The defensive area is 510 square yards
Step-by-step explanation:
Let L be the length of the rectangle
and
W be the width
According to given statement
L = 2W-15
Perimeter = 270
[tex]P = 2W+2L\\270 = 2W + 2(2W-15)\\270 = 2W + 4W -30\\270 +30 = 6W\\6W = 300[/tex]
Dividing both sides by 6
[tex]\frac{6W}{6} =\frac{300}{6}\\W = 50\ yards\\\\and\\L = 2(50) -15\\= 100-15\\=85\ yards[/tex]
We have to find the area first to find the defensive area
So,
[tex]Area = L* W\\= 85 * 50\\=4250\ square\ yards[/tex]
Defensive area = 3/25 of total area
So,
[tex]=\frac{3}{25}*4250\\=3 *170\\=510\ square\ yards[/tex]
Hence,
The defensive area is 510 square yards
Keywords: Area, Rectangle
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