intersect at point $O$

  

In $\Delta AOB\text{and}\Delta COD,$

  $\angle AOB=\angle COD$ … (Vertically opposite angles)

  $\angle OAB=\angle OCD$ … (Alternate angles)

  $\angle OBA=\angle ODC$ … (Alternate angles)

  $\therefore \Delta AOB\sim \Delta COD$ … (By AAA similarity test)

As we know that, the ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

  $\begin{array}{l}\therefore \frac{ar\left(\Delta AOB\right)}{ar\left(\Delta COD\right)}=\frac{A{B}^2}{C{D}^2}\\ \therefore \frac{ar\left(\Delta AOB\right)}{ar\left(\Delta COD\right)}=\frac{8^2}{{18}^2}\\ \therefore \frac{ar\left(\Delta AOB\right)}{ar\left(\Delta COD\right)}=\frac{16}{81}\end{array}$

Hence, ratio of the area of region I to that of region II is 16:81.

b.

To determine

To find: The ratio of the areas of triangle I to that of triangle II.

Answer to Problem 28RP

The ratio of the area of region I to that of region II is 49:81.

Explanation of Solution

Given Information:

Sides of the triangle I are

  $a=6\text{and}b=7$

Sides of the triangle II are

  $c=9\text{and}d=15$

Formula used:

Ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Calculation:

Sides of the triangle I are $\left(a\right)=6\text{and}\left(b\right)=7$

Sides of the triangle II are $\left(c\right)=9\text{and}\left(d\right)=15$

Let the triangles intersect at point $O$

  

In $\Delta AOB\text{and}\Delta COD,$

  $\angle AOB=\angle COD$ … (Vertically opposite angles)

  $\angle OAB=\angle OCD$ … (Alternate angles)

  $\angle OBA=\angle ODC$ … (Alternate angles)

  $\therefore \Delta AOB\sim \Delta COD$ … (By AAA similarity test)

As we know that, the ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

  $\begin{array}{l}\therefore \frac{ar\left(\Delta AOB\right)}{ar\left(\Delta COD\right)}=\frac{O{B}^2}{O{D}^2}\\ \therefore \frac{ar\left(\Delta AOB\right)}{ar\left(\Delta COD\right)}=\frac{7^2}{9^2}\\ \therefore \frac{ar\left(\Delta AOB\right)}{ar\left(\Delta COD\right)}=\frac{49}{81}\end{array}$

Hence, ratio of the area of region I to that of region II is 49:81.

c.

To determine

To calculate: The ratio of the areas of the two triangles.

Answer to Problem 28RP

The ratio of the area of region I to that of region II is 1:!.

Explanation of Solution

Given Information:

Sides of the triangle I are $\left(a\right)=6\text{and}\left(b\right)=10$

One side of the triangle II is $\left(c\right)=10$

Two angles of triangle I are equal to the corresponding angles of the triangle II.

Formula used:

Ratio of areas of two congruent triangle is equal to 1:1.

Calculation:

Sides of the triangle I are $\left(a\right)=6\text{and}\left(b\right)=10$

One side of the triangle II is $\left(c\right)=10$

  

In $\Delta ABE\text{and}\Delta BCD,$

  $\angle BAE=\angle DBC$ … (Given)

  $\angle AEB=\angle BCD$ … (Given)

  $AE=BC=10$ … (Given)

  $\therefore \Delta BAE\cong \Delta DBC$ … (By ASA congruency test)

As the triangles are congruent, so the triangles will have equal areas.

  $\begin{array}{l}\therefore ar\left(\Delta BAE\right)=ar\left(\Delta DBC\right)\\ \therefore \frac{ar\left(\Delta BAE\right)}{ar\left(\Delta DBC\right)}=\frac{1}{1}\end{array}$

Hence, the ratio of the area of region I to that of region II is 1:1.