Alternative solution method: ΔADE and ΔABC are similar by pairs of corresponding sides being proportional, pair AD and AB and pair AE and AC, and the angles between them being equal, in this case, common angle DAE. Side AB is twice as long as side AD, so all dimensions of ΔABC are twice those of ΔADE. Therefore, line segment BC is twice as long as line segment DE. ΔGDE and ΔGCB are next proved to be similar with dimensions of ΔGCB double those of ΔGDE. Thus, both the base and height of ΔGCB are double those of ΔGDE and ΔGCB has 4 times the area, or 24 square units. Construct line segment parallel to BC through point G and label intersection with AB as point F and intersection with AC as point H. ΔAFH and ΔABC are similar, with dimensions ΔAFH being 2/3 those of ΔABC. Height ΔGBC, when BC is considered to be its base, is the difference of heights for ΔAFH and ΔABC, or 1/3 the height of ΔABC, when BC is considered to be its base. Thus, ΔABC has 3 times the height of ΔGBC, with common base BC. Therefore ΔABC has 3 times the area of ΔGBC, or 72 square units.
Jam nje mesuese matematike ne pension nga Shqiperia. Po jap nje menyre tjeter zgjidhjeje per kete probleme.Meqe DE eshte vije e mesme atehere S ADE=S DEB=S ADC= 1/4 *S ABC Meqe BE dhe CD jane mesore, nga vetia e mesoreve dihet qe BG=2GE S BDG= 2*S DEG=2*6 = 12 ATEHERE S BDE =18 PRA S ABC= 18*4 =72
Great 👍🏿
Alternative solution method: ΔADE and ΔABC are similar by pairs of corresponding sides being proportional, pair AD and AB and pair AE and AC, and the angles between them being equal, in this case, common angle DAE. Side AB is twice as long as side AD, so all dimensions of ΔABC are twice those of ΔADE. Therefore, line segment BC is twice as long as line segment DE. ΔGDE and ΔGCB are next proved to be similar with dimensions of ΔGCB double those of ΔGDE. Thus, both the base and height of ΔGCB are double those of ΔGDE and ΔGCB has 4 times the area, or 24 square units. Construct line segment parallel to BC through point G and label intersection with AB as point F and intersection with AC as point H. ΔAFH and ΔABC are similar, with dimensions ΔAFH being 2/3 those of ΔABC. Height ΔGBC, when BC is considered to be its base, is the difference of heights for ΔAFH and ΔABC, or 1/3 the height of ΔABC, when BC is considered to be its base. Thus, ΔABC has 3 times the height of ΔGBC, with common base BC. Therefore ΔABC has 3 times the area of ΔGBC, or 72 square units.
Jam nje mesuese matematike ne pension nga Shqiperia. Po jap nje menyre tjeter zgjidhjeje per kete probleme.Meqe DE eshte vije e mesme atehere S ADE=S DEB=S ADC=
1/4 *S ABC
Meqe BE dhe CD jane mesore, nga vetia e mesoreve dihet qe BG=2GE
S BDG= 2*S DEG=2*6 = 12
ATEHERE S BDE =18
PRA S ABC= 18*4 =72