The base of the triangle equals (2x) the mid-segment, so....
12/2 = 2x/2
6= x
A rectangle and a triangle have equal areas. The length of the rectangle is 12 inches, and its width is 8 inches. If the base of the triangle is 32 inches, what is the length, in inches, of the altitude drawn to the base? WRONG WRONG WRONG NO NO NO ::"::":""::":"""::"":":""::::::::::::: ::"::": ::":: :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::"
The square of the length of the base plus the square of the length of the height will equal the square of the length of the hypotenuse of your right triangle, per Pythagoras. Square the hypotenuse, subtract the square of the height, and then find the positive square root of that and you'll have the base of your right triangle.
9 meters You answered it yourself
The area of any triangle is1/2 of (the length of the triangle's base) times (the triangle's height).
the length of the base would be 16
The length of a midsegment is half that of the parallel side of the triangle; assuming the midsegment is parallel to the [given] base, then its length is 27 ÷ 2 = 13.5 units.
The midsegment is the average of the top base and bottom base. Take B1+B2 and divide by 2.
A trapezoid midsegment is parallel to the set of parallel lines in a trapezoid and is equal to the average of the lengths of the bases
an exact answer would be twelve times the square root of three
The base is 2.5ft.
The answer depends on what information you do have about the triangle.
A rectangle and a triangle have equal areas. The length of the rectangle is 12 inches, and its width is 8 inches. If the base of the triangle is 32 inches, what is the length, in inches, of the altitude drawn to the base?
Is indeterminate.
Are of a triangle is (length of the base) times (height)/2 .
The Trapezoid midsegment conjecture- the midsegment of a trapezoid is parallel to the bases and is equal to the length to the average of the lengths of the bases. This is Some what Algebra....... what you do is take your length 90 and midsegment 85 into a prob like this (90+X)/2=85 times by two on both sides to cancel out the two. after that you end up with 90+X=85 next you have to "isolate" the X by subtracting 90 from both sides you would get 90+X=85 -90 -90 to get X= -5 the other side would be -5 so it doesnt work to check it plug the number back into the equation (90+-5)/2=85
The area of a triangle (At) is one half the length of the base (b) times the height (h).Atriangle = 0.5bhThe height of a triangle is the length of the line drawn perpendicular (at right angles to) to the base from the angle opposite the base.
The measurement of the angle of the triangle...supposing it is a triangle.