Chapter 11.1, Problem 16PPS

a.

To determine

To Find: To find the area and perimeter of the given triangle.

Answer to Problem 16PPS

The area is 4square inch and perimeter is $4(1+\sqrt{2})$ inch.

Explanation of Solution

Given information: Given tangram with a square of 4 inch.

Formula used: Area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}$

Calculation:

The given square is divided into 16 sub squares

Also, area of the given square is

  $4\times 4=16\text{square}\text{inch}$

Let the side of the sub square be x inch $$

Thus,

  $\begin{array}{l}16x^2=16\\ x^2=\frac{16}{16}\\ x^2=1\\ x=\pm 1\\ \because \text{side}\text{can}\text{not}\text{be}\text{negative}\\ \therefore x=1\text{inch}\end{array}$

Now,

The base of the triangle covers 4 sub squares

  $\therefore \text{base}=4\text{inch}$

And the height of the triangle covers 2 sub squares

  $\therefore \text{height}=2\text{inch}$

Thus, area of the triangle is

  $\begin{array}{l}=\frac{1}{2}\times 4\times 2\\ =4\text{square}\text{inch}\end{array}$

Hence, the area of the required triangle is 4square inch.

Also, the perimeter of the triangle is

  $=\text{base}\text{side}+4\times \text{diagonal}\text{of}\text{sub}\text{squares}$

Where, diagonal is obtained by Pythagoras theorem as follows

  $\begin{array}{l}=\sqrt{1^2+1^2}\\ =\sqrt{2}\end{array}$

Hence the perimeter is

  $\begin{array}{l}=4+4\times \sqrt{2}\\ =4(1+\sqrt{2})\end{array}$

b.

To determine

To Find: To find the area and perimeter of the given parallelogram.

Answer to Problem 16PPS

The area is 2square inch and the perimeter is $2(1+\sqrt{2})$ inch.

Explanation of Solution

Given information: Given tangram with a square of 4 inch.

Formula used: Area of parallelogram $=\text{base}\times \text{height}$

Calculation:

The given square is divided into 16 sub squares

Also, area of the given square is

  $4\times 4=16\text{square}\text{inch}$

Let the side of the sub square be x inch $$

Thus,

  $\begin{array}{l}16x^2=16\\ x^2=\frac{16}{16}\\ x^2=1\\ x=\pm 1\\ \because \text{side}\text{can}\text{not}\text{be}\text{negative}\\ \therefore x=1\text{inch}\end{array}$

Now,

The base of the parallelogram covers the 2 sides of sub squares

  $\therefore \text{base}=2\text{inch}$

And the height of the parallelogram covers the 1 side of sub squares

  $\therefore \text{height}=1\text{inch}$

Thus, area of the parallelogram is

  $\begin{array}{l}=2\times 1\\ =2\text{square}\text{inch}\end{array}$

Hence, the area of the required parallelogram is 2square inch.

Also, the perimeter of the parallelogram is

  $2\times \text{(sum}\text{of}\text{opposite}\text{sides)}$

Hence the perimeter is

  $\begin{array}{l}=2\times (1+\sqrt{2})\\ =2(1+\sqrt{2})\end{array}$