. Leslie is blowing up one of her favorite photographs to hang on her wall. If the photo's original length and height were 8 inches and 15 inches, and the new height is 4 feet, how many feet must the new picture be

Respuesta :

Answer:

[tex]L_2 = 2\frac{2}{15}\ ft[/tex]

Step-by-step explanation:

Given

Original

[tex]Length(L_1) = 8\ in[/tex]

[tex]Height (H_1) = 15\ in[/tex]

New

[tex]Height (H_2) = 4\ ft[/tex]

Required

Determine the new length (L2)

Since the original photographs was blown up; the original and the new photographs would have the same ratio

i.e.

[tex]Ratio = L_1 : H_1[/tex]

and

[tex]Ratio = L_2 : H_2[/tex]

Equate both ratios

[tex]L_1 : H_1 = L_2 : H_2[/tex]

Rewrite as fraction

[tex]\frac{L_1}{H_1} = \frac{L_2}{H_2}[/tex]

Substitute values for L1, H1 and H2

[tex]\frac{8}{15} = \frac{L_2}{4}[/tex]

Multiply both sides by 4

[tex]4 * \frac{8}{15} = \frac{L_2}{4} * 4[/tex]

[tex]4 * \frac{8}{15} = L_2[/tex]

[tex]\frac{4 * 8}{15} = L_2[/tex]

[tex]\frac{32}{15} = L_2[/tex]

[tex]2\frac{2}{15} = L_2[/tex]

[tex]L_2 = 2\frac{2}{15}\ ft[/tex]

The new length must be [tex]2\frac{2}{15}\ ft[/tex]

The new picture must be [tex]2\frac{2}{15}[/tex] feet or 2.1334 feet in length.

We know, [tex]1\ inch= \frac{1}{12}\ feet[/tex]

Given to us:

Initial length, [tex]L_i =8 inches[/tex];

Initial height, [tex]H_i =15\ inches[/tex];

Final height, [tex]H_f = 4\ feet[/tex];

We know that a when a picture is blown up(enlarged), it is always done in a specific ratio, therefore

Ratio, [tex]r=\dfrac{Length}{Height}[/tex]

So,

[tex]\dfrac{Initial\ Length}{Initial\ Height} =\dfrac{Final\ Length}{Final\ Height}[/tex],

[tex]\dfrac{8}{15} =\dfrac{Final\ Length}{4}\\\\{Final\ Length}=\dfrac{8}{15} \times 4 \\\\{Final\ Length}=\dfrac{32}{15}\\\\{Final\ Length}=2\frac{2}{15}\ feet[/tex]

Hence, the new picture must be [tex]2\frac{2}{15}[/tex] feet or 2.1334 feet in length.

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