9514 1404 393
Answer:
1. HA is equivalent to AAS when the triangle is a right triangle.
2. AM = BM, so the triangles are congruent by HL. CPCTC
3. The triangles are congruent by HL. CPCTC
Step-by-step explanation:
1. The acute angle of the triangle together with the right angle comprise two angles of the triangle. When two corresponding angles and a corresponding side (the side opposite the right angle) are congruent, the right triangles are congruent by the AAS theorem. (This can be referred to as the HA theorem.)
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2. CM = DM; MA = MB; ∠A = ∠C = 90°, so all of the requirements for the HL theorem are met. ΔCMA ≅ ΔDMB, so AC ≅ BD by CPCTC.
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3. TS = TV, TR = TR, ∠S = ∠V = 90°, so all requirements for the HL theorem are met. ΔTSR ≅ ΔTVR, so RS ≅ RV by CPCTC.
It has not been indicated whether the figure in the questions is a triangle or a quadrilateral. Irrespective of the shape, this can be solved. The two possible shapes and angles have been indicated in the attached image.
Now, from the information given we can infer that there is a line BD that cuts angle ABC in two parts: angle ABD and angle DBC
⇒ Angle ABC = Angle ABD + Angle DBC
Also, we know that angle ABC is 1 degree less than 3 times the angle ABD, and that angle DBC is 47 degree
Let angle ABD be x
⇒ Angle ABC = 3x-1
Also, Angle ABC = Angle ABD + Angle DBC
Substituting the values in the above equations
⇒ 3x-1 = x+47
⇒ 2x = 48
⇒ x = 24
So angle ABD = 24 degree, and angle ABC = 3(24)-1 = 71-1 = 71 degree
Answer:
$2/min
Step-by-step explanation:
14 ÷ 7 = 2
Therefore, $2 per minute.
Answer:
C = 6
D = 10
Step-by-step explanation:
C + 5 = 11
C + 5 -5 = 11 -5
C = 6
D + 4 = 14
D + 4 -4 = 14 -4
D = 10