Question

finding the area of a polygon
find the area of each polygon. each square on the grid represents
1 square centimeter. the answers are mixed up at the bottom of
the page. cross out the answers as you complete the problems.
1 diagram, 2 diagram, 3 diagram, 4 diagram with 4 cm, 2 cm, 4 cm, 5 diagram with 4 cm, 4 cm, 6 cm, 2 cm, 6 diagram with 4 cm, 3 cm, 3 cm
answers
15 cm² 24 cm² 21 cm²
10 cm² 22 cm² 27 cm²
grade 6 • lesson 1

Explanation:

Problem 4:

Step1: Area of rectangle

The top part is a rectangle with length \( 4 \, \text{cm} \) and width \( 2 \, \text{cm} \). Area of rectangle is \( \text{length} \times \text{width} \), so \( 4 \times 2 = 8 \, \text{cm}^2 \).

Step2: Area of parallelogram

The bottom part is a parallelogram with base \( 4 \, \text{cm} \) and height \( 4 \, \text{cm} \). Area of parallelogram is \( \text{base} \times \text{height} \), so \( 4 \times 4 = 16 \, \text{cm}^2 \).

Step3: Total area

Add the areas of rectangle and parallelogram: \( 8 + 16 = 24 \, \text{cm}^2 \).

Step1: Area of rectangle (full part)

The left part is a rectangle with length \( 4 \, \text{cm} \) and width \( 4 \, \text{cm} \), area \( 4 \times 4 = 16 \, \text{cm}^2 \).

Step2: Area of trapezoid (right part)

The right part is a trapezoid with bases \( 2 \, \text{cm} \) and \( 4 \, \text{cm} \), height \( 6 - 4 = 2 \, \text{cm} \). Area of trapezoid is \( \frac{(a + b)}{2} \times h \), so \( \frac{(2 + 4)}{2} \times 2 = 6 \, \text{cm}^2 \).

Step3: Total area

Add the areas: \( 16 + 6 = 22 \, \text{cm}^2 \).

Step1: Area of two triangles (upper)

The upper part can be considered as two triangles? Wait, actually, the figure can be split. Wait, the base of the large triangle? Wait, no, let's see: The figure has a rhombus - like top? Wait, no, the height from the top to the middle is \( 3 \, \text{cm} \), and the bottom part: Wait, maybe split into two parts. Wait, the total figure: Let's calculate the area of the two large triangles and subtract the middle? No, better: The figure can be seen as a combination. Wait, the base of the large triangle (if we consider the two outer triangles) and the middle. Wait, actually, let's use the formula. Wait, the figure has a base - like structure. Wait, the area can be calculated as the area of the two triangles with base \( 4 \, \text{cm} \) and height \( 3 \, \text{cm} \) each? No, wait, the middle part is a square? Wait, no, the diagram shows a figure with a top diamond - like shape and bottom. Wait, maybe the total area is calculated as follows: The area of the two triangles (each with base \( 4 \, \text{cm} \) and height \( 3 \, \text{cm} \)) plus the area of the middle square? No, wait, maybe I made a mistake. Wait, let's re - examine. The figure has a top part with side \( 4 \, \text{cm} \) and two triangles below with height \( 3 \, \text{cm} \). Wait, actually, the area can be calculated as the area of the two triangles (each with base \( 4 \, \text{cm} \) and height \( 3 \, \text{cm} \)) plus the area of the square? No, wait, the correct way: Let's consider the figure as composed of two congruent triangles (the outer ones) and a middle square? No, maybe the area is \( \frac{1}{2}\times4\times3\times2+ 3\times3\)? Wait, no, that would be \( 12 + 9=21 \, \text{cm}^2 \). Wait, let's check: The two triangles: each has area \( \frac{1}{2}\times4\times3 = 6 \), two of them is \( 12 \), and the middle square (or rhombus) with side \( 3 \, \text{cm} \) (since the height is \( 3 \, \text{cm} \))? Wait, no, the middle part is a square with side \( 3 \, \text{cm} \), area \( 3\times3 = 9 \). So total area \( 12 + 9 = 21 \, \text{cm}^2 \).

Step1: Area of two triangles

Each triangle has base \( 4 \, \text{cm} \) and height \( 3 \, \text{cm} \). Area of one triangle: \( \frac{1}{2}\times4\times3=6 \, \text{cm}^2 \). Two triangles: \( 2\times6 = 12 \, \text{cm}^2 \).

Step2: Area of middle square

The middle square has side \( 3 \, \text{cm} \), area \( 3\times3 = 9 \, \text{cm}^2 \).

Step3: Total area

Add the areas: \( 12+9 = 21 \, \text{cm}^2 \).

Answer:

\( 24 \, \text{cm}^2 \)

Problem 5: