I do not know shear formation but can give help.
starting with AB = 2. PQ =6 Area CPQ = 18 (9*2)
altofACBonAB =2 2*2*1/2 =2
Alt of CPQ on PQ = 6 6*6*1/2 =18
PA = 8
Need a hint for this one. I feel I am missing something simple.
Triangle ABP is mapped onto triangle ABQ by a shear. The lines BP and AQ intersect at C. Given that BC:CP = 1:3, and that the area of triangle ACB is 2 sq. cm, find the area of,
(a) Triangle ACP,
(b) Triangle CPQ
(c) Quadrilateral ABQP
I have solved part B, using similar triangles, and ratio of area to square of corresponding sides, to get 18 sq cm.
But I am foggy on the parts (b) and (c). I see that Triangle ABP and Triangle ABQ have the same area, under a shear transform with shear factor 3. How to proceed further?