QueenNerdy889
17.01.2021 •
Mathematics
martin has two rectangles that are the same size. One rectangle is cut into 1/2 size parts the other rectangle is cut into 1/3 size parts he wants to cut the rectangles so that they have the same size parts. How can he cut each rectangle?
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Ответ:
P=AB+BC+AC
AB=AR+RB
AR=14; RB=7
AB=14+7→AB=21
BC=PB+PC
PC=10; PB=?
BC=PB+10
AC=18
If AC is parallel to RP,and they are cut by the secant AB, the correspondent angles must be congruents, then angle CAB must be congruent with angle PRB.
The triangles ABC and RBP are similars, because they have to congruent angles:
Angle CAB is congruent with angle PRB
Angle ABC is common
Then theirs sides must be proportionals:
PB/RB=CB/AB
RB=7
BC=PB+10
AB=21
Replacing the known values:
PB/7=(PB+10)/21
Solving for PB. Multiplying both sides of the equation by 21:
21(PB/7)=21[(PB+10)/21]
3PB=PB+10
Subtracting PB both sides of the equation:
3PB-PB=PB+10-PB
2PB=10
Dividing both sides of the equation by 2:
(2PB)/2=(10)/2
PB=5
BC=PB+10
BC=5+10
BC=15
Then the perimeter is of triangle ABC is:
P=AB+BC+AC
P=21+15+18
P=54
The perimeter of triangle ABC is 54 units