A pentagon has interior angles of measures x ,(x-4),(2x-10),(2x-5) (2x+15). find the measure of each angle
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Answered by
46
sum of interior angles of a pentagon = 540
⇔ x + (x-4) + (2x-10) + (2x-5) + (2x+15) = 540
⇔ 8x - 4 = 540
⇔ 8x = 540+4 = 544
⇔ x = 544/8 = 68
Angles are:
x = 68
x-4 = 68-4 = 64
2x-10 = 2*68 -10 = 136 -10= 126
2x - 5 = 2*68 -5 = 136-5 = 131
2x +15 = 2*68+15 = 136+15 = 151
Angles are 68,64,126,131 and 151. All are in degrees.
⇔ x + (x-4) + (2x-10) + (2x-5) + (2x+15) = 540
⇔ 8x - 4 = 540
⇔ 8x = 540+4 = 544
⇔ x = 544/8 = 68
Angles are:
x = 68
x-4 = 68-4 = 64
2x-10 = 2*68 -10 = 136 -10= 126
2x - 5 = 2*68 -5 = 136-5 = 131
2x +15 = 2*68+15 = 136+15 = 151
Angles are 68,64,126,131 and 151. All are in degrees.
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Answered by
32
We know that Sum of interior angles of a pentagon is 540 degree
ATQ x + (x-4) + (2x-10) + (2x-5) + (2x+15) = 540
⇒ 8x - 4 = 540
⇒ 8x = 544
⇒ x = 544/8 = 68
So substituting value of x we get angles as
1. x = 68 degree
2. x-4 = 68-4 = 64 degree
3. 2x-10 = 2*68 -10 = 126 degree
4. 2x - 5 = 2*68 -5 = 131 degree
5. 2x +15 = 2*68+15 = 151 degree
ATQ x + (x-4) + (2x-10) + (2x-5) + (2x+15) = 540
⇒ 8x - 4 = 540
⇒ 8x = 544
⇒ x = 544/8 = 68
So substituting value of x we get angles as
1. x = 68 degree
2. x-4 = 68-4 = 64 degree
3. 2x-10 = 2*68 -10 = 126 degree
4. 2x - 5 = 2*68 -5 = 131 degree
5. 2x +15 = 2*68+15 = 151 degree
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