The ratio between the number of sides of two regular polygons is 3:4 and the ratio between the sum of their interior angle is 2:3. Find the no. of sides in each polygon.
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Hey there,
let
s1 = the sum of the angles of the first polygon.
s2 = the sum of the angles of the second polygon.
n1 = the number of sides of the first polygon.
n2 = the number of sides of the second polygon.
the sum of the angles of a polygon is equal to 180 * (n-2)
for the first polygon,
s1 = 180 * (n1 - 2)
for the second polygon,
s2 = 180 * (n2 - 2)
the ratio of the sum of the angles of the polygons is 2/3.
this means that
s1/s2 = 2/3
s1 = (2 * s2) / 3.
since
s1 is equal to 180 * (n1 - 2),
and
since s2 is equal to 180 * (n2 - 2),
substitute in the equation for s1 to get:
s1 = (2 * s2) / 3 becomes:
180 * (n1 - 2) = (2 * 180 * (n2 - 2)) / 3.
divide both sides of this equation by 180
(n1 - 2) = (2 * 180 * (n2 - 2)) / (3 * 180).
since 180 in the numerator and denominator cancel out, the equation becomes:
(n1 - 2) = (2 * (n2 - 2)) / 3.
add 2 to both sides of this equation to get:
n1 = (2 * (n2 - 2)) / 3 + 2.
since 2 is equal to 6/3, the equation becomes:
n1 = (2 * (n2 - 2)) / 3 + 6/3.
you can now consolidate under the common denominator to get:
n1 = (2 * (n2 - 2) + 6) /3.
simplify by expanding the distribution to get:
n1 = (2 * n2 - 4 + 6) / 3.
combine like terms to get:
n1 = (2 * n2 + 2) / 3.
you are given that the ratio of n1/n2 = 3/4
from that ratio, you can solve for n1 to get:
n1 = (3 * n2) / 4
you can now replace n2 with its equivalent to get:
n1 = (2 * n2 + 2) / 3 becomes:
(3 * n2) / 4 = (2 * n2 + 2) / 3.
multiply both sides of this equation by 12 to get:
3 * (3 * n2) = (2 * n2 + 2) * 4.
simplify to get:
9 * n2 = 8 * n2 + 8
subtract 8 * n2 from both sides of this equation to get:
n2 = 8
since the ratio of n1 / n2 = 3/4, then the equation of:
n1 / n2 = 3/4 becomes:
n1 / 8 = 3/4
multiply both sides of this equation by 8 to get:
n1 = (3 * 8) / 4 which becomes:
n1 = 24 / 4 which becomes:
n1 = 6
you now have:
n1 = 6
n2 = 8
the ratio of n1/n2 is equal to 6/8 which is equal to 3/4.
s1 = 180 * (n1 - 2) = 180 * (6 - 2) = 180 * 4 = 720
s2 = 180 * (n2 - 2) = 180 * (8 - 2) = 180 * 6 = 1080
the ratio of s1/s2 is equal to 720/1080 which is equal to 2/3.
the ratio of the number of sides of the two polygons is equal to 3/4 as indicated in the problem statement.
the ratio of the sum of the angles of the two polygons is equal to 2/3 as indicated in the problem statement.
the solution to the problem is that the smaller polygon has 6 sides and the larger polygon has 8 sides.
Hope this helps!
let
s1 = the sum of the angles of the first polygon.
s2 = the sum of the angles of the second polygon.
n1 = the number of sides of the first polygon.
n2 = the number of sides of the second polygon.
the sum of the angles of a polygon is equal to 180 * (n-2)
for the first polygon,
s1 = 180 * (n1 - 2)
for the second polygon,
s2 = 180 * (n2 - 2)
the ratio of the sum of the angles of the polygons is 2/3.
this means that
s1/s2 = 2/3
s1 = (2 * s2) / 3.
since
s1 is equal to 180 * (n1 - 2),
and
since s2 is equal to 180 * (n2 - 2),
substitute in the equation for s1 to get:
s1 = (2 * s2) / 3 becomes:
180 * (n1 - 2) = (2 * 180 * (n2 - 2)) / 3.
divide both sides of this equation by 180
(n1 - 2) = (2 * 180 * (n2 - 2)) / (3 * 180).
since 180 in the numerator and denominator cancel out, the equation becomes:
(n1 - 2) = (2 * (n2 - 2)) / 3.
add 2 to both sides of this equation to get:
n1 = (2 * (n2 - 2)) / 3 + 2.
since 2 is equal to 6/3, the equation becomes:
n1 = (2 * (n2 - 2)) / 3 + 6/3.
you can now consolidate under the common denominator to get:
n1 = (2 * (n2 - 2) + 6) /3.
simplify by expanding the distribution to get:
n1 = (2 * n2 - 4 + 6) / 3.
combine like terms to get:
n1 = (2 * n2 + 2) / 3.
you are given that the ratio of n1/n2 = 3/4
from that ratio, you can solve for n1 to get:
n1 = (3 * n2) / 4
you can now replace n2 with its equivalent to get:
n1 = (2 * n2 + 2) / 3 becomes:
(3 * n2) / 4 = (2 * n2 + 2) / 3.
multiply both sides of this equation by 12 to get:
3 * (3 * n2) = (2 * n2 + 2) * 4.
simplify to get:
9 * n2 = 8 * n2 + 8
subtract 8 * n2 from both sides of this equation to get:
n2 = 8
since the ratio of n1 / n2 = 3/4, then the equation of:
n1 / n2 = 3/4 becomes:
n1 / 8 = 3/4
multiply both sides of this equation by 8 to get:
n1 = (3 * 8) / 4 which becomes:
n1 = 24 / 4 which becomes:
n1 = 6
you now have:
n1 = 6
n2 = 8
the ratio of n1/n2 is equal to 6/8 which is equal to 3/4.
s1 = 180 * (n1 - 2) = 180 * (6 - 2) = 180 * 4 = 720
s2 = 180 * (n2 - 2) = 180 * (8 - 2) = 180 * 6 = 1080
the ratio of s1/s2 is equal to 720/1080 which is equal to 2/3.
the ratio of the number of sides of the two polygons is equal to 3/4 as indicated in the problem statement.
the ratio of the sum of the angles of the two polygons is equal to 2/3 as indicated in the problem statement.
the solution to the problem is that the smaller polygon has 6 sides and the larger polygon has 8 sides.
Hope this helps!
smartcow1:
sorry it's very long
could it be quite shorter
Answered by
6
Step-by-step explanation:
The ratio of number of sides of two regular polygons is 3:4 and the ratio of measures of their each interior angle is 8:9. What is the sum of the number of diagonals of both the polygons equal to?
Ratio of sides of two regular polygons = 3 : 4
Let sides of first polygon = 3n
and sides of second polygon = 4n
Sum of interior angles of first polygon
= (2 × 3n – 4) × 90° = (6n – 4) × 90°
And sum of interior angle of second polygon
= (2 × 4n – 4) × 90° = (8n – 4) × 90°
∴ ((6n – 4) × 90°)/((8n – 4) × 90°) = 2/3
⇒ (6n – 4)/(8n – 4) = 2/3
⇒ 18n – 12 = 16n – 8
⇒ 18n – 16n = -8 + 12
⇒ 2n = 4
⇒ n = 2
∴ No. of sides of first polygon
= 3n = 3 × 2 = 6
And no. of sides of second polygon
= 4n = 4n × 2 = 8
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