Question

Parallelogram EFGH has vertices at E(-7, 1), F(-5, 3), G(-5, 6), and H(-7, 4). Right triangle ABC with a right angle at A has vertices at A(4, 2), B(1, 4), and C(6, 5). Which statement about the areas of the figures is true? *

A.The area of the triangle is 0.5 units more than the area of the parallelogram

B.The area of the triangle is 2.4 units less than the area of the parallelogram

C.The area of the triangle is 0.5 units less than the area of the parallelogram

D.The area of the triangle is 2.4 units more than the area of the parallelogram

Answer 1

Comparing the calculated areas, the difference is not present in the options given. Therefore, without typos or additional context, none of the choices (A, B, C, or D) are correct.

Step-by-step explanation:

The question involves calculating the areas of a parallelogram and a right triangle and comparing them. For the parallelogram EFGH, since opposite sides are equal and parallel, we can find the base and the height by looking at the coordinates. The base EF (or HG) is the difference in the x-coordinates of E and F, which is 2 units. The height is the difference in the y-coordinates of E and H, which is 3 units. So, the area of the parallelogram is base × height = 2 × 3 = 6 square units.

For the right triangle ABC with a right angle at A, the legs AB and AC can serve as the base and height. AB is the difference in x-coordinates of A and B, which is 3 units, and AC is the difference in y-coordinates of A and C, which is 3 units. The area of the triangle is ½ × base × height = ½ × 3 × 3 = 4.5 square units.

Comparing the areas, the area of the triangle ABC is 1.5 units less than the area of parallelogram EFGH, which is not an option provided in the question. Therefore, based on the given coordinates for the parallelogram and triangle, none of the given statements (A, B, C, or D) are correct as they do not match the calculated areas.

Answer 2

check the picture below.

for the parallelogram, we can simply get the area of those 2 blue triangles and the yellow rectangle, sum them up and that's the area of the parallelogram.

for the triangle on the right, we know the right-angle is at vertex A(4,2), so the base will be the distance from (4,2) to (1,4), and its height is the distance form (4,2) to (6,5)


`\bf \stackrel{\textit{area of the parallelogram}}{\stackrel{triangle}{\cfrac{1}{2}(2)(2)}~~~~+~~~~\stackrel{rectangle}{2\cdot 1}~~~~+~~~~\stackrel{triangle}{\cfrac{1}{2}(2)(2)}}\implies 2+2+2\implies 6`

now for the area of the triangle.


`\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{4})\qquad \qquad % distance value d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ base=√((1-4)^2+(4-2)^2)\implies base=√((-3)^2+2^2) \\\\\\ base=√(9+4)\implies \boxed{base=√(13)}`


`\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{5})\qquad \qquad height=√((6-4)^2+(5-2)^2) \\\\\\ height=√(2^2+3^2)\implies height=√(4+9)\implies \boxed{height=√(13)}\\\\ -------------------------------\\\\ \stackrel{\textit{area of the triangle}}{\cfrac{1}{2}(√(13))(√(13))}\implies \cfrac{1}{2}(√(13))^2\implies \cfrac{1}{2}\cdot 13\implies \cfrac{13}{2}\implies 6.5`

so, the triangle is 6.5 square units and the parallelogram has 6 square units, surely you know which statement is true.