(Teaser) The total number of sides in two regular polygons is 13, and total number of diagonals is 25. How many sides are in each polygon?

T he two polygons have 8 sides and 5 sides.  

Step-by-step explanation:

A purely algebraic solution seems as if it would be very messy. Use trial and error, along with the formula for the number of diagonals in a n-sided polygon.

number of diagonals in a polygon with n sides: n(n-3)/2

10 sides and 3 sides: 10(7)/2 + 3(0)/2 = 35+0 = 35 nope

9 sides and 4 sides: 9(6)/2 + 4(1)/2 = 27+2 = 29 nope

8 sides and 5 sides: 8(5)/2 + 5(2)/2 = 20+5 = 25 yep!

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

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