In the figure, `/_DFE=90^(@)`, seg `FG bot` side `DE`, `DG=8`, `FG=12` then complete the following activity to find the length of seg `DE`.

In `DeltaDFE`, `/_DFE=90^(@)`,
seg `FG bot` hypotenuse `DE`
`:.` by theorem of geometric mean,
`FG^(2)=squarexxEG`
`:.12^(2)=squarexxEG`
`EG=(12xx12)/(square)`
`:.EG=square`
`DE=DG+GE=8+square=square`

In triangle DEF, /_DFE=90^@, FG_|_ED and D-G-E. If GD=8, FG=12, then find EG.

In triangle DEF, /_DFE=90^@, FG_|_ED and D-G-E. If GD=8, FG=12, then find FD.

In triangle DEF, /_DFE=90^@, FG_|_ED and D-G-E. If GD=8, FG=12, then find EF.

In the figure, /_LMN=/_LKN=90^(@) seg MK bot seg LN . Complete the following activity to prove R is the midpoint of seg MK . Proof: In DeltaLMN , /_LMN=90^(@) seg MR bot hypotenuse LN :. by property of geometric mean, MR^(2)=squarexxRN In DeltaLKN , /_LKN=90^(@) seg KR bot hypotenuse LN :. by property of geometric mean, KR^(2)=LRxxsquare From (1) and (2) , we get MR^(2)=square :. MR=square . :. R is the midpoint of seg MK .

In the adjoining figure, angle DFE = 90^(@), FG bot ED if GD = 8, FG = 12 then find (i) EG (ii) FD (iii) EF.

In the adjoining figure, angleDFE=90^(@),FGbotED.IfGD=8,FG=12, find (i) EG (ii) FD, and (iii) EF

In the adjoining figure, angleDFE=90^(@),FGbotED.IfGD=8,FG=12, find (i) EG (ii) FD, and (iii) EF

In the adjoining figure , "seg " PQ||" seg " DE, A(DeltaPQF)=20 sq units . PF = 2 , then find A(squareDPQE) by completing the following activity.

In the figure, AD=17 , AB=10 , BC=15 . /_ABC=/_BCD=90^(@) seg AE bot side CD then find the length of (i) AE (ii) DE (iii) DC .