QUADRILATERALS are 4 sided polygons having parallelograms (square, rectangle, rhombus), trapezoids, and trapeziums as its members. The SUM OF ITS INTERIOR ANGLES is equal to 360°.
The above problem is a quadrilateral inside a circle, that is what we refer as CYCLIC QUADRILATERAL.
We know that a quadrilateral has a sum of its interior angles equal to 360°, that means we add all the angles of the quadrilateral FAIT.
AFT + FTI + TIA + FAI = 360°
But since we already know 2 of them
AFT = 75°
FTI = 98°
We can now solve for the sum of TIA and FAI
TIA + FAI = 360° - 75° - 98°
TIA + FAI = 187°
FAI = 187° - TIA (eq. 1)
As we notice, the quadrilateral is a TRAPEZIUM because it has no parallel sides, no equal sides, and no equal angles. Thus, everything is unequal.
Recall that one CHARACTERISTIC OF A TRAPEZIUM is that half the sum of two of its opposite angles is equal to half the sum of the remaining two opposite angles.
In the problem we have
(AFT + TIA)/2 = (FTI + FAI)/2
Simplifying both sides by eliminating 2
AFT + TIA = FTI + FAI (eq. 2)
Recall that we already have an equation for the sum of TIA and FAI.
Answers & Comments
Answer:
Step-by-step explanation:
QUADRILATERALS are 4 sided polygons having parallelograms (square, rectangle, rhombus), trapezoids, and trapeziums as its members. The SUM OF ITS INTERIOR ANGLES is equal to 360°.
The above problem is a quadrilateral inside a circle, that is what we refer as CYCLIC QUADRILATERAL.
We know that a quadrilateral has a sum of its interior angles equal to 360°, that means we add all the angles of the quadrilateral FAIT.
AFT + FTI + TIA + FAI = 360°
But since we already know 2 of them
AFT = 75°
FTI = 98°
We can now solve for the sum of TIA and FAI
TIA + FAI = 360° - 75° - 98°
TIA + FAI = 187°
FAI = 187° - TIA (eq. 1)
As we notice, the quadrilateral is a TRAPEZIUM because it has no parallel sides, no equal sides, and no equal angles. Thus, everything is unequal.
Recall that one CHARACTERISTIC OF A TRAPEZIUM is that half the sum of two of its opposite angles is equal to half the sum of the remaining two opposite angles.
In the problem we have
(AFT + TIA)/2 = (FTI + FAI)/2
Simplifying both sides by eliminating 2
AFT + TIA = FTI + FAI (eq. 2)
Recall that we already have an equation for the sum of TIA and FAI.
So substituting equation 1 to 2
75° + TIA = 98° + (187° - TIA)
2(TIA) = 98° + 187° - 75°
TIA = 105°
Therefore,
FAI = 187° - TIA = 82°
Thus, FAI=82° and TIA=105°
Verified answer
Question:-
Quadrilateral VIRU is inscribed in circles, M∠VIR = 85° and M∠IRU = 65°. Solve for x if M∠RUV = (3x + 5)°.
Options:
a) 22.
b) 26.
c) 28.
d) 30.
Given:-
M∠VIR = 85°.
M∠IRU = 65°.
M∠RUV = (3x + 5)°.
To solve:-
Solution:-
M∠VIR = 85°.
M∠IRU = 65°.
M∠RUV = (3x + 5)°.
Now,
∠RUV + ∠VIR = 180°.
⇒ 3x + 5 + 85 = 180°.
⇒ 3x + 90° = 180°.
⇒ 3x = 180° – 90° = 90°.
⇒ x =
[tex] \sf \large{ \cancel{ \frac{90}{3} }}[/tex]= 30°.
Answer:-
[tex]{ \boxed{ \sf \huge \color{red} \therefore \: x = 30 \degree.}}[/tex]
Hope you have satisfied. ⚘