p and q are respectively midpoints of side ab and bc of a triangle abc and R is a midpoint of ap show that
1. area of triangle prq = 1\2area of arc
2. area of triangle rqc= 3\8 area of ABC
3. area of triangle pbq = area of arc
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(i) area (ΔPRQ)=1/2 area(APQ)[since RQ is a median and median divides triangle into equal areas]
=area(ΔABC)=............(1)
area (ΔARC)=1/2 area(ΔAPC) [in the ΔACP , CR is the median]
=1/2*1/2 area(ΔABC) [since CP is the median of ABC]
=1/4*x=x/4 ............(2)
from (1) and (2) area(ΔPRQ)=1/2 area(ΔARC)
(ii)
area(ΔRQC)=1/2 area(ΔARC) [since RQ is the median of triangle ARC]
=1/2*1/2area(ΔAPC)=1/4*1/2 area(ΔABC)=1/8x
(iii)
area(ΔPBQ)=area(ΔPQC)[the triangle with same base and between same set of parallel lines ]
=1/2 area(ΔAPC) [since PQ is the median of triangle APC]
=1/2*1/2 area(ΔABC)
=1/4 area(ΔABC)=x/4
from (2) area (ΔARC)=x/4
hence area(ΔPBQ)=area(ΔARC)
=area(ΔABC)=............(1)
area (ΔARC)=1/2 area(ΔAPC) [in the ΔACP , CR is the median]
=1/2*1/2 area(ΔABC) [since CP is the median of ABC]
=1/4*x=x/4 ............(2)
from (1) and (2) area(ΔPRQ)=1/2 area(ΔARC)
(ii)
area(ΔRQC)=1/2 area(ΔARC) [since RQ is the median of triangle ARC]
=1/2*1/2area(ΔAPC)=1/4*1/2 area(ΔABC)=1/8x
(iii)
area(ΔPBQ)=area(ΔPQC)[the triangle with same base and between same set of parallel lines ]
=1/2 area(ΔAPC) [since PQ is the median of triangle APC]
=1/2*1/2 area(ΔABC)
=1/4 area(ΔABC)=x/4
from (2) area (ΔARC)=x/4
hence area(ΔPBQ)=area(ΔARC)
robin9388:
Hehe sorry for making u confuse
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