Answer:
thanks
Step-by-step explanation:
Answer:
Step-by-step explanation: its c. x+ 5
Since you haven't provided the data to answer the problem, I have my notes here that might guide you solve the problem on your own:
Now, consider a triangle that’s graphed in the coordinate plane. You can always use the distance formula, find the lengths of the three sides, and then apply Heron’s formula. But there’s an even better choice, based on the determinant of a matrix.
Here’s a formula to use, based on the counterclockwise entry of the coordinates of the vertices of the triangle
(x1<span>, </span>y1), (x2<span>, </span>y2), (x3<span>, </span>y3<span>) or (2, 1), (8, 9), (1, 8): </span>A<span> = </span>x1y2<span> + </span>x2y3<span> + </span>x3y1<span> – </span>x1y3<span> – </span>x2y1<span> – </span>x3y2<span>.</span>
Answer:
900% or 9 times the size
Step-by-step explanation:
Step 1: Problem
Step 2: Concept
The area of the figure is the sum of the area of a triangle and a semi-circle.
Step 3: Method
Perimeters
Height of the triangle = 15cm
Base of the triangle = 2 x Radius of the semi-circle = 2 x 8 = 16cm
Radius r = 8cm
![\begin{gathered} \text{Area of the triangle = }\frac{Base\text{ x Height }}{2} \\ \\ =\text{ }\frac{16\text{ x 15}}{2} \\ =\text{ }\frac{240}{2} \\ =120cm^2 \\ \text{Area of the semi-circle = }\pi r^2 \\ \pi\text{ = 3.14} \\ =\text{ 3.14 }\times8^2 \\ =\text{ 3.14 x 64} \\ =200.96cm^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7BArea%20of%20the%20triangle%20%3D%20%7D%5Cfrac%7BBase%5Ctext%7B%20x%20Height%20%7D%7D%7B2%7D%20%5C%5C%20%20%5C%5C%20%3D%5Ctext%7B%20%7D%5Cfrac%7B16%5Ctext%7B%20x%2015%7D%7D%7B2%7D%20%5C%5C%20%3D%5Ctext%7B%20%7D%5Cfrac%7B240%7D%7B2%7D%20%5C%5C%20%3D120cm%5E2%20%5C%5C%20%5Ctext%7BArea%20of%20the%20semi-circle%20%3D%20%7D%5Cpi%20r%5E2%20%5C%5C%20%5Cpi%5Ctext%7B%20%3D%203.14%7D%20%5C%5C%20%3D%5Ctext%7B%203.14%20%20%20%7D%5Ctimes8%5E2%20%5C%5C%20%3D%5Ctext%7B%203.14%20x%2064%7D%20%5C%5C%20%3D200.96cm%5E2%20%5Cend%7Bgathered%7D)
Step 4: Final answer
Area of the figure = Area of triangle + Area of semi-circle
= 120 + 200.96
= 320.96