Calculate the area of the parallelogram with the given vertices. (0, 0), (2, 6), (11, 8), (9, 2)

Question

answer to example question

Answer 1

I need an helping hands in real analysis please can u help me out.

 

regrds

Olayinka

Answer 2

Label each vertex: O(0,0), A(2,6), B(11,8), and C(9,2).

Extend AB to the left that crosses y-axis at point D.

Draw a vertical line (x=9) passing thru vertex C that crosses AB at point E, and x-axis at point F(9,0).

Thus, OD∥EF and DE∥OC, so quadrilateral ODEC is a parallelogram as well.

CE⊥OF, so OF is the altitude of parallelogram ODEC, and OF=9 units.

In ΔOAD and ΔCBE, ∠OAD=∠CBE,∠ODA=∠CEB, and AO=BC, so ΔOAD≡ΔCBE (AAS) 

Thus, area of parallelogram OABC = area of parallelogram ODEC

The slope of line AB is: m=(8-6)/(11-2)=2/9, so the y-intersept of line AB is 6-(2/9)x2=50/9.

Thus, OD=50/9 units

And area of parallelogram ODEC=OD (base) x OF (altitude)=(50/9)x 9=50 (unit²) 

The answer is: area of parallelogrm OABC is 50 (unit²)    

Answer 3

Label each vertex: O(0,0), A(2,6), B(11,8) and C(9,2).

Draw a horizontal (y=8) passing thru B that crosses y-axis (x=0) at point D(0.8).

Draw a vertcal (x=11) passing thru B that crosses x-axis (y=0) at point E(11,0).

Connect A toD, and C to E.

In ΔOAD and ΔBAD, let the foot of altitude from A to OD be G(0,6), and the one from A to BD be H(2,8).

In ΔOAD and ΔBCE, OD∥BE and OA∥BC, so ∠AOD=∠CBE.   OD=BE and OA=BC

Thus, ΔOAD≡ΔBCE (SAS)   In the same manner, ΔBAD≡ΔOCE (SAS)  

Therfore, area of parallelogram OABC=area of rectangle ODBE - 2(ΔOAD + ΔBAD)

Here, area of rectangle ODBE=ODxDB=8x11=88 (unit²),

ΔOAD=½xODxAG=½x8x2=8, and ΔBAD=½xBDxAH=½x11x2=11   

Therefore, area of parallelogram OABC=88 - 2(8+11)=88 - 38=50 (unit²)

The answer is: area of parallelogram OABC is 50 (unit²)