Step 1: Understanding the Concept:
In a right circular cone, the radius (\(r\)), the perpendicular height (\(h\)), and the slant height (\(l\)) form a right-angled triangle. The slant height (\(l\)) is the hypotenuse of this triangle.
Step 2: Key Formula or Approach:
We can use the Pythagorean theorem to relate the radius, height, and slant height of the cone:
\[ l^2 = r^2 + h^2 \]
Taking the square root gives the formula for the slant height:
\[ l = \sqrt{r^2 + h^2} \]
Step 3: Detailed Explanation:
We are given the following values:
Radius, \(r = 5\) cm
Perpendicular height, \(h = 12\) cm
Substitute these values into the formula for slant height:
\[ l = \sqrt{5^2 + 12^2} \]
\[ l = \sqrt{25 + 144} \]
\[ l = \sqrt{169} \]
\[ l = 13 \]
So, the slant height is 13 cm.
Step 4: Final Answer:
The slant height of the cone is 13 cm.
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\))
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
सरस्वती विद्यालय, कोल्हापुर में मनाए गए 'शिक्षक दिवस' समारोह का 70 से 80 शब्दों में वृत्तांत लेखन कीजिए।
(वृत्तांत में स्थल, काल, घटना का उल्लेख होना अनिवार्य है)
निम्नलिखित जानकारी के आधार पर 50 से 60 शब्दों में विज्ञापन तैयार कीजिए :