Question

In the parallelogram ABCD, DE and DF are perpendiculars. If AB= 12 cm, DE= 7cm,
DF = 14 cm, find the length of BC.
{ IF YOU ANSWER THIS CORRECT I WILL MAKE YOU BRAINLIEST }​

Answer 1

Step-by-step explanation:

In this diagram, I've extended side AB to point E, where it intersects the perpendicular that passes through point C. I've also added the perpendicular BF. I've labelled the lengths of the perpendiculars as h , and the lengths of the segments EB and CF as a .

Note that each of these makes a useful right triangle. In particular, triangle ACE is a right triangle whose hypotenuse is AC, one of the diagonals we were given: AC = 20. We also know that AB = 12, so AE = 12+a . Finally, EC = h . The pythagorean theorem gives us:

(12+a)2+h2=202

144+24a+a2+h2=400

a2+h2+24a=256

The other interesting triangle is BDF. This is also a right triangle, and its hypotenuse is BD, the other diagonal we were given: BD = 16. We also know that CD = 12, so DF = 12−a , and of course BF = h . Again, the pythagorean theorem gives us:

(12−a)2+h2=162

144−24a+a2+h2=256

a2+h2−24a=112

Subtracting these two equations gives us:

48a=144

a=3

You can quickly find h as well:

h2+(12−3)2=162

h2+81=256

h2=175

h=57–√

So the area of the parallelogram is just AB times h, or

12∗57–√=607–√

Answer 2

Answer:

solution is 6cm because we have to know the formula of parallel