The Total Surface Area (TSA) of a frustum of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The total surface area $(TSA)$ of a frustum of a cone consists of the lateral surface area plus the areas of the two circular bases.$1$. The later

Vedclass · Accepted Answer

$\pi l(r_{1}+r_{2})+\pi r_{1}^{2}+\pi r_{2}^{2}$

Vedclass · Answer

(A) The total surface area $(TSA)$ of a frustum of a cone consists of the lateral surface area plus the areas of the two circular bases.$1$. The lateral surface area of a frustum is given by $\pi l(r_{1} + r_{2})$,where $l$ is the slant height and $r_{1}, r_{2}$ are the radii of the two circular bases.$2$. The area of the top circular base is $\pi r_{1}^{2}$.$3$. The area of the bottom circular base is $\pi r_{2}^{2}$.$4$. Therefore,the total surface area $(TSA)$ is the sum of these three parts: $TSA = \pi l(r_{1} + r_{2}) + \pi r_{1}^{2} + \pi r_{2}^{2}$.

The Total Surface Area $(TSA)$ of a frustum of a cone is equal to $\ldots \ldots \ldots \ldots .$

Similar Questions

Total surface area of a cone exceeds its curved surface area by $\ldots \ldots \ldots . . .$

Total surface area of a cuboid measuring $15\,cm \times 12\,cm \times 10\,cm$ is $\ldots \ldots \ldots \,cm^{2}$.

The curved surface area of a cone is $85 \pi \, cm^{2}$. If its diameter is $10 \, cm$,its slant height is $\dots \dots \dots \dots \, cm$.

The ratio of the radii of two spheres is $3:4$. Then,the ratio of their volumes is $\ldots \ldots \ldots$

The volume of a sphere is $\ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module