Thread: Shear transformation problem

Hey,

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?

Thanks.

Hi mathguy80,
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



bjh

Hey bjhopper,

I don't follow how you arrived at AB = 2, Can you please clarify? Thanks.

Hi mathguy80,
Given area ofACB =2.Altitude of ACB on AB = a
a*AB *1/2 =2 a*AB =4 a and AB are each 2


bjh

Hi bjhopper,

Sorry if I am really dense today. But I still don't follow how, given altitude * AB = 4, leads to altitude and AB both equal to 2.

Couldn't AB = 1, altitude = 4, (ie:- 1x4 = 4), or vice versa also satisfy the same condition?

Hi math guy 80,
You are right AB =1 or 2.In either case all required areas will be the same.


bjh

are the angles PAB and APQ right angles, or are they unknown?