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2.2.1 Polygons II, PT3 Practice

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2.2.1 Polygons II, PT3 Practice

Question 1:
Diagram below shows a pentagon PQRST. TPU and RSV are straight lines.
Find the value of x.

Solution:
Sum of interior angles of a pentagon =( 5−2 )× 180 o =3× 180 o = 540 o ∠TSR= 180 o − 70 o           = 110 o ∠TPQ= 180 o − 85 o           = 95 o x o = 540 o −( 110 o + 105 o + 115 o + 95 o )     = 540 o − 425 o     = 115 o    x=115


Question 2:
In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines.
Find the value of x + y.

Solution:
∠ Q P U = 180 o − 160 o = 20 o Reflex ∠ P U T = 360 o − 80 o = 280 o ∠ U T S = 180 o − 120 o = 60 o ∠ T S R = 180 o − 35 o = 145 o Sum of interior angles of a hexagon = ( 6 − 2 ) × 180 o = 720 o x o + y o + 145 o + 60 o + 280 o + 20 o = 720 o x o + y o = 720 o − 505 o = 215 o x + y = 215


Question 3:
Diagram below shows a regular hexagon PQRSTU. PUV is a straight line.
Find the value of x + y.

Solution:
Size of each interior angle of a regular hexagon = ( 6 − 2 ) × 180 o 6 = 120 o x o = 180 o − 120 o 2 = 30 o y o = 180 o − 120 o = 60 o x o + y o = 30 o + 60 o = 90 o x + y = 90


Question 4:
In the diagram below, KLMNP is a regular pentagon. LKS and MNQ are straight lines.
Find the value of x + y.
 
Solution:
Size of each interior angle of a regular pentagon = ( 5 − 2 ) × 180 o 5 = 108 o ∠ P K S = ∠ P N Q = 180 o − 108 o = 72 o Reflex angle ∠ K P N = 360 o − 108 o = 252 o

Sum of interior angles of a hexagon = ( 6 − 2 ) × 180 o = 720 o ∴ x o + y o + 72 o + 252 o + 72 o + 100 o = 720 o x o + y o = 720 o − 496 o = 224 o x + y = 224


Question 5:
In Diagram below, PQR is an isosceles triangle and PRU is a straight line.
Find the value of x + y.

Solution:
x o = 180 o − 20 o − 20 o = 140 o ∠ P R S = 180 o − 110 o = 70 o y o + 85 o + 75 o + 70 o = 360 o y o + 230 o = 360 o y o = 130 o x o + y o = 140 o + 130 o = 270 o x + y = 270