Agalar help mee agalar yardım

510°

Step-by-step explanation:

1st.  solution.

Draw the perpendicular EP ⊥FB.

So from quadrilateral FAEP we find the∡AEP  ( sum of all angles in quadrilateral=360°)

∡AEP= 360°-∡FAE-∡AFP-∡FPE=360°-20°-90°-90°=160°

Draw CK ⊥FB

From triangle CKB we can find ∡BCK=180°-10°-90°=80°

From pentagon PEDCK (sum of PEDCK angles= 540°) we can find the sum of ∡PED+∡DCK= 540°-∡EPK-∡EDC-∡CKP= 540°-90°-90°-90°=270°

∡DEA+∡DCB=∡AEP+∡PED+∡DCK+∡KCB=160°+270°+80°=510°

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584