Question

Pls, help me with this math question

Answer 1

Answer:

Perimeter $\sqrt{26}+\sqrt{5}+\sqrt{10}+\sqrt{17}$ units.  Area 12 square units.

Step-by-step explanation:

Perimeter: total distance around the figure.

Distance Formula:  the distance between points $\left(x_1,y_1\right) \text{ and } \left(x_2,y_2\right)$ is

$d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$AB=\sqrt{(6-1)^2+(2-1)^2}=\sqrt{25+1}=\sqrt{26}$

$BC=\sqrt{(5-6)^2+(4-2)^2}=\sqrt{1+4}=\sqrt{5}$$CD=\sqrt{(2-5)^2+(5-4)^2}=\sqrt{9+1}=\sqrt{10}$$DA=\sqrt{(1-2)^2+(1-5)^2}=\sqrt{1+16}=\sqrt{17}$

The perimeter is the sum of all those segment lengths.

One way to find the area of the figure is to surround it with a rectangle, insert some lines so that the areas you do not want can be found and subtracted from the rectangle's area.  (See attached image.)

The area of the large rectangle around the figure is 5 x 4 = 20 square units.

The triangles have areas 1/2 (base) (height):

A. (1/2)(1)(4) = 2 square units

B. (1/2)(3)(1) = 1.5 square units

D. (1/2)(1)(2) = 1 square unit

E. (1/2)(5)(1) = 2.5 square units

Square C.  (1)(1) = 1 square unit

Total of all the area you don't want to include:

2 + 1.5 + 1 + 2.5 + 1 = 8 square units

Subtract 8 from the surrounding rectangle's area of 20, and you get the area of the figure is 20 - 8 = 12 square units.