Chapter 3:Pair of Linear Equations in Two Variables
Chapter 5: Arithmetic Progression
Chapter 8: Introduction to Trigonometry
Chapter 9:Some Applications of Trigonometry
Chapter 12:Area Related to Circles
Surface Areas and Volumes
Given
The Volume of each cube is = 64 cm3
let's side of the cube =a
a3 = 64 cm3
∴ a = 4 cm
the side of the cube = a = 4 cm
Also, when we joined end to end of each cube then the length and breadth of the resulting cuboid will 8 cm and 4 cm.
So, the surface area of the cuboid = 2(lb+bh+lh)
= 2(8 × 4 + 4 × 4 + 4 × 8) cm2
= 2(32+16+32) cm2
= (2×80) cm2 = 160 cm2
The diameter of the hemisphere = D = 14 cm
The radius of the hemisphere = r = 7 cm
the height of the cylinder h = 13 - 7 = 6 cm
the radius of the hollow hemisphere = 7 cm
Now, the inner surface area of the vessel = Coverd Surface area of the cylindrical part + Coverd Surface area of hemispherical part
(2πrh+2πr2) cm2 = 2πr(h+r) cm2
2×(22/7)×7(6+7) cm2 = 572 cm2
Given that radius of the cone (r) = 3.5 cm
r= 7/2 cm
The total height of the toy = 15.5 cm.
So, the height of the cone = 15.5-3.5 = 12 cm
Total surface area of the toy =curved surface area of cone +curved surface area of the hemisphere = πrl + 2πr2
=(22/7)×(7/2)×(25/2) + 2×(22/7)×(7/2)2
= (275/2)+77 cm2
= (275+154)/2 cm2
= 429/2 cm2 = 214.5cm2
So, the total surface area of the toy is 214.5cm2
the side of cube = Diameter of hemisphere
So side of cube is 7 cm. and the radius will be 7/2 cm.
We know,
The total surface area of solid = surface area of cubical block +curved surface area of hemisphere – Area of base of hemisphere
∴ TSA of solid = 6×(side)2+2πr2-πr2
= 6×(side)2+πr2
= 6×(7)2+(22/7)×(7/2)×(7/2)
= (6×49)+(77/2)
= 294+38.5 = 332.5 cm2
So, the surface area of the solid is 332.5 cm2
Now, the diameter of hemisphere = Edge of the cube = l
So, the radius of hemisphere = l/2
∴ The total surface area of solid = surface area of cube + CSA of hemisphere – Area of base of hemisphere
TSA of remaining solid = 6 (edge)2+2πr2 - πr2
= 6l2 + πr2
= 6l2 + π(l/2)2
= 6l2 + πl2/4
= l2/4(24+π) sq. units
Two hemisphere and one cylinder are included
Here, the diameter of the capsule = 5 mm
∴ Radius = 5/2 = 2.5 mm
Now, the length of the capsule = 14 mm
So, the length of the cylinder = 14-(2.5+2.5) = 9 mm
∴ The surface area of a hemisphere = 2πr2 = 2×(22/7)×2.5×2.5
= 275/7 mm2
Now, the surface area of the cylinder = 2πrh
= 2×(22/7)×2.5×9
(22/7)×45 = 990/7 mm2
Thus, the required surface area of medicine capsule
= 2 × surface area of hemisphere + surface area of the cylinder
= (2×275/7) × 990/7
(550/7) + (990/7) = 1540/7 = 220 mm2
we know that
Diameter = 4 m
Slant height of the cone l = 2.8 m
Radius of the cone r = Radius of cylinder = 4/2 = 2 m
Height of the cylinder h = 2.1 m
So, the required surface area of tent = surface area of cone + surface area of cylinder
= πrl+2πrh
= πr(l+2h)
= (22/7)×2(2.8+2×2.1)
= (44/7)(2.8+4.2)
= (44/7)×7 = 44 m2
∴ The cost of the canvas of the tent at the rate of ₹500 per m2 will be
= Surface area × cost per m2
44×500 = ₹22000
So, Rs. 22000 will be the total cost of the canvas.
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm2.
we know that
The diameter of the cylinder = diameter of conical cavity = 1.4 cm
So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7
Also, the height of the cylinder = height of the conical cavity = 2.4 cm
Now, the Total surface area of remaining solid = surface area of conical cavity + Total surface area of the cylinder
= πrl+(2πrh+πr2)
= πr(l+2h+r)
= (22/7)× 0.7(2.5+4.8+0.7)
= 2.2×8 = 17.6 cm2
So, the total surface area of the remaining solid is 17.6 cm2