What is the length of the hypotenuse? If necessary, round to the nearest tenth.
2 answers:
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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.)