Anthony's lovker is (x+4) ft . long qnd (x+2) ft wide
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I think instead of "conjunction" it should say "conjecture"
Anyways, draw out a quadrilateral (see figure 1 attached). It can be any figure with 4 sides. Label the angles as A, B, C and D. The conjecture to prove is that A+B+C+D = 360. In other words, the sum of the four angles of any quadrilateral is always 360 degrees.
Draw a segment from A to C. This segment will cut the quadrilateral into two triangles. Re-label angle A to angle E and F. Do the same for angle C (label it G and H). Essentially, angle A = E+F and angle C = G+H.
See figure 2. Notice how I've color-coded things. The blue angles correspond to one triangle while the red angles correspond to the other triangle.
Since we know that three angles of a triangle always add to 180, we can say
B+G+E = 180
F+D+H = 180
Add the two equations (add left side separately; do the same for the right side)
Doing so leads to
B+G+E+F+D+H = 180+180
(E+F) + B + (G+H) + D = 360
A+B+C+D = 360
Which proves the conjecture
Anyways, draw out a quadrilateral (see figure 1 attached). It can be any figure with 4 sides. Label the angles as A, B, C and D. The conjecture to prove is that A+B+C+D = 360. In other words, the sum of the four angles of any quadrilateral is always 360 degrees.
Draw a segment from A to C. This segment will cut the quadrilateral into two triangles. Re-label angle A to angle E and F. Do the same for angle C (label it G and H). Essentially, angle A = E+F and angle C = G+H.
See figure 2. Notice how I've color-coded things. The blue angles correspond to one triangle while the red angles correspond to the other triangle.
Since we know that three angles of a triangle always add to 180, we can say
B+G+E = 180
F+D+H = 180
Add the two equations (add left side separately; do the same for the right side)
Doing so leads to
B+G+E+F+D+H = 180+180
(E+F) + B + (G+H) + D = 360
A+B+C+D = 360
Which proves the conjecture