Triangle $ABC$ has vertices $A(0, 8)$, $B(2, 0)$, $C(8, 0)$. A horizontal line with equation $y=t$ intersects line segment $ \overline{AB} $ at $T$ and line segment $ \overline{AC} $ at $U$, forming $\triangle ATU$ with area 13.5. Compute $t$.
+128448 Triangle ABC has vertices A(0, 8), B(2, 0), C(8, 0)
A horizontal line with equation y= t intersects line segment {AB} at T and line segment {AC} at U, forming triangle ATU with area 13.5. Compute t.
The area of the given triangle is (1/2)BC * height = (1/2)(6)(8) = 24
Since TU is parallel to BC.....then triangle ATU is similar to triangle ABC
The square of the scale factor * area of ABC = area of ATU
Scale factor ^2 = area of ATU / Area of ABC
Scale factor = √ [13.5/24] = 3/4
So.....the height of triangle ATU = 3/4 the height of triangle ABC
So..... the line segment TU will be located at y = 8 - (3/4)8 = 8 - 6 = 2
See the pic below :
Proof of ATU area : (1/2)*TU* height of 6 = (1/2) 4.5 (6) = 4.5 * 3 = 13.5
