Question

find area of square????
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Answer 1

Step-by-step explanation:

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A triangle with sides AB = 3, BC = 4, and AC = 5 has an inscribed square PQRS with side QR along the side AC. What is the area of the square?

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Because ABC is a 3-4-5 triangle, it is a right triangle.

Consequently, triangles ABC, PBS, SRC, and AQP are similar triangles. In each triangle, the longer leg to the shorter leg has a ratio of 4/3 (and the shorter leg to longer leg is a ratio of 3/4).

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Suppose the square has a side length equal to s. Then QR = s. In triangle SRC, SR = s, so RC = (4/3)s. In triangle AQP, PQ = s, so AQ = (3/4)s.

Thus we have:

AC = AQ + QR + RC

AC = (3/4)s + s + (4/3)s

5 = s(3/4 + 1 + 4/3)

s = 60/37

side of square= 60/37

or. 1.621621621

or. = 1.63

I hope you like my ${\tt{\purple{\underline{\underline{\huge{Answer}}}}}}$

Answer 2

$\huge\fbox\fcolorbox{aqua}{green}{Answer}$

Because ABC is a 3-4-5 triangle, it is a right triangle.

Consequently, triangles ABC, PBS, SRC, and AQP are similar triangles. In each triangle, the longer leg to the shorter leg has a ratio of 4/3 (and the shorter leg to longer leg is a ratio of 3/4).

Suppose the square has a side length equal to s.

Then QR = s. In triangle SRC, SR = s, so RC = (4/3)s.

In triangle AQP, PQ = s, so AQ = (3/4)s.

Thus we have:

AC = AQ + QR + RC

AC = (3/4)s + s + (4/3)s

5 = s(3/4 + 1 + 4/3)

s = 60/37

side of square= 60/37

or. 1.621621621

or. = 1.63

$\huge\fbox\fcolorbox{orange}{aqua}{Hᴏᴘᴇ ɪᴛ ʜᴇʟᴘs}$