Given that DE is a midsegment for ABC and the length of DE is 8 units. What is the length of BC?

1 answer:

4 0

Answer:

BC = 16 units

Step-by-step explanation:

Since DE is the midsegment of the triangle, based on the midsegment theorem, thus:

DE = ½(BC)

DE = 8 units

Plug in the value

8 = ½(BC)

Multiply both sides by 2

2*8 = BC

BC = 16 units

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Leni [432]

2(3)-3(-2)=6-(-6)=6+6=12

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3 years ago

Lelu [443]

Answer:

$\frac{n^{2}}{2a}$

Step-by-step explanation:

Just split it up first $\frac{9an^{3}}{18a^{2}n}=\frac{9}{18}\times \frac{a}{a^2} \times\frac{n^3}{n}$
Then 9 divided by 18 can be simplified into the fraction 1/2
What can go into $a$ and $a^2$ ? a itself right, so if you divide a by a you get 1, if you divide $a^2$ by a you get a
Same as the a fraction, if you divide $n^3$ by n you get $n^2$
$\frac{9an^{3}}{18a^{2}n}=\frac{9}{18}\times \frac{a}{a^2} \times\frac{n^3}{n}=\frac{1}{2}\times \frac{1}{a} \times\frac{n^2}{1}=\frac{n^{2}}{2a}$