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Solution: Here,        Length of the base of rectangular pyramid (l) = 8 cm        Breadth of the base of rectangular pyramid (b) = 5 cm        Area of the base of the pyramid (A) = .....

Solution: Given,           Side of the square base (a)     = 8 cm. Now,           Area of the base (A)     = a2               .....

Solution: Here,          The length of the base of the pyramid = the side of the cube = a = 12 cm          The height of the pyramid = h = 8 cm. Now,          The volume of .....

Solution: Here,          Lngth of the base of pyramid (a) = 16 cm          Let, l and h be the slant height and height of the pyramid. We have,         TSA of the pyramid = 800 .....

Solution:          Height of the pyramid (h) = 12 cm         Area of the square base (A) = 64 cm2 We know,          Volume of the pyramid (V) = 1/3⨉ Area of base ⨉ .....

Solution: Given,          Diameter of the base (d) = 8 cm          Radius of the base (r) = d/2 = 8 cm/2 = 4 cm          Height of the cone (h) = 3 cm. According to .....

Solution: Given,         Radius = r = 7cm        Height = h = 9cm Now, The volume of the cone is given by,        Volume= 1/3 πr2h     .....

Solution: Given, Radius of the base (r) = 5 cm     Height of the cone (h) = 12 cm According to Pythagoras theorem, Slant height of the cone (l) = √(r2+ h2 )                .....

Solution: Given,        Diameter of the base (d) = 10 cm        Radius of the base (r) = d/2 = 10 cm/2 = 5 cm        Height of the cone (h) = 24 cm.        According to .....

Solution: Given,         Area of the base of pyramid (A) = 35 cm2         Area of the triangular faces along the length = 20.8 cm2         Area of 2 triangular faces along the length (Al) .....

Solution: Given,         Radius of the base (r) = 5.6 cm         Slant height of the cone (l) = 10 cm. We know,         Total surface area of the cone (TSA) = πr (r + l)     .....

Solution: Here,         Slant height = l,         Vertical height = h,         Length of base = a = ? Now,     The length of the base of the pyramid is given by,     .....

Solution: Given,         Diameter of the base (d) = 3 cm         Radius of the base (r) = d/2 = 3 cm/2 = 1.5 cm         Slant height of the cone (l) = 5 cm. We know, Curved surface area .....

Solution: Given,       Diameter of the base (d) = 10 cm       Radius of the base (r) = d/2 = 10 cm/2 = 5 cm       height of the cone (l) = 13 cm. We know, Curved surface area of the cone (CSA) = .....

Solution: Given, Radius of the base (r) = 8.4 cm Slant height of the cone (l) = 15 cm We know, Curved surface area of the cone (CSA) = πrl                     = 22/7 X 8.4 .....

Given,         Length of the base (l) = 8 cm         Breadth of the base (b) = 6 cm. Area of the base (A1)   = lb                         .....

Solution: Given,        Height = h = 24cm        Slant height = l = 25cm.        Radius = r =? We know,           Radius of the cone .....

Solution: Here,          Volume of the pyramid (V) = 230 cm3           Base area of the pyramid (A) = 115 cm2 We know,          Volume of the pyramid (V)    .....

Solution: Here,         Area of base (A1) = 48 cm2        The area of each triangular face (A2) =24.5 cm2 Now,  Total surface area of a pyramid is given by,   .....

Solution: We have, The total surface area of prism is given by, TSA = area of 2 triangular base + Perimeter of base × Height of prism         = (2x +yz) .....

Solution: The volume of prism is given by, Volume = 1/3 × Area of square base × height of prism               = 81cm2×18cm /3               = 486 .....

Let, a be the base of pyramid, l be the slant height and h be the height of  it. We know, Lateral Surface Area of pyramid = 4 ⨉ Area of one triangle  Or, 540 = 4⨉ 1/2 ⨉ 30 ⨉ a  Or,  a = 12 cm. Now, (slanted .....

Solution:        The length of the base of the pyramid (a) = 12 cm Here,          Total surface area of the pyramid = 384 cm2 Or,    a2 +2ah = 384 Or,    122 + 2×12×h = .....

We have given, Base of pyramid = a =12cm. Height of pyramid  = h = 8cm. We know,      l2 = (a/2)2 + h2  or,l2 = 62 + 82 = 102  Therefore, slanted height (l) = 10 cm. We know,        .....

Solution: Given,         Diameter of the base (d) = 10 cm         Radius of the base (r) = d/2 = 10 cm/2 = 5 cm         Slant height of the cone (l) = 13 cm. We know,       .....

Solution: Here,      AM = MB = ½ ⨉ 6 cm = 3 cm      BN = NC = ½ ⨉ 8.4 cm = 4.2 cm      OM = 5.8 cm. In right angled triangle OMB, OB = √(OM2+ MB2  )       = .....

Solution: Here,        Height of right angled triangular base (h) = 8 cm        Base length of the triangular base (b) = 6 cm Now,          Area of triangular base (A) = ½⨉ .....

Solution: Here,        Length of the base of pyramid (l) = 8 cm        Breadth of the base of pyramid (b) = 8 cm        Area of the base of the pyramid (A) = lb         .....

Solution:           Let, a = AB = 3cm , b = BC = 4cm. Here, the volume of the prism=480 cm3 or,    Area of base ⨉  height = 480cm3  or,      abh2 = 480cm3  or, .....

Solution:       Let, c=10cm , a=8 cm . Now,        b = c2-a2         = 102-82         = 6 cm. Here, the volume of the prism = 480 cm3  or,    Area .....

Solution: Given,         Length of the base (a) = 24 cm         Area of the square base (A)     = a2                           .....

Solution: Given,         Diameter of the base (d) = 10 cm         Radius of the base (r) = d/2 = 10 cm/2 = 5 cm         Height of the cone (h) = 24 cm, According to Pythagoras .....

Solution: Given, Diameter of the base (d) = 10 cm Radius of the base (r) = d/2 = 10 cm/2 = 5 cm Slant height of the cone (l) = 17.7 cm We know, Curved surface area of the cone (CSA) = πrl            .....

Solution: Let, h and a be the height and the length of base of the pyramid. Then,       a = 2h. Now,          The volume of the pyramid = 288 cm3 Or,   1/3×a2×h = 288cm3 Or,  .....

Solution: Here,        Slant height(h)=10cm and length of square base (a)=16cm. Now,         The area of triangular surfaces of pyramid is given by,              Are  .....

Solution: Given,         Length of base = a         Vertical height = h         Slant height = l =? We know,         The slant height of .....

Solution: Given,       Diameter of the base (d) = 28 cm       Radius of the base (r) = d/2 = 28 cm/2 = 14 cm       Slant height of the cone (l) = 20 cm. We know, Curved surface area of the cone (CSA) = .....

Solution: Given,          Diameter of the base (d) = 42 cm          Radius of the base (r) = d/2 = 42 cm/2 = 21 cm          Slant height of the cone (l) = 28 cm. We .....

Solution: Given,         Length of the base of pyramid (a) = 8 cm         Breadth of the base of pyramid (b) = 6 cm         Area of the base of pyramid (A)     = ab   .....

Solution: Given,         Length of the square base (a) = 10 cm         Area of the base of pyramid (A) =a2                             .....

Solution: Given,       Length of the base of the pyramid (l) = 8 cm       Breadth of the base of the pyramid (b) = 6 cm       Area of the base of pyramid (A)     = lb       .....

Solution: Here,           Length of the base of rectangular pyramid (l) = 12 cm           Breadth of the base of rectangular pyramid (b) = 7 cm           Area of the base of .....

Solution: Let,        a=AB=16cm, b=AC=DF=20cm, c=CB, h=AD=45cm. Now,       In right angled triangle ABC,           c = b2-a2             .....

Solution: Here,        The length of equilateral triangle (a) = 6cm        The height of prism(h) = 63cm. Now, the volume of the prism is given by,        Volume =Area of base ⨉ .....

Solution: Here,         The height of the pyramid (h) = 8cm          The length of the square base(a) = 12cm. Now, we have formula, Volume of the pyramid is given by,         .....

Let the slanted height be l and the vertical height be h. We know,      TSA = 4 ⨉ Area of triangle + Area of base square Or,4 ⨉ 1/2 ⨉ l  ⨉ 14 + 142 = 896, Or, 28l = 896 - 196 Or, l = 700 / 28 Or, l  .....

Solution: Let,         a=AB=6cm, b=AC=DF=10cm, c=CB, h=AD=30cm Now,          In right angled triangle ABC,               c = b2-a2       .....

Solution: Given,        Length of the base (a) = 10 cm       Area of the square base (A)     = a2                             .....

Solution: Here,          The height of the pyramid (h1) = 12 cm          The area of the base of cuboid = the area of the base of pyramid = A = 10 cm2          The height of .....

Given,          Radius = r =7cm,          Slant height = l = 25cm. We know that,           (Slanted Height)2 = radius2 + (vertical height)2     Or, 252 = h2 + .....