Question

Nautical flags are used to represent letters of alphabets. The flag for the letter O consists of a yellow right triangle and a red right triangle joined together along their hypotenuse to form a square. The joint hypotenuse of the two triangles is three inches longer than a side of the square. Find the length of a side of the flag. Round your answer to the nearest tenth.

Answer 1

Answer:

The length of a side of the flag is 7.2inches

Step-by-step explanation:

Let length = x

Let hypotenuse = x+3

Using Pythagoras' Theorem:

x^2 + x^2 = (x+3)^2

x^2 + x^2 = x^2 + 6x + 9

2x^2 = x^2 + 6x + 9

×^2 - 6x - 9 = 0

$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a}$

x=[-(-6)+sqrt.(-6)^2-4(1)(-9)]/2(1)

×=[6+sqrt.72]/2

×=7.2inches

×=[-(-6)+sqrt.(-6)^2-4(1)(-9)]/2(1)

x=[6-sqrt.72]/2

×=-1.2(rej)

(PS. sqrt is square root)

(Correct me if i am wrong)

Answer 2

Answer:

7.2 in

Step-by-step explanation:

Let length of side of flag=x

Hypotenuse of right triangle=x+3

According to question information

$(x+3)^2=x^2+x^2$

Using Pythagoras theorem

$(hypotenuse)^2=(base)^2+(perpendicular\;side)^2$

$x^2+6x+9=2x^2$

Using identity: $(x+y)^2=x^2+y^2+2xy$

$2x^2-x^2-6x-9=0$

$x^2-6x-9=0$

Using quadratic formula :$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{6\pm\sqrt{(-6)^2-4(1)(-9)}}{2(1)}$

$x=\frac{6\pm\sqrt{36+36}}{2}$

$x=\frac{6\pm\sqrt{72}}{2}$

$x=\frac{6\pm6\sqrt2}{2}$

$x=\frac{6+6\sqrt2}{2}=3+3\sqrt2=7.2$

$x=\frac{6-6\sqrt2}{2}=3-3\sqrt2$

$x=-1.24$

It is not possible because the length of side is always positive.

Hence, the side of flag=7.2 in