This one too Solve for x z=xy
2 answers:
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<u>Given</u>:
Given that the graph OACE.
The coordinates of the vertices OACE are O(0,0), A(2m, 2n), C(2p, 2r) and E(2t, 0)
We need to determine the midpoint of EC.
<u>Midpoint of EC:</u>
The midpoint of EC can be determined using the formula,
Substituting the coordinates E(2t,0) and C(2p, 2r), we get;
Simplifying, we get;
Dividing, we get;
Thus, the midpoint of EC is (t + p, r)
Hence, Option A is the correct answer.
This type of problems are solved by partitioning the figure into rectangles, whose areas we can find easily.
Picture 1)
From the point denoted by E, draw two altitudes towards the sides AG and AC.
Since ED is 3, then BC is 3. Thus, AB=4-3=1.
Since EF is 1, then HG is 1, so AH=3-1=2.
The area of the figure is equal to the area of rectangle ACDH + the area of the square HEFG.
That is, the area of the figure is 4*2+1*1=8+1=9 (square in).
Picture 2)
As in the first part, draw the altitudes EB and EH from point E.
GF is 5, so AB is 5. This means that BC=7-5=2 (in).
FE is 3, so GH is 3, that is HA=6-3=3.
Thus, Area(figure)=Area(ABFG)+Area(BCDE)=5*6+2*3=30+6=36 (square in.)
Answer: 9 (square in); 36 (square in.)
Picture 1)
From the point denoted by E, draw two altitudes towards the sides AG and AC.
Since ED is 3, then BC is 3. Thus, AB=4-3=1.
Since EF is 1, then HG is 1, so AH=3-1=2.
The area of the figure is equal to the area of rectangle ACDH + the area of the square HEFG.
That is, the area of the figure is 4*2+1*1=8+1=9 (square in).
Picture 2)
As in the first part, draw the altitudes EB and EH from point E.
GF is 5, so AB is 5. This means that BC=7-5=2 (in).
FE is 3, so GH is 3, that is HA=6-3=3.
Thus, Area(figure)=Area(ABFG)+Area(BCDE)=5*6+2*3=30+6=36 (square in.)
Answer: 9 (square in); 36 (square in.)