First, we find the area of the circle.
A = pi * r^2
A = pi * (1 cm)^2
A = pi cm^2
The area of the circle is pi cm^2.
The area of the triangle is also pi cm^2.
Now we use the area of a triangle.
A = (1/2)bh
(1/2)bh = A
(1/2)(3 cm)h = pi cm^2
(3 cm)h = 2pi cm^2
h = (2/3)pi cm
The exact height is 
If you want an approximate height, then it is 2.09 cm.
1. 2
2. 2
3. 5
4. 1
5. 8
6. 2
7. -2
8. 2
9. 3
10. -1
11. -7
12. -2
Answer:
Step-by-step explanation:
Answer:
a=k1![\sqrt[3]{B}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7BB%7D)
B=125
Step-by-step explanation:
Given :
a=3
B=64
According to question
a ∝
therefore
a=k1 ........Eq(1)
K1=![\frac{a}{\sqrt[3]{B} }](https://tex.z-dn.net/?f=%5Cfrac%7Ba%7D%7B%5Csqrt%5B3%5D%7BB%7D%20%7D)
......Eq(2)
Putting the value of a and B we get in Eq(2) we get
Putting the value of k1 in Eq(1)
.......................Eq(3)
putting the value of a=15/4 IN Eq(3) we get
![\frac{15}{4}\ =\frac{3}{4} \sqrt[3]{B} \\\\\sqrt[3]{B}\ =\ 5\\Cubing\ both\ side\ we\ get\\B=125](https://tex.z-dn.net/?f=%5Cfrac%7B15%7D%7B4%7D%5C%20%3D%5Cfrac%7B3%7D%7B4%7D%20%5Csqrt%5B3%5D%7BB%7D%20%20%5C%5C%5C%5C%5Csqrt%5B3%5D%7BB%7D%5C%20%3D%5C%205%5C%5CCubing%5C%20%20both%5C%20%20side%5C%20%20we%5C%20%20get%5C%5CB%3D125)
