The anwser is 6 possible combinations.
What is the area of rectangle PQRS with vertices at P(0, 2), Q(0, 10), R(4, 10), and S(4, 2)?
1 answer:
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Answer:
The option is C i.e 115°, 65°. proof is given below.
Step-by-step explanation:
Given:
ABCD is a quadrilateral.
m∠ A = 100 + 5x
m∠ B = 77 - 4y
m∠ C = 106 + 3x
m∠ D = 47 + 6y
To Prove:
ABCD is a parallelogram if opposing angles are congruent by finding the measures of angles.
m∠ A = m∠ C and
m∠ B = m∠ D
Proof:
ABCD is a quadrilateral and is a parallelogram if opposing angles are congruent.
∴ m∠ A = m∠ C
On substituting the given values we get
∴ 100 + 5x = 106 +3x
∴
m∠ A = 100 + 5x = 100 + 5 × 3 =100 + 15 = 115°
m∠ C = 106 + 3x = 106 + 3 ×3 =106 + 9 = 115°
∴ m∠ A = m∠ C = 115°
Similarly,
∴ m∠ B = m∠ D
77 - 4y = 47 + 6y
10y = 77 - 47
10y =30
∴
m∠ B = 77 - 4y =77 - 4 × 3 = 77 - 12 = 65°
m∠ D = 47 + 6y = 47 + 6 × 3 = 47 + 18 = 65°
∴ m∠ B = m∠ D = 65°
Therefore the option is C i.e 115°, 65°