+0

# HELPELPELPEL

0
131
1

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 $t$.

Oct 14, 2019

#1
+1

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   Oct 14, 2019
edited by CPhill  Oct 14, 2019