Home > Surface Areas and Volumes > CBSE_NCERT_EXEMPLAR_CH11 > b5c71268-b938-5409-b6d3-c7bf35e5749f
Not the exact question you were looking for?

Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.

Surface Areas and Volumes
NCERT Exemplar Ch 11
CBSE_NCERT_EXEMPLAR_CH11
Grade 10

Question:

A toy is in the form of a cone of radius $3.5\text{ cm}$ mounted on a hemisphere of same radius. The total height of the toy is $15.5\text{ cm}$. Find the total surface area of the toy.

Step-by-Step Solution

Key Concept: Radius $r = 3.5\text{ cm}$. Height of cone $h = 15.5 - 3.5 = 12\text{ cm}$. Slant height $l = \sqrt{12^2 + 3.5^2} = 12.5\text{ cm}$. TSA $= \pi r l + 2\pi r^2$.
Stepwise Solution:

Height of cone $h = 15.5 - 3.5 = 12\text{ cm}$. [0.5 Mark]

Slant height $l = \sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5\text{ cm}$. [0.5 Mark]

TSA of toy $= \pi r l + 2\pi r^2 = \pi r(l + 2r) = \dfrac{22}{7} \times 3.5 \times (12.5 + 7) = 11 \times 19.5 = 214.5\text{ cm}^2$. [1.0 Mark]

Marking Scheme:

• Finding cone height $12\text{ cm}$ and slant height $l = 12.5\text{ cm}$: 1.0 Mark
• Evaluating TSA $= 214.5\text{ cm}^2$: 1.0 Mark

Correct Answer:
Mathbee AI Mentor (Free Demo)

Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.

Master Surface Areas and Volumes with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free