
The semicircle shown at left has center X and diameter W Z. The radius XY of the semicircle has length 2. The chord Y Z has length 2. What is the area of the shaded sector formed by obtuse angle WXY?

RADIUS = 2
CHORD = 2
RADIUS --> XY , XZ , WX
( BEZ THEY TOUCH CIRCUMFERENCE OF THE CIRCLES AFTER STARTING FROM CENTRE OF THE CIRCLE)

THE AREA OF THE SHADED SECTOR FORMED BY OBTUSE ANGLE WXY.

AREA COVERED BY THE ANGLE IN A SEMI SPHERE


Total Area Of The Semi Sphere:-

Area Under Unshaded Part .
Given a triangle with each side 2 units.
This proves that it's is a equilateral triangle which means it's all angles r of 60° or π/3 Radian
So AREA :-


Total Area - Area Under Unshaded Part


Answer:
42
Step-by-step explanation:
divide the total by the cost of each invite
Cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.
<u>Solution:
</u>
Need to calculate
and then multiply the result by ![\sqrt[3]{14903}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D)
Let us first evaluate ![\sqrt[3]{1728}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1728%7D)
![\Rightarrow \sqrt[3]{1728}=\sqrt[3]{12 \times 12 \times 12}=12](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csqrt%5B3%5D%7B1728%7D%3D%5Csqrt%5B3%5D%7B12%20%5Ctimes%2012%20%5Ctimes%2012%7D%3D12)
As need to multiply 12 by ![\sqrt[3]{14903}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D)
![\Rightarrow 12 \times \sqrt[3]{14903}](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D)
On solving
, we get
![\sqrt[3]{14903}=24.608](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D%3D24.608)
![\Rightarrow 12 \times \sqrt[3]{14903}=12 \times 24.608=295.306](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D%3D12%20%5Ctimes%2024.608%3D295.306)
Hence cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.