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show all work and annotations for full credit. **#1.) 1.02 - 1.03: the …

Question

show all work and annotations for full credit. *#1.) 1.02 - 1.03: the base of a triangle is (x + 3) in. and its height is 4 in. if the area is 8 in², determine the measure of its base. a.) 1 in. b.) 4 in. c.) -1 in. d.) 16 in. #2.) 1.04:a square has four congruent sides. given one side has endpoints (7, -4) and (7,2), determine the area of the square. a. 36 units² b. 6 units² c. 4 units² d. 12 units² #3.) 1.09: write the equation of the line that is the perpendicular bisector of the line segment having endpoints (9,0) and (3, -4). write your final answer in slope - intercept form then graph both lines below.

Explanation:

Response
#1

Step1: Use triangle - area formula

The area formula for a triangle is $A=\frac{1}{2}bh$. We know $A = 8$ in², $h=4$ in, and $b=(x + 3)$ in. Substitute into the formula: $8=\frac{1}{2}(x + 3)\times4$.

Step2: Simplify the equation

First, simplify the right - hand side of the equation. $\frac{1}{2}(x + 3)\times4=2(x + 3)=2x+6$. So the equation becomes $2x + 6=8$.

Step3: Solve for x

Subtract 6 from both sides: $2x=8 - 6=2$. Then divide both sides by 2: $x = 1$.

Step4: Find the base

The base $b=x + 3$. Substitute $x = 1$ into the base formula, $b=1+3=4$ in.

Step1: Calculate the length of the side

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For the points $(7,-4)$ and $(7,2)$, $x_1 = 7$, $x_2 = 7$, $y_1=-4$, $y_2 = 2$. Then $d=\sqrt{(7 - 7)^2+(2+4)^2}=\sqrt{0 + 36}=6$ units.

Step2: Calculate the area of the square

The area of a square is $A=s^2$, where $s$ is the side - length. Since $s = 6$ units, $A=6^2=36$ units².

Step1: Find the mid - point of the line segment

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For the points $(9,0)$ and $(3,-4)$, the mid - point $M=(\frac{9 + 3}{2},\frac{0-4}{2})=(6,-2)$.

Step2: Find the slope of the line segment

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For the points $(9,0)$ and $(3,-4)$, $m_1=\frac{0 + 4}{9 - 3}=\frac{4}{6}=\frac{2}{3}$.

Step3: Find the slope of the perpendicular bisector

If two lines are perpendicular, the product of their slopes is $-1$. Let the slope of the perpendicular bisector be $m_2$. Then $m_1m_2=-1$. Since $m_1=\frac{2}{3}$, $m_2=-\frac{3}{2}$.

Step4: Write the equation of the line in point - slope form

The point - slope form of a line is $y - y_0=m(x - x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope. Using the point $(6,-2)$ and $m=-\frac{3}{2}$, we have $y+2=-\frac{3}{2}(x - 6)$.

Step5: Convert to slope - intercept form

Expand the right - hand side: $y+2=-\frac{3}{2}x+9$. Then subtract 2 from both sides: $y=-\frac{3}{2}x + 7$.

Answer:

b. 4 in.

#2