5 Marks
34
(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 $\text{m}^2$.
Concept:
1
The area of canvas required to make a tent is the sum of the Curved Surface Areas (CSA) of the shapes involved. (We do not include the base area because tents don't have canvas floors unless stated).
$\text{Total Canvas} = \text{CSA of Cylinder} + \text{CSA of Cone} + \text{Wastage}$.
First, find the height of the cone by subtracting the cylinder's height from the total height. Then calculate the slant height ($l$) of the cone using Pythagoras ($l^2 = r^2 + h^2$). Apply formulas $2\pi rh$ (cylinder) and $\pi rl$ (cone).
$\text{Total Canvas} = \text{CSA of Cylinder} + \text{CSA of Cone} + \text{Wastage}$.
First, find the height of the cone by subtracting the cylinder's height from the total height. Then calculate the slant height ($l$) of the cone using Pythagoras ($l^2 = r^2 + h^2$). Apply formulas $2\pi rh$ (cylinder) and $\pi rl$ (cone).
Hint:
1
Cone height ($h_2$) = $13.5 - 3 = 10.5\text{ m}$.
2
Slant height ($l$) = $\sqrt{14^2 + 10.5^2}$.
3
Add $\pi rl + 2\pi rh + 26$ to get the total area. Multiply by 250 for cost.
Mistakes to Avoid:
!
Forgetting the wastage: The question explicitly adds $26\text{ m}^2$ for stitching and wastage. If you forget to add this to your surface area before multiplying by the cost, your final answer will be incorrect.
!
Using total height for the cone: Make sure you subtract the 3m cylindrical base from the 13.5m total to find the height of the cone ($10.5\text{m}$) before calculating the slant height.
Official CBSE Solution:
1. Dimensions:
Radius ($r$) = $14\text{ m}$. Cylindrical height ($h$) = $3\text{ m}$.
Height of Conical part = Total Height - Cylinder Height
$= 13.5 - 3 = 10.5\text{ m}$.
2. Slant Height ($l$) of cone:
$l = \sqrt{r^2 + h^2} = \sqrt{(14)^2 + (10.5)^2}$
$l = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5\text{ m}$.
3. Surface Areas:
CSA of Conical part = $\pi rl = \frac{22}{7} \times 14 \times 17.5 = 770\text{ m}^2$.
CSA of Cylindrical part = $2\pi rh = 2 \times \frac{22}{7} \times 14 \times 3 = 264\text{ m}^2$.
4. Total Area and Cost:
Total area of canvas required = CSA of cone + CSA of cylinder + Wastage
$= 770 + 264 + 26 = 1060\text{ m}^2$.
Cost of canvas @ ₹250/m² = $250 \times 1060 = \text{₹}265000$.
Ans: Canvas Area = $1060\text{ m}^2$. Total Cost = ₹2,65,000.
🎯 CBSE Marking Scheme (5 Marks):
• Finding height of conical part ($10.5\text{ m}$): ½ Mark
• Calculating slant height ($17.5\text{ m}$): 1 Mark
• Calculating CSA of cone ($770\text{ m}^2$): 1 Mark
• Calculating CSA of cylinder ($264\text{ m}^2$): 1 Mark
• Finding total area including wastage ($1060\text{ m}^2$): ½ Mark
• Finding total cost (₹265000): 1 Mark
• Finding height of conical part ($10.5\text{ m}$): ½ Mark
• Calculating slant height ($17.5\text{ m}$): 1 Mark
• Calculating CSA of cone ($770\text{ m}^2$): 1 Mark
• Calculating CSA of cylinder ($264\text{ m}^2$): 1 Mark
• Finding total area including wastage ($1060\text{ m}^2$): ½ Mark
• Finding total cost (₹265000): 1 Mark
5 Marks
34
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 $\text{cm}^2$.
Concept:
1
The total volume of the toy is the sum of the volume of the cone ($\frac{1}{3}\pi r^2 h$) and the hemisphere ($\frac{2}{3}\pi r^3$). Set the sum of these formulas equal to the given total volume to solve for the unknown height $h$. Once $h$ is found, calculate the Curved Surface Area (CSA) of the hemisphere ($2\pi r^2$) and multiply by the rate to find the painting cost.
Hint:
1
Convert the mixed fraction: $166 \frac{5}{6} = \frac{1001}{6}$.
2
Factor out $\frac{1}{3}\pi r^2$ from the volume equation to make the math easier: $\frac{1001}{6} = \frac{1}{3}\pi r^2 (h + 2r)$.
Mistakes to Avoid:
!
Painting the wrong parts: The question asks for the cost of painting the *hemispherical part* ONLY. Do not calculate the total surface area of the toy. Only calculate the CSA of the hemisphere ($2\pi r^2$) and multiply by ₹15.
Official CBSE Solution:
Let '$h$' cm be the height of the conical part.
Volume of wood = Volume of conical part + Volume of hemispherical part
$166 \frac{5}{6} = \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3$
Substitute values ($r=3.5$):
$\frac{1001}{6} = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times h + \frac{2}{3} \times \frac{22}{7} \times (3.5)^3$
$\frac{1001}{6} = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{4} [h + 2(3.5)]$
$\frac{1001}{6} = \frac{77}{6} [h + 7]$
Cancel the 6 in the denominator and solve for $h$:
$1001 = 77(h + 7) \implies \frac{1001}{77} = h + 7$
$13 = h + 7 \implies h = 6\text{ cm}$.
Calculate Cost of Painting Hemisphere:
CSA of hemispherical part $= 2\pi r^2$
$= 2 \times \frac{22}{7} \times (3.5)^2 = 77\text{ cm}^2$
Cost @ ₹15/cm² $= 15 \times 77 = \text{₹}1155$
Ans: Height = 6 cm. Cost = ₹1155.
🎯 CBSE Marking Scheme (5 Marks):
• Correctly equating total volume to sum of formulas: 1 Mark
• Substituting values correctly: 1 Mark
• Solving and finding height ($h=6\text{ cm}$): 1 Mark
• Calculating CSA of hemisphere ($77\text{ cm}^2$): 1 Mark
• Calculating final cost (₹1155): 1 Mark
• Correctly equating total volume to sum of formulas: 1 Mark
• Substituting values correctly: 1 Mark
• Solving and finding height ($h=6\text{ cm}$): 1 Mark
• Calculating CSA of hemisphere ($77\text{ cm}^2$): 1 Mark
• Calculating final cost (₹1155): 1 Mark
5 Marks
35
(A) Manish is standing on level ground and observes a kite flying 200 m away from him at an angle of elevation of $30^\circ$. Mahesh, standing on the roof of a 50 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.
Concept:
1
This requires splitting the problem into two right triangles. First, use Manish's triangle to find the total height of the kite from the ground ($BE$). Then, subtract the height of Mahesh's building ($50\text{m}$) from $BE$ to find the height of the kite *relative* to Mahesh ($BR$). Finally, use Mahesh's triangle to find his distance to the kite (the hypotenuse $BD$).
Hint:
1
In $\triangle ABE$: $\sin(30^\circ) = \frac{BE}{200}$. Solve for $BE$.
2
$BR = BE - 50$ (height of building).
3
In $\triangle BRD$: $\sin(45^\circ) = \frac{BR}{BD}$. Solve for $BD$.
Mistakes to Avoid:
!
Not subtracting the building height: If you use the full height of the kite ($100\text{m}$) in Mahesh's triangle, your final answer will be wrong. Mahesh is already $50\text{m}$ in the air, so the kite is only $50\text{m}$ above his horizontal eye-level line.
Official CBSE Solution:
Let Manish be at point 'A' and Mahesh be at point 'D'. Let the kite be at 'B'.
Distance of kite from Manish ($AB$) = $200\text{ m}$.
In rt. angled $\triangle AEB$:
$\frac{BE}{AB} = \sin 30^\circ$
$\frac{BE}{200} = \frac{1}{2} \implies BE = 100\text{ m}$
Calculate height of kite above Mahesh's eye level ($BR$):
$BR = BE - RE$
Since $RE$ is equal to building height $DC$, $RE = 50\text{ m}$.
$BR = 100 - 50 = 50\text{ m}$.
In rt. angled $\triangle BRD$:
$\frac{BR}{BD} = \sin 45^\circ$
$\frac{50}{BD} = \frac{1}{\sqrt{2}} \implies BD = 50\sqrt{2}\text{ m}$
Ans: Distance of kite from Mahesh is $50\sqrt{2}\text{ m}$.
🎯 CBSE Marking Scheme (5 Marks):
• Correct figure: 1 Mark
• Applying trig in $\triangle AEB$ to find $BE=100$: 1½ Marks
• Finding $BR=50$: 1 Mark
• Applying trig in $\triangle BRD$ to find $BD$: 1 Mark
• Final answer ($50\sqrt{2}$): ½ Mark
• Correct figure: 1 Mark
• Applying trig in $\triangle AEB$ to find $BE=100$: 1½ Marks
• Finding $BR=50$: 1 Mark
• Applying trig in $\triangle BRD$ to find $BD$: 1 Mark
• Final answer ($50\sqrt{2}$): ½ Mark
5 Marks
35
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 m apart and one ship is exactly behind the other then find the height of the lighthouse.
Concept:
1
Because angles of depression alternate with angles of elevation, the ships view the lighthouse at $45^\circ$ and $30^\circ$. Form two right triangles sharing the height $h$. Set up equations for $\tan(45^\circ)$ and $\tan(30^\circ)$. Solve them simultaneously to find $h$. Rationalize the denominator at the end.
Hint:
1
From $\triangle ABC$ ($45^\circ$): $\tan(45^\circ) = \frac{h}{BC} \implies 1 = \frac{h}{BC} \implies BC = h$.
2
From $\triangle ABD$ ($30^\circ$): $\tan(30^\circ) = \frac{h}{BC + 200}$. Substitute $BC = h$.
3
Solve $\frac{1}{\sqrt{3}} = \frac{h}{h+200}$ for $h$.
Mistakes to Avoid:
!
Not rationalizing the denominator: If you get $h = \frac{200}{\sqrt{3}-1}$ and stop, you lose marks. Always multiply numerator and denominator by the conjugate $(\sqrt{3}+1)$.
Official CBSE Solution:
Let the height of the lighthouse be '$h$' m.
In rt. angled $\triangle ABC$:
$\frac{h}{BC} = \tan 45^\circ = 1$
$\implies BC = h$ --- (i)
In rt. angled $\triangle ABD$:
$\frac{h}{BC + CD} = \tan 30^\circ$
$\frac{h}{h + 200} = \frac{1}{\sqrt{3}}$
Cross-multiplying and solving for $h$:
$h\sqrt{3} = h + 200$
$h(\sqrt{3} - 1) = 200$
$h = \frac{200}{\sqrt{3} - 1}$
Rationalizing the denominator:
$h = \frac{200}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$
$h = \frac{200(\sqrt{3} + 1)}{3 - 1} = \frac{200(\sqrt{3} + 1)}{2}$
$h = 100(\sqrt{3} + 1)$
Ans: Height of the lighthouse is $100(\sqrt{3} + 1)\text{ m}$.
🎯 CBSE Marking Scheme (5 Marks):
• Correct figure: 1 Mark
• In $\triangle ABC$, finding $BC = h$: 1 Mark
• In $\triangle ABD$, setting up $\frac{1}{\sqrt{3}} = \frac{h}{h+200}$: 1½ Marks
• Solving to $h = \frac{200}{\sqrt{3}-1}$: ½ Mark
• Rationalizing and final answer $100(\sqrt{3}+1)$: 1 Mark
• Correct figure: 1 Mark
• In $\triangle ABC$, finding $BC = h$: 1 Mark
• In $\triangle ABD$, setting up $\frac{1}{\sqrt{3}} = \frac{h}{h+200}$: 1½ Marks
• Solving to $h = \frac{200}{\sqrt{3}-1}$: ½ Mark
• Rationalizing and final answer $100(\sqrt{3}+1)$: 1 Mark
