Method using no construction: 1. Triangle AFG similar to triangle AME (AAA) with side ratio 5:2 Hence area ratio 25:4 Hence area of triangle AFG = 25 2. In triangle AFG, triange ABG has same height as triangle BFG with base side ratio 4:1 Hence area of triangle ABG = 20, area of triangle BFG = 5 3. Triangle ADM is congruent to triangle ABG (SAS). Hence area of ADM = 20 4. Triangle GHC is congruent to triangle BFG (AAS). Hence area of GHC = 5 5. Area of square = 4 x area of triangle BFG = 4 x 20 = 80 6. Area of yellow region = area of square - area of ADM - area of ABG - area of GHC + area of AEM = 80 - 20 - 20 - 5 + 4 = 39.
@NixonChanMath6 ай бұрын
respect!
@Ernie-work6 ай бұрын
有用,想要program,謝謝
@NixonChanMath6 ай бұрын
多謝支持!請盡快到 bit.ly/Nx-GetProg 入紙拎Prog! 😃
@thomasshi97266 ай бұрын
Area DEGH=DMFH-MFGE=60-21=39
@camping57576 ай бұрын
我識method 1 加用面績與高做,都好快做完。
@NixonChanMath6 ай бұрын
叻
@camping57576 ай бұрын
@@NixonChanMath I used 10 seconds to finish it, and I always spent 20-25 seconds on finishing this question type.