Find the perimeter and area of ADLI. ADLI is the largest rectangle, 15 x 11, of which all the other rectangles are a part.
The perimeter is 2( l + w).
P = 2(15 + 11)
P = 2(26)
P = 52 units
The area of a rectangle is lw
A = 15(11)
A = 165 square units
Rectangle CDLK is shaded green. Its dimensions are 11 x (15 - 13) or 11 X 2.
Its perimeter is 2 ( 11 + 2) or 2(13) or 26 units. Its area is 11(2) or 22 square units.
GHCDLJ is a hexagon composed of the purple and green rectangles. First we have to find the length of GH. KL is congruent to CD, JK is congruent to GH. In the above exercise we found that CD = 2. We were given the length of JL. It is 10.
Let BC = x.
JL = GH + CD Segment addition postulate
10 = x + 2
x = 10 -2 or 8.
Therefore GH = 8.
Next we find the length of CJ. DL is congruent to CK, DL = 11, so CK = 11.
CH = 4.
CH + HK = CK Segment addition postulate
4 + x = 11
x = 11 - 4
x = 7
So HK = 7
The perimeter of GHCDLJ = GH + HC + CD + DL + LJ + JG
P = 8 + 4 + 2 + 11 + 10 + 7
P = 42 units.
Now find the area of the purple rectangle and the area of the green rectangle and add them together to find the area of the hexagon GHCDLJ.
ABCD is a rectangle. Therefore BC and AD are parallel and congruent; AB and CD are parallel and congruent. The distance from point A to point B is equal to the distance from point D to point C.
Therefore, points B and C share the same y coordinate, p, and points D and C share the same x coordinate, q. The coordinates of point C are (q, p)
DEF if an isoceles triangle. The perpendicular bisector of an isoceles triangle is the angle bisector of the vertex angle. Therefore, <DEP is congruent to <FEP. EF = ED (given in the definition of an isoceles triangle), EP = EP (reflective property of congruence) Therefore, by SAS, triangle DEP is congruent to triangle FEP. DP is congruent to PF by CPCTC. The distance from D to F is twice the distance from D to P. Point F is collinear with point D and lies on the x axis where y = 0. So, the cordinates of point F are (2p, 0).