Chapter 11.7, Problem 9PSB

a.

To determine

To find: the ratio of areas of two triangles.

Answer to Problem 9PSB

The ratio of area I to area II is $64:225$

Explanation of Solution

Given Information:

All the three angles of triangle I are equal to the corresponding angles of triangle II.

Measure of two sides of triangle I are $6$ and $8$ .

Measure of one side of triangle II is $15$ .

Formula used:

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

Calculation:

Let the first triangle be $ABC$ and the second be $DEF$ .

  

In $\Delta ABC$ and $\Delta DEF$ , we have

  $\angle A=\angle D$ … (Given)

  $\angle B=\angle E$ … (Given)

  $\angle C=\angle F$ … (Given)

  $\therefore \Delta ABC\sim \Delta DEF$ … (By Angle-Angle-Angle similarity test)

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

  $\begin{array}{l}\frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{A{B}^2}{D{E}^2}=\frac{B{C}^2}{E{F}^2}=\frac{A{C}^2}{D{F}^2}\\ \therefore \frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{B{C}^2}{E{F}^2}=\frac{8^2}{{15}^2}=\frac{64}{225}\\ \therefore \frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{64}{225}\end{array}$

Hence, ratio of area of triangle I to area of triangle II is $64:225$ .

b.

To determine

To calculate: the ratio of area of triangle to area of rectangle

Answer to Problem 9PSB

The ratio of area of triangle to area of rectangle is $1:2.$

Explanation of Solution

Given Information:

In triangle, base( b ) = 10 and height( h ) = 14.

In rectangle, length( l ) = 14 and breadth( b₁ ) = 10

Formula used:

Area of a triangle $=\frac{1}{2}\times base\times height$

  $=\frac{1}{2}\times b\times h$

Area of a rectangle $=length\times breadth$

  $=l\times b_1$

Calculation:

In the triangle, we have b = 10 and h = 14

We know that, Area of a triangle $=\frac{1}{2}\times b\times h$

  $\begin{array}{l}\therefore Areaoftriangle=\frac{1}{2}\times 10\times 14=5\times 14\\ \therefore Areaoftriangle=70\end{array}$

Now, in the rectangle we have, l = 14 and b₁ = 10

We know that, Area of a rectangle $=l\times b_1$

  $\begin{array}{l}\therefore Areaofrectangle=14\times 10\\ \therefore Areaofrectangle=140\end{array}$

So, the ratio of area of I to area of II is given by

  $\frac{Areaoftriangle}{Areaof\text{rectangle}}=\frac{70}{140}=\frac{1}{2}$

Hence, the ratio of area I to area II is $1:2.$

c.

To determine

To calculate: the ratio of areas of two triangles with same height.

Answer to Problem 9PSB

The ratio of areas of two triangles with same height is $1:1.$

Explanation of Solution

Given Information:

In triangle I and triangle II, bases are equal and have same height.

Formula used:

Area of a triangle $=\frac{1}{2}\times base\times height$

  $=\frac{1}{2}\times b\times h$

Here, $b$ is the base of the triangle

  $h$ is the height of the triangle

Calculation:

Let triangle I be ABC and triangle II be ADE .

Let’s draw a perpendicular from A to BE intersecting it at F which act as the altitude or height for the big triangle and for the three small triangles.

  

BC = CD = DE … (Given)

We know that, Area of a triangle $=\frac{1}{2}\times b\times h$

Area of triangle I $=\frac{1}{2}\times BC\times AF$

Also, Area of triangle II $=\frac{1}{2}\times DE\times AF=\frac{1}{2}\times BC\times AF$

  $\therefore \frac{AreaoftriangleI}{AreaoftriangleII}$

  $=\frac{\frac{1}{2}\times BC\times AF}{\frac{1}{2}\times BC\times AF}=1$

Hence, ratio of area I to area II is $1:1.$

d.

To determine

To calculate: the ratio of areas of two similar triangles when two sides of each triangle are given.

Answer to Problem 9PSB

The ratio of areas of the given two similar triangles is 4:9.

Explanation of Solution

Given Information:

In the figure, the two opposite sides are parallel where the intersecting lines act as the transversals.

Formula used:

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

Calculation:

Let the two sides be AB and CD where AC and BD intersect at P .

  

As AB || CD and AC is the transversal, we have

  $\angle BAP=\angle DCP$ … (Alternate angles) (i)

As AB || CD and BD is the transversal, we have

  $\angle ABP=\angle PDC$ … (Alternate angles) (ii)

  $\angle APB=\angle CPD$ … (Vertically opposite angles) (iii)

Therefore, all three angles of $\Delta APB$ are equal to the corresponding angles of $\Delta CPD$

  $\therefore \Delta APB\sim \Delta CPD$ … (By Angle-Angle-Angle similarity test)

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

  $\begin{array}{l}\frac{ar(\Delta APB)}{ar(\Delta CPD)}=\frac{A{P}^2}{C{P}^2}=\frac{P{B}^2}{P{D}^2}=\frac{A{B}^2}{C{D}^2}\\ \therefore \frac{ar(\Delta APB)}{ar(\Delta CPD)}=\frac{A{B}^2}{C{D}^2}=\frac{{12}^2}{{18}^2}=\frac{12\times 12}{18\times 18}=\frac{4}{9}\\ \therefore \frac{ar(\Delta APB)}{ar(\Delta CPD)}=\frac{4}{9}\end{array}$

Hence, ratio of area I to area II is $4:9.$