## Yasminelove1 Group Title What is the approximate perimeter of the composite figure shown? Use three point one four for pi. one year ago one year ago

1. Yasminelove1 Group Title

2. ash2326 Group Title

Can you find the area of right triangle with base 5 mi and height 5 mi?

3. Yasminelove1 Group Title

no

4. ash2326 Group Title

Do you know the formula for area of right triangle and\or circle?

5. Yasminelove1 Group Title

A=1/2xbxh or A=bxh/2

6. Yasminelove1 Group Title

Circle Area = π × r2 Circumference = 2 × π × r r = radius

7. ash2326 Group Title

Good, you're right here for the triangle b=5 mi h=5 mi $\frac 1 2 b\times h$ Just plugin the values, you'll get the area of triangle

8. Yasminelove1 Group Title

wat do you mean by that

9. ash2326 Group Title

put b=5 and h=5, then solve it to find the area of triangle

10. Yasminelove1 Group Title

ok

11. Yasminelove1 Group Title

so ur saying 5x5x3.14

12. ash2326 Group Title

Nope, $\text{Area of figure}=\text{Area of triangle+ Area of circle}$ $\text{Area of triangle}=\frac 1 2 \times 5 \times 5\ mi^2$

13. Yasminelove1 Group Title

oh ok i get 12 1/2 tho

14. ash2326 Group Title

Yes, you're right. Now we need to find the area of semi circle, It's diameter is 7 mi Can you find the area?

15. Yasminelove1 Group Title

A semi-circle is half of a circle and the area of a circle of radius r is pi r2. We know that the area of a circle is: area of a circle = pi r2 A semi-circle is half of a circle. Therefore the area of a semi-circle will be: area of a semi-circle = 1/2 pi r2 We know that the radius is 9ft. Pi is approximately 3.14 or 22/7. Now you just have to plug in the numbers and you will find the area!

16. ash2326 Group Title

17. Yasminelove1 Group Title

First, let's look at the 2 dimensional region (base of solid) Each cross-section perpendicular to the x-axis is a "strip" of length e^x - 0 = e^x Now we can look at 3 dimensional solid Each cross-section perpendicular to the x-axis is a semi-circle with diameter = e^x So each cross-section has area = 1/2 (π r²) = 1/2 π (e^x/2)² = π/8 * e^(2x) Each very thin cross section of width dx has volume = π/8 * e^(2x) dx Integrating from x = -1 to x = 1, we get: Volume = ∫₋₁¹ π/8 * e^(2x) dx . . . . . . = π/16 * e^(2x) |₋₁¹ . . . . . . = π/16 (e^(2) - e^(-2)) . . . . . . = 1.4242647516

18. ash2326 Group Title

From where are you posting this??? It's not relevant here

19. amistre64 Group Title

@preetha we need to get shadow to clean up the database .....

20. amistre64 Group Title

?

22. amistre64 Group Title

please check the users profile page for some rather inappropriate content.

23. amistre64 Group Title

they also seem to have a habit of editing the content out of their questions :/