In the instructions it states ABC is reflected over line BA. So the rigid transformation is "a reflection across the line containing BA"
Answer:
![\large\boxed{A_\triangle=25\sqrt{15}\ cm^2}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7BA_%5Ctriangle%3D25%5Csqrt%7B15%7D%5C%20cm%5E2%7D)
Step-by-step explanation:
Look at the picture.
The formula of an area of a triangle:
![A_\triangle=\dfrac{bh}{2}](https://tex.z-dn.net/?f=A_%5Ctriangle%3D%5Cdfrac%7Bbh%7D%7B2%7D)
<em>b</em><em> - base</em>
<em>h</em><em> - height</em>
<em />
We need a length of a height.
Use the Pythagorean theorem:
![leg^2+leg^2=hypotenuse^2](https://tex.z-dn.net/?f=leg%5E2%2Bleg%5E2%3Dhypotenuse%5E2)
We have:
![leg=5,\ leg=h,\ hypotenuse=20](https://tex.z-dn.net/?f=leg%3D5%2C%5C%20leg%3Dh%2C%5C%20hypotenuse%3D20)
Substitute:
![5^2+h^2=20^2](https://tex.z-dn.net/?f=5%5E2%2Bh%5E2%3D20%5E2)
<em>subtract 25 from both sides</em>
![h^2=375\to h=\sqrt{375}\\\\h=\sqrt{(25)(15)}\\\\h=\sqrt{25}\cdot\sqrt{15}\\\\h=5\sqrt{15}\ cm](https://tex.z-dn.net/?f=h%5E2%3D375%5Cto%20h%3D%5Csqrt%7B375%7D%5C%5C%5C%5Ch%3D%5Csqrt%7B%2825%29%2815%29%7D%5C%5C%5C%5Ch%3D%5Csqrt%7B25%7D%5Ccdot%5Csqrt%7B15%7D%5C%5C%5C%5Ch%3D5%5Csqrt%7B15%7D%5C%20cm)
Calculate the area:
![A_\triangle=\dfrac{(10)(5\sqrt{15})}{2}=\dfrac{50\sqrt{15}}{2}=25\sqrt{15}\ cm^2](https://tex.z-dn.net/?f=A_%5Ctriangle%3D%5Cdfrac%7B%2810%29%285%5Csqrt%7B15%7D%29%7D%7B2%7D%3D%5Cdfrac%7B50%5Csqrt%7B15%7D%7D%7B2%7D%3D25%5Csqrt%7B15%7D%5C%20cm%5E2)
Answer:
like this?
Step-by-step explanation:
Answer:
The answers are H and A, let me know in a comment if you want me to explain it
Your answer should be 12+16x