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The length of a rectangular painting is 3 inches longer than its width. If the diagonal is 15 inches long, what is the length of the painting?
A.
9 inches
B.
12 inches
C.
13 inches
D.
18 inches

Respuesta :

kudzordzifrancis

ANSWER



12 inches



EXPLANATION



Let the length of the rectangular painting be [tex]l[/tex] inches, then the width will be [tex]l-3[/tex] inches.



Since the diagonal is the hypotenuse, we can apply the Pythagoras Theorem to find the length.



[tex](l-3)^2+l^2=15^2[/tex]



We expand to obtain;



[tex]l^2-6l+9+l^2=225[/tex]



We simplify to obtain a quadratic equation in [tex]l[/tex].



[tex]2l^2-6l-216=0[/tex]



We divide through by 2 to obtain.



[tex]l^2-3l-108=0[/tex]



We factor to obtain



[tex]l^2-12l+9l+108=0[/tex]



[tex]l(l-12)+9(l-12)=0[/tex]



[tex](l-12)(l+9)=0[/tex]



[tex]\Rightarrow l=12\:or\:l=-9=0[/tex]



Since we are dealing with length we discard the negative value.



Hence the length is 12 inches

Ver imagen kudzordzifrancis
alinakincsem

Answer:

B. 12 inches

Step-by-step explanation:

We know that the length of the rectangular painting is 3 inches longer than its width so we can write it as:

l = w + 3

and the diagonal is 15 inches long.

We can use use the Pythagoras Theorem to find the length of the painting.

(w)² + (w + 3)² = (15)²

w² + w² + 6w + 9 = 225

2w² + 6w - 216 = 0

w² + 3w - 108 = 0

(w + 12)(w - 9) = 0

w = -12 (ignore), w = 9  

So, l = w + 3

l = 12

Therefore, the length of the rectangle is 12 inches.

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