Answer:
This cannot be simplified any further. That is the simplest you'll get.
Answer:
someone will be with you in a moment
Step-by-step explanation:
These angles should add to 180 degrees because the angles are on a straight line so you should use the formula x=180-90-38. Therefore, x should be 52 degrees

- Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>
- To find - <u>Area </u><u>of </u><u>trapezium</u>
Refer the figure attached ~
In the given figure ,
AB = 25 cm
BC = AD = 15 cm
CD = 13 cm
<u>Construction</u><u> </u><u>-</u>

Now , we can clearly see that AECD is a parallelogram !
AE = CD = 13 cm
Now ,

Now , In ∆ BCE ,

Now , by Heron's formula

Also ,

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

hope helpful :D
Answer:
The volume of the toy is 
Step-by-step explanation:
step 1
Find the volume of the hemisphere
The volume of the hemisphere is given by the formula

In this problem, the wide of the toy is equal to the diameter of the hemisphere
so

----> the radius is half the diameter
substitute

step 2
Find the volume of the cone
The volume of the cone is given by

we know that
The radius of the cone is the same that the radius of the hemisphere
so

The height of the cone is equal to subtract the radius of the hemisphere from the height of the toy

substitute the given values

step 3
Find the volume of the toy
we know that
The volume of the toy, is equal to the volume of the cone plus the volume of the hemisphere.
so


assume

