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Xelga [282]
3 years ago
6

The perimeter of a rectangular garden is 24. If the length is half the width, what is the width?

Mathematics
2 answers:
Anna11 [10]3 years ago
4 0

Let length be x

  • Width be 2x

ATQ

\\ \sf\longmapsto 2(L+B)=24

\\ \sf\longmapsto 2(x+2x)=2=

\\ \sf\longmapsto 2(3x)=24

\\ \sf\longmapsto 6x=24

\\ \sf\longmapsto x=\dfrac{24}{6}

\\ \sf\longmapsto x=4

Ber [7]3 years ago
3 0

Step-by-step explanation:

let the length be X and breadth be 2x

p=2(l+b)

24=2(X+2x)

24=2x+4x

24=6x

24÷6=x

4=X

so,we know that

X=4 and 2x=2 X 4= 8

therefore the value of breadth or width is 8

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Leto [7]

Area inside the semi-circle and outside the triangle is  (91.125π - 120) in²

Solution:

Base of the triangle = 10 in

Height of the triangle = 24 in

Area of the triangle = \frac{1}{2} bh

                                $=\frac{1}{2} \times 10\times24

Area of the triangle = 120 in²

Using Pythagoras theorem,

\text{Hypotenuse}^2=\text{base}^2+\text{height}^2

\text{Hypotenuse}^2=10^2+24^2

\text{Hypotenuse}^2=100+576

\text{Hypotenuse}^2=676

Taking square root on both sides, we get

Hypotenuse = 23 inch = diameter

Radius = 23 ÷ 2 = 11.5 in

Area of the semi-circle = \frac{1}{2}\pi r^2

                                      $=\frac{1}{2} \pi \times (13.5)^2

Area of the semi-circle = 91.125π in²

Area of the shaded portion = (91.125π - 120)  in²

Area inside the semi-circle and outside the triangle is  (91.125π - 120)  in².

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