Mathematics,Class 10, Ex-12.1
1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
STEP 1: FIND SIDE OF CUBE
Volume = a³ = 64 cm³
Therefore, a = 4 cm.
STEP 2: DIMENSIONS OF CUBOID
Length (l) = 4 + 4 = 8 cm
Breadth (b) = 4 cm, Height (h) = 4 cm.
STEP 3: CALCULATION
Surface Area = 2(lb + bh + hl)
= 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32)
= 2(80) = 160 cm².
2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
STEP 1: RADII
Diameter = 14 cm, so Radius (r) = 7 cm.
STEP 2: HEIGHTS
Total Height = 13 cm
Cylinder height (h) = 13 – 7 = 6 cm.
STEP 3: INNER AREA
Area = 2πrh + 2πr² = 2πr(h+r)
= 2 × 22/7 × 7 × (6 + 7) = 572 cm².
3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
STEP 1: FIND CONE HEIGHT
Radius (r) = 3.5 cm
Height of cone (h) = 15.5 – 3.5 = 12 cm.
STEP 2: SLANT HEIGHT
l = √(h² + r²) = √(12² + 3.5²) = 12.5 cm.
STEP 3: TOTAL AREA
Area = πrl + 2πr²
= 22/7 × 3.5 × 12.5 + 2 × 22/7 × (3.5)² = 214.5 cm².
4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
STEP 1: MAX DIAMETER
Greatest diameter = side of cube = 7 cm.
STEP 2: AREA LOGIC
Total Area = (Area of 6 faces) + (CSA of Hemisphere) – (Area of circular base of hemisphere).
STEP 3: RESULT
Area = 6(7²) + 2π(3.5)² – π(3.5)²
= 294 + 38.5 = 332.5 cm².
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
STEP 1: SURFACE AREA FORMULA
Remaining Area = 6l² + 2π(l/2)² – π(l/2)².
STEP 2: SIMPLIFICATION
Area = 6l² + πl²/4.
STEP 3: FINAL ANSWER
Area = l²/4 (24 + π) square units.
6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
STEP 1: DIMENSIONS
Radius (r) = 2.5 mm
Height of cylinder (h) = 14 – 5 = 9 mm.
STEP 2: SURFACE COMPONENTS
Total Area = CSA of cylinder + 2 × CSA of hemisphere.
STEP 3: RESULT
Area = 2πr(h + 2r) = 2 × 22/7 × 2.5 × 14 = 220 mm².
7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)
STEP 1: FIND SURFACE AREA
CSA = 2πrh + πrl = πr(2h + l)
= 22/7 × 2 × (2 × 2.1 + 2.8) = 44 m².
STEP 2: COST CALCULATION
Rate = ₹ 500/m²
Cost = Area × Rate = 44 × 500.
STEP 3: TOTAL
Total Cost = ₹ 22,000.
8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
STEP 1: FIND SLANT HEIGHT
r = 0.7 cm, h = 2.4 cm
l = √(0.7² + 2.4²) = 2.5 cm.
STEP 2: COMBINED AREA
Area = CSA Cylinder + CSA Cone + Base Area
= 2πrh + πrl + πr².
STEP 3: CALCULATION
Area = 17.6 cm²
To the nearest cm², Area = 18 cm².
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, if the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
STEP 1: AREA COMPONENTS
Total Area = CSA of cylinder + 2 × CSA of hemisphere.
STEP 2: SUBSTITUTION
Area = 2πrh + 2(2πr²) = 2πr(h + 2r)
= 2 × 22/7 × 3.5 × (10 + 7).
STEP 3: RESULT
Area = 22 × 17 = 374 cm².
Mathematics – Class X (Important Question)
Q. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Step 1: Given Information
Let the circle touch sides BC, AB, and AC at points D, E, and F respectively.
Radius of circle (r) = 4 cm.
BD = 8 cm and DC = 6 cm.
Step 2: Theorem of Tangents
Theorem: Lengths of tangents from an external point to a circle are equal.
1. CD = CF = 6 cm (Tangents from C)
2. BD = BE = 8 cm (Tangents from B)
3. Let AE = AF = x cm (Tangents from A)
Step 3: Calculating Sides & s
Side a (BC) = 6 + 8 = 14 cm
Side b (AC) = 6 + x cm
Side c (AB) = 8 + x cm
Semi-perimeter (s) = (14 + 6 + x + 8 + x) / 2
s = (28 + 2x) / 2 = (14 + x) cm
Step 4: Area using Heron’s Formula
Area = √[s(s-a)(s-b)(s-c)]
s – b = (14+x) – (6+x) = 8
s – c = (14+x) – (8+x) = 6
Area of ΔABC = √[(14+x) · x · 8 · 6]
Area = √[48x(14+x)] … (i)
Step 5: Area using Sub-triangles
Area(ΔABC) = Area(ΔOBC) + Area(ΔOCA) + Area(ΔOAB)
Area = 1/2 · 4 · (BC + AC + AB)
Area = 2 · (14 + 6+x + 8+x)
Area = 2 · (28 + 2x) = 4(14+x) … (ii)
Step 6: Final Calculation
Equating (i) and (ii):
48x(14+x) = 16(14+x)² (Squaring sides)
3x = 14 + x
2x = 14
x = 7 cm
AC = 6 + 7 = 13 cm