Question

solve for ( x ). leave your answer in simplest radical form.
(image of a geometric figure with a right triangle on the left with hypotenuse 8, base 6, and a right triangle on the right with leg 3 and hypotenuse ( x ), sharing a common vertical leg.)
answer attempt 1 out of 2
( x = ) input box submit answer

Explanation:

Step1: Find the height of the left triangle

In the left right - triangle with hypotenuse \( c = 8 \) and one leg \( a=6 \), we use the Pythagorean theorem \( a^{2}+b^{2}=c^{2} \), where \( b \) is the height (let's call it \( h \)). So \( 6^{2}+h^{2}=8^{2} \), which simplifies to \( 36 + h^{2}=64 \). Then \( h^{2}=64 - 36=28 \), so \( h=\sqrt{28} = 2\sqrt{7}\). But we can also use this height in the right - triangle on the right. Wait, actually, the two right - triangles share the same height (the vertical segment). Let's denote the height as \( h \). For the left right - triangle, by Pythagoras: \( h^{2}+6^{2}=8^{2}\), so \( h^{2}=64 - 36 = 28\).

Step2: Solve for \( x \) in the right - triangle on the right

In the right - triangle on the right, we have one leg \( 3 \) and the other leg is the same height \( h \) (from the left triangle), and hypotenuse \( x \). By the Pythagorean theorem, \( x^{2}=h^{2}+3^{2}\). We know \( h^{2} = 28\), so \( x^{2}=28 + 9=37 \). Then \( x=\sqrt{37}\). Wait, no, wait. Wait, maybe I made a mistake. Wait, the left triangle: legs 6 and \( h \), hypotenuse 8. So \( h=\sqrt{8^{2}-6^{2}}=\sqrt{64 - 36}=\sqrt{28}=2\sqrt{7}\). Then the right triangle: legs \( h \) and 3, hypotenuse \( x \). So \( x=\sqrt{h^{2}+3^{2}}=\sqrt{28 + 9}=\sqrt{37}\)? Wait, no, wait, maybe the two right - triangles: the left one has legs 6 and \( h \), hypotenuse 8. The right one has legs \( h \) and 3, hypotenuse \( x \). Wait, but let's re - check. Wait, the left triangle: \( 6^{2}+h^{2}=8^{2}\), so \( h^{2}=64 - 36 = 28\). Then the right triangle: \( 3^{2}+h^{2}=x^{2}\), so \( x^{2}=9 + 28=37\), so \( x = \sqrt{37}\). Wait, but maybe there is a simpler way. Wait, actually, the two right - triangles: the left triangle: \( 8^{2}-6^{2}=h^{2}\), the right triangle: \( x^{2}-3^{2}=h^{2}\). So \( 8^{2}-6^{2}=x^{2}-3^{2}\). So \( 64 - 36=x^{2}-9\), \( 28=x^{2}-9\), \( x^{2}=28 + 9 = 37\), so \( x=\sqrt{37}\).

Answer:

\( \sqrt{37} \)