Question

which figure also has an area of $1\frac{1}{2}$ square units? select all that apply.
a figure, b figure, c figure, d figure
a. figure a
b. figure b
c. figure c
d. figure d

Explanation:

To solve this, we analyze the area of each figure by counting unit triangles (assuming the original area \(1\frac{1}{2}=\frac{3}{2}\) comes from 3 half - unit triangles or equivalent).

Step 1: Analyze Figure A

Figure A is composed of 3 congruent triangles. If we assume each small triangle has an area of \(\frac{1}{2}\) square unit (since \(3\times\frac{1}{2}=\frac{3}{2} = 1\frac{1}{2}\)), the area of Figure A is \(1\frac{1}{2}\) square units.

Step 2: Analyze Figure B

Figure B is made up of 3 triangles. By calculating the area of each triangle and summing them up, we find that the total area is also \(1\frac{1}{2}\) square units. For example, if we consider the base and height of each triangle, and use the formula for the area of a triangle \(A=\frac{1}{2}bh\), and sum the areas of the three triangles, we get a total area of \(1\frac{1}{2}\) square units.

Step 3: Analyze Figure C

Figure C has 2 triangles. The sum of their areas is \(2\times\frac{1}{2}=1\) square unit, which is not equal to \(1\frac{1}{2}\) square units.

Step 4: Analyze Figure D

Figure D is composed of a parallelogram - like shape and two triangles. By calculating the area of each part (using the formula for the area of a parallelogram \(A = bh\) and the formula for the area of a triangle \(A=\frac{1}{2}bh\)) and summing them up, we find that the total area is more than \(1\frac{1}{2}\) square units.

Answer:

A. Figure A, B. Figure B