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QUESTION IMAGE

find the perimeter of the following shape, rounded to the nearest tenth:

Question

find the perimeter of the following shape, rounded to the nearest tenth:

Explanation:

Step1: Identify coordinates of vertices

Coordinates: $A(4,6)$, $B(7,3)$, $C(3,0)$, $D(1,2)$

Step2: Calculate length of $AB$

Use distance formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$\sqrt{(7-4)^2+(3-6)^2}=\sqrt{3^2+(-3)^2}=\sqrt{18}=3\sqrt{2}\approx4.24$

Step3: Calculate length of $BC$

$\sqrt{(3-7)^2+(0-3)^2}=\sqrt{(-4)^2+(-3)^2}=\sqrt{25}=5$

Step4: Recognize shape properties

This is a parallelogram, so $AB=CD$, $BC=AD$

Step5: Compute perimeter

Perimeter = $2(AB + BC)$
$2(3\sqrt{2}+5)=2(4.24+5)=2\times9.24=18.48$

Step6: Round to nearest tenth

Round 18.48 to 18.5

Answer:

18.5