Saved Bookmarks
| 1. |
In the figure seg `PQ||` seg DE, `A(DeltaPQF)=20` units `PF=2DP`,then find `A(square DPQE)` by completing the following activity: Activity: `A(DeltaPQF)=20` sq units, `PF=2DP`. Let us assume `DP=x` `:.PF=2x` `DF=DE+square=square+square=3x` In `DeltaFDE` and `DeltaFPQ`. `/_FDE~=/_square` ..........(Corresponding angles) `/_FED~=/_square` .....(Corresponding angles) `:.DeltaFDE~DeltaFPQ` .....(AA test) `:.(A(DeltaFDE))/(A(DeltaFPQ))=(square)/(square)=((3x)^(2))/((2x)^(2))=9/4` `A(DeltaFDE)=9/4A(DeltaFPQ)=9/4xxsquare=square` `A(squareDPQE)=A(DeltaFDE)-A(DeltaFPQ)` `=square-square` `=square` |
|
Answer» `A(DeltaPQF)=20` sq units, `PF=2DP`. Let us assume `DP=x` `:.PF=2x` `DF=DP+PF=x+2x=3x` In `DeltaFDE` and `DeltaFPQ` `/_Fde~=/_FPQ` ……(Corresponding angles) `/_FED~=/_FQP` …..(Corresponding angles) `:.DeltaFDE~DeltaFPQ` ...(AA test) `:.(A(DeltaFDE))/(A(DeltaFPQ))=(DF^(2))/(PF^(2))=((3x^(2)))/((2x)^(2))=9/4` `A(DeltaFDE)=9/4A(DeltaFPQ)=9/4xx20=45` sq units `A(square DPQE)=A(DeltaFDE)-A(DeltaFPQ)` `=45-20` `=25` sq units. |
|