plz answer this fast plz i have to answer it do teacher....Question 19....
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BP:PC=1:2=>
the area of tri.DQC/
the area of tri.PQC=3:1=>
area of tri.DQC=3(20)=60 cm^2
Area of tri.CDP=
area of tri.DQC-area of tri. PQC=
60-20=40 cm^2.
Area of ABCD=BC*its height
Area of CDP=0.5PC*its height
=>
Area of ABCD/area of CDP=
(3/2)/(0.5)area of CDP=>
Area of ABCD=3(40)=120 cm^2
the area of tri.DQC/
the area of tri.PQC=3:1=>
area of tri.DQC=3(20)=60 cm^2
Area of tri.CDP=
area of tri.DQC-area of tri. PQC=
60-20=40 cm^2.
Area of ABCD=BC*its height
Area of CDP=0.5PC*its height
=>
Area of ABCD/area of CDP=
(3/2)/(0.5)area of CDP=>
Area of ABCD=3(40)=120 cm^2
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