Answer:
Explanation:
Area Of Triangle And Rectangle
Given a triangle of base b and height h (perpendicular to b), the area can be computed by
![\displaystyle A=(bh)/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/pv5r7k96kvfdog6gv4tqbwqc6848z48qc6.png)
A rectangle of the same dimensions has an area of
![A=bh](https://img.qammunity.org/2021/formulas/mathematics/high-school/qu2hlsd4h6dsol8g456d6dwuk4nd06d6gz.png)
We have a triangle of base 3 cm and a height of 4 cm. Its area is
![\displaystyle A=((3)(4))/(2)=6\ cm^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/kbl95tt6gccxkw9tmuo4gh51axm7qyl943.png)
That triangle moves upward at 4 cm per second for 3 seconds. It means that the triangle 'sweeps' upwards three times its height forming a rectangle of base 3 cm and height 12 cm. The area of the swept area is
![A=(3)(12)=36\ cm^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/8qldcetwwi213njq6biv2gzmp6585le2s3.png)
The triangle stays in the top of this rectangle, so its area is part of the total swept area:
Total swept area = 6 + 36 =
![42\ cm^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/s040ayxibx307ptfvmj9u6oktziworvpzu.png)