Chapter 11.7, Problem 24WE

(a)

To determine

To Find: The ratio of area of $IandIII$ .

Answer to Problem 24WE

  $\frac{1}{4}$

Explanation of Solution

Given:

  

In triangles $IandIII$

Both triangles are similar.

Therefore

  $\frac{ar(I)}{ar(II)}={\left(\frac{5}{10}\right)}^2=\frac{1}{4}$

Hence, $\frac{1}{4}$ .

(c)

To determine

To Find: The ratio of area of $IandIV$ .

Answer to Problem 24WE

  $\frac{1}{2}$

Explanation of Solution

Given:

  

In triangles $IandIV$

Height of both triangles are same.

Therefore

If two triangles have equal heights, then the ratio of their areas equal to the ratio of their bases.

Thus

  $\frac{b_1}{b_2}=\frac{5}{10}=\frac{1}{2}$

Hence,

  $\frac{1}{2}$ .

(d)

To determine

To Find: The ratio of area of $IIandIV$ .

Answer to Problem 24WE

  $1$

Explanation of Solution

Given:

  

In triangles $IIandIV$

  $\frac{ar(II)}{ar(IV)}=\frac{\frac{ar(I)}{ar(IV)}}{\frac{ar(I)}{ar(II)}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$

(e)

To determine

To Find: The ratio of area of $Iandtrapezoid$ .

Answer to Problem 24WE

  $\frac{1}{9}$

Explanation of Solution

Given:

  

In triangle $Iandtrapezoid$

  $\begin{array}{l}\frac{ar(I)}{ar(trapezoid)}=\frac{ar(I)}{1+\frac{ar(II)}{ar(I)}+\frac{ar(III)}{ar(I)}+\frac{ar(IV)}{ar(I)}}\\ =\frac{1}{1+2+4+2}=\frac{1}{9}\end{array}$