Here, 1st expression = 3x4 – 15x2                          = 3x2(x2 – 5). 2nd expression = 6x2 – 30                  .....

Here, 1st expression = 2x3 – 16                       = 2(x3 – 8)                        = 2(x3 – 23)   .....

Here, Expression I:                     =x2-y2                     =(x-y) (x+y). Expression II:             .....

Here, Expression I:                      =  (x-y)2+4xy                      = x2-2xy+y2+4xy             .....

Here,

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Here, 1st expression:                        = a2 – (b + c)2                        = (a + b +c) (a – b – .....

Here, 1st expression:                        = x3 – 1                        = x3 - 13           .....

Here, 1st expression  = (x2 + 4xy – 21y2)                          = x2 + (7 – 3)xy – 21y                     .....

Here, 1st expression:                        = 8x3 + y3                        = (2x)3 + y3           .....

Here, Expression I = 6a2-a-1                     =6a2-3a+2a-1                     =3a2a-1+12a-1               .....

Here, First expression                           = x2 – y2                           = (x + y) (x – .....

Here, First expression                        = x5y3                        = x.x.x.x.x.y.y.y Second expression     .....

Here, 1st expression = 4x4 + 16x3 – 20x2                        = 4x2(x2 + 4x – 5)                        = 4x2{x2 + .....

Here, First expression                           = a5b3 – a3b5                           =a3b3(a2 – .....

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Here, expression I = 2x-22                     =2x-2x-2 Expression II = 2x2- 8=2x2-4                      =2x2-22       .....

Here,

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Solution:

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Here, Expression I:                     = (a+b)2-4ab                     = a2+2ab+b2-4ab                 .....

Here, Expression I:                     =a2-b2                     =(a-b)(a+b) Expression II:               .....

As we know, HCF is common factors. In case of two different prime numbers, there is no common factors. So,      HCF = .....

Here, 1st expression,  = 3x2 – 22x + 19                           = 3x2 – (19 + 3)x + 19                       .....

Here, 1st expression = (1 – x2) (1 – y2) + 4xy                        = 1 – y2 – x2+ x2y2+ 4xy                     .....

Here,

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Here, 1st expression  = 16 (a2 + 2a – 3)                          = 16{a2 + (3 – 1)a – 3}                     .....

Here, 1st expression:                        = (x2 + 4xy – 21y2)                        = x2 + (7 – 3)xy – .....

Here, 1st expression = 3a4 – 8a3 + 4a2 = a2(3a2 – 8a + 4) = a2{3a2 – (6 + 2)a + 4 = a2{3a2 – 6a – 2a + 4} = a2{3a (a – 2) – 2(a – 2)} = a2 (a – 2) (3a – 2) 2nd expression = a3 – 4a = .....

Here, 1st expression = 9x2 – 4y2 – 8yz – 4z2                      =9x2– (4y2+ 8yz + 4z2)                     .....

Here, 1st expression = 9m2 – 4n2 – 4nr – r2                                  = (3m)2 – [(2n)2 + 2.2n.r + (r)2]         .....

Solution: Here,          LSA = 30cm2          Height = 5cm. Now, The lateral surface area of prism is given by,           LSA = Perimeter .....

Solution: Given,           Length of base =  a = 20cm           Height = h = 24cm. The volume of prism is given by, Volume = (1/3) × Area of triangular base .....

Solution: Given,         Length of side = a = 5cm         Height = h = 4cm Now, The volume of prism is given by,        Volume = Area of triangular base × height of prism   .....

Solution: Given,           Area of base = A =25cm2           Height = h = 4cm           Volume = ? Now, The volume of the prism is given by,       V = Area of .....

Solution: Given,          Length of basen = a = 15cm          Slant heightg = l = 18cm.          TSA = ? We know, The total surface area of the square based pyramid is .....

Solution: Given,         TSA = 96cm2          Length of base = a =6cm          Slant height = l =? We know, The total surface area of the square .....

Solution: Here,      Area of one triangular surface of pyramid (A) = x cm2         LSA = ? Now, Lateral surface area is given by,       LSA = 4A = 4x .....

Solution: Given,         Radius of the base (r) = 8.4 cm         Slant height of the cone (l) = 15 cm. We know,         Total surface area of the cone (TSA) = πr (r + l)    .....

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 know,       .....

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 of the base (r) = 5.6 cm       Slant height of the cone (l) = 10 cm. We know, Curved surface area of the cone (CSA) = πrl                 .....

Solution: Given,        Volume of the pyramid (V) = 216 cm3        Height of the pyramid (h) = 18 cm We know,       Volume of pyramid (V) = 1/3 ⨉ Area of base ⨉ height of the .....

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,     .....

Solution: Given,         Area of the base of pyramid (A) = 64 cm2         Area of the triangular faces on each side = 24 cm2         So, area of the 4 triangular faces (LSA)  = 4 ⨉ .....

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, Total surface area .....

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,       .....

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,          Radius of the circular base (r) = 7 cm          Height of the cone (h) = 12 cm. We know,         Volume of the cone (V) = 1/3 πr2h     .....

Solution: Here,         Volume of the pyramid (V) = 350 cm3         Base area of the pyramid (A) = 105 cm2 We know,            Volume of the pyramid (V) = 1/3 ⨉ Area of base .....

Solution: Here,          Volume of the pyramid (V) = 240 cm3          Height of the pyramid (h) = 18 cm. We know,           Volume of the pyramid (V) = 1/3 ⨉ Area of base .....

Solution: Here,           Length of the square base of pyramid (a) = 9 cm           Area of the base of the pyramid (A) = a2                     .....