(i) Let ABC be a conical tent.
Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let the slant height of the tent be l.
In \(∆\)ABO, AB2 = AO2 + BO2
l2 = h2 + r2
= (10 m)2 + (24 m)2
I= \(\sqrt{676}\)
= 26 m
Therefore, the slant height of the tent is 26 m.
(ii) curved surface area of the cone = \(\pi rl\)
= \(\frac{22}{7}\) × 24 m × 26 m
= \(\frac{13728}{7}\)m²
The cost of the canvas required to make the tent, at \(₹\) 70 per m² = 70 × Curved surface area of the cone
= \(\frac{13728}{7}\) × 70
= \(₹\)137280
Thus, slant height of the tent is 26 m and the cost of the canvas is ₹ 137280.
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?