Question

. 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

Answer 1

Answer:

$L_2 = 2\frac{2}{15}\ ft$

Step-by-step explanation:

Given

Original

$Length(L_1) = 8\ in$

$Height (H_1) = 15\ in$

New

$Height (H_2) = 4\ ft$

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.

$Ratio = L_1 : H_1$

and

$Ratio = L_2 : H_2$

Equate both ratios

$L_1 : H_1 = L_2 : H_2$

Rewrite as fraction

$\frac{L_1}{H_1} = \frac{L_2}{H_2}$

Substitute values for L1, H1 and H2

$\frac{8}{15} = \frac{L_2}{4}$

Multiply both sides by 4

$4 * \frac{8}{15} = \frac{L_2}{4} * 4$

$4 * \frac{8}{15} = L_2$

$\frac{4 * 8}{15} = L_2$

$\frac{32}{15} = L_2$

$2\frac{2}{15} = L_2$

$L_2 = 2\frac{2}{15}\ ft$

The new length must be $2\frac{2}{15}\ ft$

Answer 2

The new picture must be $2\frac{2}{15}$ feet or 2.1334 feet in length.

We know, $1\ inch= \frac{1}{12}\ feet$

Given to us:

Initial length, $L_i =8 inches$;

Initial height, $H_i =15\ inches$;

Final height, $H_f = 4\ feet$;

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

Ratio, $r=\dfrac{Length}{Height}$

So,

$\dfrac{Initial\ Length}{Initial\ Height} =\dfrac{Final\ Length}{Final\ Height}$,

$\dfrac{8}{15} =\dfrac{Final\ Length}{4}\\\\{Final\ Length}=\dfrac{8}{15} \times 4 \\\\{Final\ Length}=\dfrac{32}{15}\\\\{Final\ Length}=2\frac{2}{15}\ feet$

Hence, the new picture must be $2\frac{2}{15}$ feet or 2.1334 feet in length.

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