1. ## Trigo

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

2. Hello cookiii

Welcome to Math Help Forum!
Originally Posted by cookiii
What is the measurement of the dimensions regular pentagon if one of the diagonals measures 10 cm? URGENTTT
The interior angles of a regular pentagon are each $108^o$. So two of the sides, together with a diagonal, form an isosceles triangle whose angles are $108^o, 36^o, 36^o$. Draw the line of symmetry of this triangle, and you'll see that the length of the side of the pentagon is $\frac{5}{\cos36^o}= 6.18$ cm (to 2 d.p.)

(If you need an exact answer, that's $5(\sqrt5-1)$ - which I'll explain if necessary.)

3. ## Trigoo

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

4. Hello cookiii
Originally Posted by cookiii
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
Have 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 $b = \tfrac12(33-21)=6$ cm?

Then $a = \frac{6}{\cos 77^o}$ and $h = 6\tan 77^o$. (I'll leave you to work those out.)

The area of the trapezium is $\tfrac12(33 + 21)h=27h$ and the area of the square is $a^2$. I hope you can complete the question now.