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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 + .....