Question

solve for ( x ) to the nearest tenth.

Explanation:

Step1: Find the hypotenuse of the lower right triangle

The lower right triangle has legs 7 and let the hypotenuse (which is also a leg of the upper right triangle) be \( y \). Using the Pythagorean theorem \( a^2 + b^2 = c^2 \), where \( a = 7 \), \( b \) is unknown, and \( c = 10 \)? Wait, no, wait. Wait, the lower triangle: legs 7 and the other leg? Wait, no, the lower triangle is a right triangle with legs 7 and the segment adjacent to the right angle, and hypotenuse 10? Wait, no, looking at the diagram: the lower triangle has legs 7 and (let's call the common side \( y \)), and hypotenuse 10. Wait, no, the upper triangle is a right triangle with leg 4 and leg \( x \), and hypotenuse \( y \). The lower triangle is a right triangle with leg 7 and leg \( y \), and hypotenuse 10? Wait, no, the lower triangle: right angle, one leg 7, another leg \( y \), hypotenuse 10. So first, find \( y \) using Pythagorean theorem: \( 7^2 + y^2 = 10^2 \)? Wait, no, wait, the hypotenuse of the lower triangle is 10? Wait, the side labeled 10 is the hypotenuse of the lower right triangle? Wait, the lower triangle: right angle, one leg 7, another leg \( y \), hypotenuse 10. So \( 7^2 + y^2 = 10^2 \)? Wait, no, that would be \( y^2 = 10^2 - 7^2 \). Wait, 10 squared is 100, 7 squared is 49, so \( y^2 = 100 - 49 = 51 \), so \( y = \sqrt{51} \). Then, the upper triangle is a right triangle with leg 4 and leg \( x \), and hypotenuse \( y = \sqrt{51} \). So using Pythagorean theorem for the upper triangle: \( 4^2 + x^2 = y^2 \). But we know \( y^2 = 51 \), so \( 16 + x^2 = 51 \). Then, \( x^2 = 51 - 16 = 35 \). Then, \( x = \sqrt{35} \approx 5.9 \) (to the nearest tenth). Wait, let's check again.

Wait, the lower triangle: right angle, legs 7 and \( y \), hypotenuse 10? Wait, no, the side labeled 10 is the hypotenuse? Wait, the diagram: the lower triangle has a right angle, one leg 7, another leg \( y \), and hypotenuse 10. So \( 7^2 + y^2 = 10^2 \)? Wait, 77=49, 1010=100, so \( y^2 = 100 - 49 = 51 \), so \( y = \sqrt{51} \approx 7.141 \). Then the upper triangle: right angle, leg 4, leg \( x \), hypotenuse \( y \). So \( 4^2 + x^2 = y^2 \). So \( 16 + x^2 = 51 \), so \( x^2 = 51 - 16 = 35 \), so \( x = \sqrt{35} \approx 5.916 \), which to the nearest tenth is 5.9.

Wait, but let's confirm the diagram. The upper triangle is a right triangle with vertical leg 4, horizontal leg \( x \), and hypotenuse (the diagonal). The lower triangle is a right triangle with vertical leg \( y \) (same as the hypotenuse of the upper triangle), horizontal leg 7, and hypotenuse 10. So yes, that's correct. So first, find the length of the common side (the hypotenuse of the upper triangle and a leg of the lower triangle) using the lower triangle: \( 7^2 + y^2 = 10^2 \) → \( y^2 = 100 - 49 = 51 \). Then, upper triangle: \( 4^2 + x^2 = y^2 \) → \( 16 + x^2 = 51 \) → \( x^2 = 35 \) → \( x = \sqrt{35} \approx 5.9 \).

Step1: Calculate the length of the common hypotenuse (let's call it \( y \)) using the lower right triangle.

The lower right triangle is a right triangle with legs 7 and \( y \), and hypotenuse 10. By the Pythagorean theorem:
\( 7^2 + y^2 = 10^2 \)
\( 49 + y^2 = 100 \)
\( y^2 = 100 - 49 = 51 \)

Step2: Calculate \( x \) using the upper right triangle.

The upper right triangle is a right triangle with legs 4 and \( x \), and hypotenuse \( y \) (where \( y^2 = 51 \)). By the Pythagorean theorem:
\( 4^2 + x^2 = y^2 \)
\( 16 + x^2 = 51 \)
\( x^2 = 51 - 16 = 35 \)
\( x = \sqrt{35} \approx 5.9 \) (rounded to the nearest tenth)

Answer:

\( \boxed{5.9} \)