Two adjacent figures are represented by the isosceles trapezoid and the square.
What is the difference in the area of these two figures ?
EXAM TOMORROW ! and I dont understand
Hello cookiii
Welcome to Math Help Forum!The interior angles of a regular pentagon are each $\displaystyle 108^o$. So two of the sides, together with a diagonal, form an isosceles triangle whose angles are $\displaystyle 108^o, 36^o, 36^o$. Draw the line of symmetry of this triangle, and you'll see that the length of the side of the pentagon is $\displaystyle \frac{5}{\cos36^o}= 6.18$ cm (to 2 d.p.)
(If you need an exact answer, that's $\displaystyle 5(\sqrt5-1)$ - which I'll explain if necessary.)
Grandad
thank you so muchh ! but i have another question because i have an exam tomorroww and i don't understand !
Two adjacent figures are represented by the isosceles trapezoid and the square.
What is the difference in the area of these two figures ?
I made it on word document so it would be easier to answer
Hello cookiiiHave a look at the attached diagram. You'll see that I've drawn in the height of the trapezium and marked the lengths of the sides of the right-angled triangle.
Can you see why $\displaystyle b = \tfrac12(33-21)=6$ cm?
Then $\displaystyle a = \frac{6}{\cos 77^o}$ and $\displaystyle h = 6\tan 77^o$. (I'll leave you to work those out.)
The area of the trapezium is $\displaystyle \tfrac12(33 + 21)h=27h$ and the area of the square is $\displaystyle a^2$. I hope you can complete the question now.
Grandad