Question

exercises
find the perimeter and area of each figure.

1. 4x - 1

3x

1. x + 4
2. 12

8
x - 5

1. 5x + 7

20

Explanation:

Step1: Recall perimeter and area formulas

Perimeter of rectangle $P = 2(l + w)$, area of rectangle $A=l\times w$, perimeter of square $P = 4s$, area of square $A=s^{2}$, for right - triangle $P=a + b + c$ and $A=\frac{1}{2}ab$ (where $a$ and $b$ are legs).

Step2: Solve for problem 25

• Perimeter: $P = 2(4x - 1+3x)=2(7x - 1)=14x-2$.
• Area: $A=(4x - 1)\times3x = 12x^{2}-3x$.

Step3: Solve for problem 26

• Since it's a square with side length $s=x + 4$.
• Perimeter: $P = 4(x + 4)=4x+16$.
• Area: $A=(x + 4)^{2}=x^{2}+8x + 16$.

Step4: Solve for problem 27

• First, find $x$ using the Pythagorean theorem $a^{2}+b^{2}=c^{2}$. So, $(x - 5)^{2}+8^{2}=12^{2}$. Expanding, $x^{2}-10x + 25+64 = 144$, $x^{2}-10x-55 = 0$. Using the quadratic formula $x=\frac{10\pm\sqrt{100+220}}{2}=\frac{10\pm\sqrt{320}}{2}=\frac{10\pm8\sqrt{5}}{2}=5\pm4\sqrt{5}$. Since $x-5>0$, $x = 5 + 4\sqrt{5}$.
• Perimeter: $P=(5 + 4\sqrt{5}-5)+8 + 12=20+4\sqrt{5}$.
• Area: $A=\frac{1}{2}\times8\times(5 + 4\sqrt{5}-5)=16\sqrt{5}$.

Step5: Solve for problem 28

• Perimeter: $P = 2(5x + 7+20)=2(5x+27)=10x + 54$.
• Area: $A=(5x + 7)\times20=100x+140$.

Answer:

Problem 25: Perimeter is $14x - 2$, Area is $12x^{2}-3x$.
Problem 26: Perimeter is $4x + 16$, Area is $x^{2}+8x + 16$.
Problem 27: Perimeter is $20 + 4\sqrt{5}$, Area is $16\sqrt{5}$.
Problem 28: Perimeter is $10x + 54$, Area is $100x+140$.