MHB Find the height and base of a trapezium

[Etrujillo]

To start with problem #5 i cut the shape
Into 2, a triangle and a square, i know that the additional leg length to the triangle can be found by subtracting base 1 and base 2=4 so i have a triangle with a hypotenuse of 8 inches, 1 leg=4 and now i have to find the length of the other leg. The length of the other leg can be found by multiplying the length of the other leg by the square root of 3 to get (4×3squared)=6.9282 the area for that triangle would be 13.84. Now i have to find the missing side of x. It seems to be a rectangle, and i know the formula for that is length×width to get the area but i noticed a squared angle in the right bottom. Thats where i get lost. As for #6, i can start with finding out the missing leg of the triangle on the left which is half of its hypotenuse so i would divide 7÷2=3.5 then id get started with calculating the shorter length of the triangle on the right which would be half its length of hypotenuse to 8÷2=4 what i think i would do next is add 3.5+4+12= 19.5 as the value of x. Can anyone please verify if I am right? If not what did i do wrong?. Thank you

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[HOI]

In problem 5, the side marked "x" is a leg of a right triangle with hypotenuse of length 8 in and the other leg of length 14- 10= 4 in. Using the Pythagorean theorem, 8^2= 64= 4^2+ x^2= 16+ x^2 so x^2= 64- 16= 48. x= \sqrt{48}= 4\sqrt{3} which is approximately 6.9 in. I am not sure why you then calculate areas. The problem posted only asks for the length x.

In problem 6, divide the side marked "x" into three portions by drawing perpendiculars from the upper vertices to the base. The leftmost portion is a leg of a right triangle with hypotenuse of length 7 ft and the other leg of length 6 ft. Use the Pythagorean theorem to get \sqrt{49- 36}= \sqrt{13} which is 3.6 to one decimal place. The middle portion has length the same as the top edge, 12 ft, and the right portion is one leg of a right triangle with hypotenuse of length 8 ft and one leg of length 6 ft. The right portion has length \sqrt{64- 36}= \sqrt{28}= 2\sqrt{7} which is 5.3 to one decimal place. x= 3.6+ 12+ 5.3= 20.9 ft.

I don't know why you think that "the shorter length of the triangle on the right which would be half its length of hypotenuse". Where did you get that idea?