bullsfan4584

28.05.2021 •

Mathematics

Agregenme o baneo la pregunta

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Ответ:

chdt510m1

08.07.2020

Mathematics

$A=(72+36\sqrt{3})\ mm^2$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

Let

a ----> the height of rectangle in mm

b ---> the base of rectangle in mm

step 1

Find the base of rectangle

b=AB+BC+CD+DE ----> by segment addition postulate

substitute

$b=3+3+3+3=12\ mm$

step 2

Find the height of rectangle

a=FB+CG+GH ---> by segment addition postulate

substitute the given values

$a=3+CG+3\\a=6+CG$

Find the length sides CG

Applying the Pythagorean Theorem

BG^2=BC^2+CG^2

substitute the given values

6^2=3^2+CG^2

CG^2=6^2-3^2

$CG=\sqrt{27}\ mm$

simplify

$CG=3\sqrt{3}\ mm$

therefore

$a=(6+3\sqrt{3})\ mm$

step 3

Find the area of rectangle

A=ab

we have

$a=(6+3\sqrt{3})\ mm$

$b=12\ mm$

substitute

$A=(6+3\sqrt{3})(12)=(72+36\sqrt{3})\ mm^2$

Please help!! MathsWatch, Surd Expressions, Clip 207c

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