Question

1. given: ptom ≡ cbad

grade 8: week of 9/22, math hw - day 2
directions: show all steps of your thinking and circle
problem #1 - unit 1 (mild)
describe how to move polygon a into the position of polygon b using a transformation.
rotate polygon a 90 degrees counter clockwise about a vertex and then translate it to match polygon b
problem #2 - unit 1 (mild)

Explanation:

Step1: Recall property of congruent polygons

Corresponding sides and angles of congruent polygons are equal. Given \(PTOM\cong CBAD\), corresponding - side and - angle relationships hold.

Step2: Find corresponding sides

For sides:

• Since \(PTOM\cong CBAD\), \(PT = CB\). Given \(CB = 41\mathrm{cm}\), so \(PT = 41\mathrm{cm}\).
• \(DA=MO\). Given \(MO = 50\mathrm{cm}\), so \(DA = 50\mathrm{cm}\).
• \(TO=BA\). Given \(BA = 79\mathrm{cm}\), so \(TO = 79\mathrm{cm}\).
• \(CD=MP\). Given \(MP = 112\mathrm{cm}\), so \(CD = 112\mathrm{cm}\).

Step3: Find corresponding angles

• \(m\angle B=m\angle T\). Given \(m\angle T = 144^{\circ}\), so \(m\angle B=144^{\circ}\).
• \(m\angle O=m\angle A\). Given \(m\angle A = 40^{\circ}\), so \(m\angle O = 40^{\circ}\).
• \(m\angle M=m\angle C\). Given \(m\angle C = 60^{\circ}\), so \(m\angle M = 60^{\circ}\).
• \(m\angle C=m\angle M = 60^{\circ}\).

Answer:

\(PT = 41\mathrm{cm}\)
\(DA = 50\mathrm{cm}\)
\(TO = 79\mathrm{cm}\)
\(CD = 112\mathrm{cm}\)
\(m\angle B = 144^{\circ}\)
\(m\angle O = 40^{\circ}\)
\(m\angle M = 60^{\circ}\)
\(m\angle C = 60^{\circ}\)