Home/ manvi zoon/Best Answers
  1. Asked: In:

    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.

    manvi zoon

    Height of conical part = Height of cylindrical part (h) = 2.4 cm Diameter of cylindrical part = 1.4 m, so, the radius of cylindrical part (r) = 0.7 m Slant height of cylindrical part (l) = √(r²+ h²) = √((0.7)² + (2.4)²) = √(0.49 + 5.76) = √(6.25) = 2.5 The total surface area of the remaining solid =Read more

    Height of conical part = Height of cylindrical part (h) = 2.4 cm
    Diameter of cylindrical part = 1.4 m, so, the radius of cylindrical part (r) = 0.7 m
    Slant height of cylindrical part (l) = √(r²+ h²)
    = √((0.7)² + (2.4)²) = √(0.49 + 5.76) = √(6.25) = 2.5
    The total surface area of the remaining solid
    = CSA of cylindrical + CSA of conical part + Area of base of cylinder
    = 2πrh + πrl + πr²
    = 2 × 22/7 × 0.7 × 2.4 + 22/7 × 0.7 × 2.5 + 22/7 × 0.7 × 0.7
    = 4.4 × 2.4 + 2.2 × 2.5 × 0.7 = 10.56 + 5.50 + 1.56 = 17.60 cm².

    See less
    • 1
  2. Asked: In:

    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.

    manvi zoon

    The maximum diameter of hemishere = Side of cubical block (a) = 7 cm Radius of hemisphere = a/2 = 3.5 cm The surface are of the soild = Surface area of cubical block + CSA of hemisphere - Area of base of hemisphere = 6a² + 2πr² - πr² = 6a² + πr² = 6 × 7² + 22/7 × 3.5² = 294 + 38.5 = 332.5 cm² Hence,Read more

    The maximum diameter of hemishere = Side of cubical block (a) = 7 cm
    Radius of hemisphere = a/2 = 3.5 cm
    The surface are of the soild
    = Surface area of cubical block + CSA of hemisphere – Area of base of hemisphere
    = 6a² + 2πr² – πr²
    = 6a² + πr²
    = 6 × 7² + 22/7 × 3.5² = 294 + 38.5 = 332.5 cm²
    Hence, the surface area of the solid is 332.5 cm²

    See less
    • 1
  3. Asked: In:

    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.

    manvi zoon

    The maximum daimeter of hemisphere = side of cubical block = l Radius of hemisphere (r) = l/2 The surface area of the remaining solid = TSA of cubical block + CSA of hemisphere - Area of base of hemisphere = 6l² + 2πr² - πr² = 6l² + πr² = 6l² + π(l/2)² = (6 + π/4)l² Hence, the surface area of the reRead more

    The maximum daimeter of hemisphere = side of cubical block = l
    Radius of hemisphere (r) = l/2
    The surface area of the remaining solid
    = TSA of cubical block + CSA of hemisphere – Area of base of hemisphere
    = 6l² + 2πr² – πr²
    = 6l² + πr²
    = 6l² + π(l/2)²
    = (6 + π/4)l²
    Hence, the surface area of the remaining solid is (6 + π/4)l².

    See less
    • 1
  4. Asked: In:

    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.

    manvi zoon

    Radius of cone = 3.5 cm Height of cone = 15.5 - 3.5 = 12 cm Radius of hemisphere = 3.5 cm Slant height of cone (l) = √r² +h² = √((3.5)² + (12)²) = √(12.25 + 144) = √(156.25) = 12.5 The total surface area of the toy = CSA of cone + CSA of hemisphere = πrl + 2πr² = 22/7 × 3.5 × 12.5 + 2 × 22/7 ×3.5² =Read more

    Radius of cone = 3.5 cm
    Height of cone = 15.5 – 3.5 = 12 cm
    Radius of hemisphere = 3.5 cm
    Slant height of cone (l)
    = √r² +h² = √((3.5)² + (12)²) = √(12.25 + 144) = √(156.25) = 12.5
    The total surface area of the toy
    = CSA of cone + CSA of hemisphere
    = πrl + 2πr²
    = 22/7 × 3.5 × 12.5 + 2 × 22/7 ×3.5²
    = 137.5 + 77
    = 214.5 cm²
    Hence, the total surface area of the toy is 572 cm²

    See less
    • 1
  6. Asked: In:

    Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and Angle B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

    manvi zoon

    Consider the following situation. If a circle is drawn through B, D, and C, BC will be its diameter as ∠BDC is measure 90°. the centre E of this circle will be the mid-point of BC. The required tangents can be constructed on the given circle as follows. Step 1 Join AE and bisect it. Let F be the midRead more

    Consider the following situation. If a circle is drawn through B, D, and C, BC will be its diameter as ∠BDC is measure 90°. the centre E of this circle will be the mid-point of BC.
    The required tangents can be constructed on the given circle as follows.
    Step 1
    Join AE and bisect it. Let F be the mid-point of AE.
    Step 2
    Taking F as centre and FE as its radius, Draw a circle which will intersect the circle at point B and G. Join AG. AB and AG are the required tangents.
    Justification
    The construction can be justified by proving that AG and AB are the tangents to the circle. For this, join EG.
    ∠AGE is an angle in the semi- circle. We know that an angle in a semi-circle is a right angle.
    ∴ ∠AGE = 90°
    ⇒ EG ⊥AG
    Since EG is the radius of the circle, AG has to be a tangent of the circle.
    Already, ∠B = 90°
    ⇒ AB ⊥ BE
    Since BE is the radius of the circle, AB has to be a tangent of the circle.

    See less
    • 1