Question

solve for ( x ) to the nearest tenth.

Explanation:

Step1: Identify the hypotenuse of the lower triangle

The lower right triangle has legs 7 and (let's call the hypotenuse \( h \)). Using the Pythagorean theorem \( h^2 = 7^2 + 6^2 \)? Wait, no, wait. Wait, the lower triangle has legs 7 and... Wait, the vertical side is 6, but the lower triangle has legs 7 and the other leg? Wait, no, looking at the diagram: the lower right triangle (with the right angle at the bottom left) has legs 7 and the hypotenuse of the upper triangle? Wait, no, actually, the two right triangles: the upper one has legs \( x \) and 6, hypotenuse (let's say \( h \)). The lower one has legs 7 and \( h \)? Wait, no, the diagram shows a quadrilateral with two right angles, and a diagonal. The right side is 10. Wait, maybe the diagonal is common. Let's re-examine:

Wait, the upper triangle: right-angled at the top left, legs \( x \) (horizontal) and 6 (vertical), hypotenuse (let's call it \( d \)). The lower triangle: right-angled at the bottom left, legs 7 (horizontal) and \( d \) (vertical), hypotenuse 10 (the right side). Wait, no, the hypotenuse of the lower triangle is 10. So by Pythagoras, for the lower triangle: \( 7^2 + d^2 = 10^2 \). Then for the upper triangle: \( x^2 + 6^2 = d^2 \). So we can solve for \( d^2 \) from the lower triangle, substitute into the upper, then solve for \( x \).

Step2: Calculate \( d^2 \) from the lower triangle

Lower triangle: legs 7 and \( d \), hypotenuse 10. So \( 7^2 + d^2 = 10^2 \)
\( 49 + d^2 = 100 \)
\( d^2 = 100 - 49 = 51 \)

Step3: Calculate \( x \) from the upper triangle

Upper triangle: legs \( x \) and 6, hypotenuse \( d \) (so \( d^2 = 51 \)). So \( x^2 + 6^2 = d^2 \)
\( x^2 + 36 = 51 \)
\( x^2 = 51 - 36 = 15 \)
\( x = \sqrt{15} \approx 3.9 \) (to the nearest tenth)

Answer:

\( \boxed{3.9} \)