e Sac 2. D is any point on the side BC of AABC (Fig. 35). Fill in the blanks with >, < or = to make the statement true. (1) AD AC + DC AB + BD (ii) AD (iii) AD 11 12 (AB + BC + CA) 2 Loin XC to show
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AP=1cm,PB=3cm,AQ=1.5cm,QC=4.5cm [ Given ]
Then, AB=AP+PB=1+3=4cm
In △APQ and △ABC,
∠A=∠A [ Common angle ]
AB
AP
=
AC
AQ
[ Each equal to
4
1
]
∴ △APQ∼△ABC [ By SAS similarity ]
∴
area(△ABC)
area(△APQ)
=
AB
2
AP
2
[ By area of similar triangle theorem ]
∴
area(△ABC)
area(△APQ)
=
4
2
1
2
∴
area(△ABC)
area(△APQ)
=
16
1
∴ area(△APQ)=
16
1
×area(△ABC)
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