SECTION - D (5 Marks Each)
Q.34
Surface Areas & Volumes
[5 Marks]
(A) A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are $3\text{ m}$ and $14\text{ m}$ respectively, and the total height of the tent is $13.5\text{ m}$, then find the area of the canvas required for making the tent, keeping a provision of $26\text{ m}^2$ of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of ₹ $250$ per $\text{m}^2$.
OR
(B) A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same radius. The radius of hemisphere is $3.5\text{ cm}$ and the total wood used in making the toy is $166 \frac{5}{6}\text{ cm}^3$. Find the height of the conical part. Also, find the cost of painting the hemispherical part of the toy at the rate of ₹ $15$ per $\text{cm}^2$.
💡 Hint (A): Area of canvas = Curved Surface Area (CSA) of Cylinder + CSA of Cone. The base of the tent is on the ground, so do NOT include $\pi r^2$. Don't forget to add the $26\text{ m}^2$ wastage BEFORE calculating the final cost!
Solution 34(A)
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Radius of cylinder and cone ($r$) $= 14\text{ m}$.
Height of cylinder ($h_1$) $= 3\text{ m}$.
Total height of tent $= 13.5\text{ m}$.
$\Rightarrow$ Height of conical part ($h_2$) $= 13.5 - 3 = 10.5\text{ m}$. [1 Mark]
Slant height of cone ($l$) $= \sqrt{r^2 + h_2^2}$
$l = \sqrt{(14)^2 + (10.5)^2} = \sqrt{196 + 110.25}$
$l = \sqrt{306.25} = 17.5\text{ m}$. [1 Mark]
Area of canvas used in tent $= \text{CSA of Cylinder} + \text{CSA of Cone}$
$= 2\pi r h_1 + \pi r l = \pi r (2h_1 + l)$
$= \frac{22}{7} \times 14 \times (2(3) + 17.5)$
$= 44 \times (6 + 17.5) = 44 \times 23.5 = 1034\text{ m}^2$. [1 Mark]
Provision for stitching and wastage $= 26\text{ m}^2$.
Total canvas to be purchased $= 1034 + 26 = 1060\text{ m}^2$. [1 Mark]
Total Cost $= \text{Area} \times \text{Rate}$
$\text{Cost} = 1060 \times 250 = ₹2,65,000$.
$\therefore \text{Cost of canvas is } ₹2,65,000$ [1 Mark]
💡 Hint (B): Total Volume = Volume of Cone + Volume of Hemisphere. Convert the mixed fraction $166 \frac{5}{6}$ into an improper fraction $\frac{1001}{6}$. Equate and solve for $h$. For the painting cost, calculate ONLY the CSA of the hemisphere ($2\pi r^2$).
Solution 34(B)
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Radius of hemisphere and cone ($r$) $= 3.5\text{ cm} = \frac{7}{2}\text{ cm}$.
Total volume of wood $= 166 \frac{5}{6} = \frac{1001}{6}\text{ cm}^3$. [1 Mark]
Let height of cone be $h$.
$\text{Total Volume} = \text{Vol of Cone} + \text{Vol of Hemisphere}$
$\frac{1001}{6} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$
$\frac{1001}{6} = \frac{1}{3}\pi r^2 (h + 2r)$ [1 Mark]
Substituting values:
$\frac{1001}{6} = \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \left(h + 2\left(\frac{7}{2}\right)\right)$
$\frac{1001}{6} = \frac{77}{6} \times (h + 7)$
Cancelling $6$ from denominators:
$1001 = 77(h + 7)$
$h + 7 = \frac{1001}{77} = 13 \Rightarrow h = 6\text{ cm}$. [1.5 Marks]
Now, we paint only the hemispherical part.
Area to be painted $= \text{CSA of Hemisphere} = 2\pi r^2$
$= 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 77\text{ cm}^2$. [1 Mark]
Cost of painting $= 77 \times 15 = ₹1155$.
$\therefore h = 6\text{ cm, Cost} = ₹1155$ [0.5 Mark]
Q.35
Heights and Distances
[5 Marks]
(A) Manish is standing on level ground and observes a kite flying $200\text{ m}$ away from him at an angle of elevation of $30^\circ$. Mahesh, standing on the roof of a $50\text{ m}$ high building on the opposite side of the kite, observes the same kite at an angle of elevation of $45^\circ$. Find the distance between the kite and Mahesh.
OR
(B) The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be $45^\circ$ and $30^\circ$. If the ships are $200\text{ m}$ apart and one ship is exactly behind the other then find the height of lighthouse.
💡 Hint (A): "Flying 200m away from him" refers to the direct line-of-sight distance (the hypotenuse). Use $\sin(30^\circ)$ to find the total height of the kite from the ground. Then, subtract the 50m building height to find the triangle height for Mahesh.
Solution 35(A)
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Let $K$ be the position of the kite and $P$ be the point on the ground directly below it.
Let Manish be at $M_1$ and Mahesh be at $M_2$ (on top of the $50\text{ m}$ building).
Given: $KM_1 = 200\text{ m}$ (Distance from Manish to Kite). $\angle KM_1P = 30^\circ$.
In right $\Delta KPM_1$:
$\sin 30^\circ = \frac{KP}{KM_1}$
$\frac{1}{2} = \frac{KP}{200} \Rightarrow KP = 100\text{ m}$. (Total height of kite). [1.5 Marks]
Draw a horizontal line from Mahesh ($M_2$) to intersect $KP$ at point $Q$.
Height of building $= 50\text{ m}$, so $QP = 50\text{ m}$.
The vertical distance from Mahesh's level to the kite is $KQ$:
$KQ = KP - QP = 100 - 50 = 50\text{ m}$. [1.5 Marks]
In right $\Delta KQM_2$:
The angle of elevation for Mahesh is $\angle KM_2Q = 45^\circ$. We need to find $KM_2$.
$\sin 45^\circ = \frac{KQ}{KM_2}$
$\frac{1}{\sqrt{2}} = \frac{50}{KM_2}$
$KM_2 = 50\sqrt{2}\text{ m}$. [1.5 Marks]
Using $\sqrt{2} \approx 1.414$:
$KM_2 = 50 \times 1.414 = 70.7\text{ m}$.
$\therefore \text{Distance between kite and Mahesh is } 50\sqrt{2}\text{ m (or } 70.7\text{ m)}$ [0.5 Mark]
💡 Hint (B): Draw two right-angled triangles sharing the same height ($h$). The closer ship has a $45^\circ$ angle of elevation, meaning its distance from the base is exactly $h$. The further ship is at distance $(h + 200)$. Use $\tan(30^\circ)$!
Solution 35(B)
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Let the height of the lighthouse $CD = h$. Let the base of the lighthouse be $D$.
Let the two ships be at points $A$ and $B$, where $A$ is closer ($45^\circ$) and $B$ is further ($30^\circ$).
Given: Distance between ships $AB = 200\text{ m}$.
Let distance $DA = x$. Then distance $DB = x + 200$. [1 Mark]
In right $\Delta CDA$:
$\tan 45^\circ = \frac{CD}{DA} \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h$. [1.5 Marks]
In right $\Delta CDB$:
$\tan 30^\circ = \frac{CD}{DB} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 200}$
Substitute $x = h$:
$\frac{1}{\sqrt{3}} = \frac{h}{h + 200}$ [1 Mark]
Cross multiplying:
$h\sqrt{3} = h + 200$
$h\sqrt{3} - h = 200$
$h(\sqrt{3} - 1) = 200$
$h = \frac{200}{\sqrt{3} - 1}$
Rationalizing the denominator:
$h = \frac{200(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{200(\sqrt{3} + 1)}{3 - 1}$
$h = \frac{200(\sqrt{3} + 1)}{2} = 100(\sqrt{3} + 1)\text{ m}$. [1 Mark]
Using $\sqrt{3} \approx 1.732$:
$h = 100(1.732 + 1) = 100(2.732) = 273.2\text{ m}$.
$\therefore \text{Height of lighthouse is } 100(\sqrt{3} + 1)\text{ m (or } 273.2\text{ m)}$ [0.5 Mark]
