Respuesta :
Answer:
The possible measures of the midsegment are 32 units and 212 units.
Step-by-step explanation:
Given:
A triangle PQR with the midsegment made by sides PQ and PR has length equal to [tex]8x^2+12x+36[/tex] and the base length QR opposite to the midsegment is [tex]4x^2+60x+120[/tex].
From midsegment theorem, we know that, the midsegment is a line parallel to the base opposite to it and half the length of the base length.
Therefore, Midsegment = [tex]\frac{1}{2}\times[/tex] Base length QR
[tex]8x^2+12x+36=\frac{1}{2}(4x^2+60x+120)\\\\8x^2+12x+36=2x^2+30x+60\\\\(8x^2-2x^2)+(12x-30x)+(36-60)=0\\\\6x^2-18x-24=0\\\\6(x^2-3x-4)=0\\\\x^2-3x-4=0\\\\x^2+x-4x-4=0\\\\x(x+1)-4(x+1)=0\\\\(x+1)(x-4)=0\\\\x=-1\ or\ x=4[/tex]
Now, midsegment can be calculated using the values of 'x'.
First, plug in -1 for 'x'. This gives,
[tex]Midsegment=8(-1)^2+12(-1)+36\\\\Midsegment=8\times 1-12+36\\\\Midsegment=8-12+36=32\ units[/tex]
Now, plug in 4 for 'x'. This gives,
[tex]Midsegment=8(4)^2+12(4)+36\\\\Midsegment=8\times 16-48+36\\\\Midsegment=128+48+36=212\ units[/tex]
Therefore, the possible measures of the midsegment are 32 units and 212 units
