The outer and inner diameters of a circular pipe are 6 cm and 4 cm respectively. If its length is 10 cm, then what is the total surface -Maths 9th

The outer and inner diameters of a circular pipe are 6 cm and 4 cm respectively. If its length is 10 cm, then what is the total surface -Maths 9th
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A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. -Maths 9th

Description : A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. -Maths 9th

Last Answer : Let r1 and r2 Inner and outer radii of cylindrical pipe r1 = 4/2 cm = 2 cm r2 = 4.4/2 cm = 2.2 cm Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm (i) curved surface ... CSA of roller = (500 31680) cm2 = 15840000 cm2 = 1584 m2. Therefore, area of playground is 1584 m2. Answer!

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Description : A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Last Answer : Inner radius of hemispherical bowl = 5cm Thickness of the bowl = 0.25 cm Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of ... 22/7) (5.25)2 = 173.25 Therefore, the outer curved surface area of the bowl is 173.25 cm2.

In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Description : In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Last Answer : Height of cylindrical pipe = Length of cylindrical pipe = 28m Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of ... = 2 (22/7) 0.025 28 m2 = 4.4m2 The area of the radiating surface of the system is 4.4m2.

The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Description : The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

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The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Description : The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Last Answer : This answer was deleted by our moderators...

. The inner wall of a furnace is at a temperature of 700°C. The composite wall is made of two substances, 10 and 20 cm thick with thermal conductivities of 0.05 and 0.1 W.m-1.°C-1 respectively. The ambient air is at 30°C and the heat transfer co-efficient between the outer surface of wall and air is 20 W.m-2 .°C-1 . The rate of heat loss from the outer surface in W.m-2 is (A) 165.4 (B) 167.5 (C) 172.5 (D) 175

Description : . The inner wall of a furnace is at a temperature of 700°C. The composite wall is made of two substances, 10 and 20 cm thick with thermal conductivities of 0.05 and 0.1 W.m-1.°C-1 respectively. The ambient air is at 30°C ... from the outer surface in W.m-2 is (A) 165.4 (B) 167.5 (C) 172.5 (D) 175

Last Answer : (A) 165.4

The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively -Maths 9th

Description : The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively -Maths 9th

Last Answer : answer:

The external length, breadth and height of a closed box are 10cm, 9cm and 7cm respectively. The total inner surface area of the box is 262 sq.cm. If the walls of the box are of uniform thickness “t” cm, then “t” equals a) 1cm b) 23/3 cm c) 28/9 cm d) 42.5 cm

Description : The external length, breadth and height of a closed box are 10cm, 9cm and 7cm respectively. The total inner surface area of the box is 262 sq.cm. If the walls of the box are of uniform thickness “t” cm, then “t” equals a) 1cm b) 23/3 cm c) 28/9 cm d) 42.5 cm

Last Answer : a) 1cm

Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

Bottom bars under the columns are extended into the interior of the footing slab to a distance greater than (A) 42 diameters from the centre of the column (B) 42 diameters from the inner edge of the column (C) 42 diameters from the outer edge of the column (D) 24 diameters from the centre of the column

Description : Bottom bars under the columns are extended into the interior of the footing slab to a distance greater than (A) 42 diameters from the centre of the column (B) 42 diameters from the inner edge of ... ) 42 diameters from the outer edge of the column (D) 24 diameters from the centre of the column

Last Answer : Answer: Option C

The ratio of strength of a hollow shaft to that of a solid shaft subjected to torsion if both are of the same material and of the same outer diameters, the inner diameter of hollow shaft being half of the outer diameter is a. 15/16 b. 16/15 c. 7/8 d. 8/7

Description : The ratio of strength of a hollow shaft to that of a solid shaft subjected to torsion if both are of the same material and of the same outer diameters, the inner diameter of hollow shaft being half of the outer diameter is a. 15/16 b. 16/15 c. 7/8 d. 8/7

Last Answer : a. 15/16

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find -Maths 9th

Description : The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find -Maths 9th

Last Answer : Radius of well = (r) = 3.5/2 m Depth of well = (h) = 10 m (i) Inner curved surface area of well = 2 πrh = 2 x 22/7 x 3.5/2 x 10 = 110 m2 (ii) Cost of plastering 1 m2 = ₹ 40 ∴ Cost of plastering 110 m2 = ₹110 X 40 = ₹4400

The volume of the metal of a cylindrical pipe is 748 cm^3. The length of the pipe is 14 cm and its external radius is 9 cm -Maths 9th

Description : The volume of the metal of a cylindrical pipe is 748 cm^3. The length of the pipe is 14 cm and its external radius is 9 cm -Maths 9th

Last Answer : answer:

Large diameter sewers subjected to external pressure alone, are reinforced A. Near the inner surface of the pipe B. Near the outer surface of the pipe C. Both A. and B. D. With elliptical cage

Description : Large diameter sewers subjected to external pressure alone, are reinforced A. Near the inner surface of the pipe B. Near the outer surface of the pipe C. Both A. and B. D. With elliptical cage

Last Answer : ANS: D

A hollow prismatic beam of circular section is subjected to a torsional moment. The maximum shear stress occurs at (a) inner wall of cross section (b) middle of thickness (c) outer surface of shaft (d) none of these

Description : A hollow prismatic beam of circular section is subjected to a torsional moment. The maximum shear stress occurs at (a) inner wall of cross section (b) middle of thickness (c) outer surface of shaft (d) none of these

Last Answer : (c) outer surface of shaft

Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Last Answer : Length of cuboid (l) = 80 cm Breadth (b) = 40 cm Height (h) = 20 cm (i) ∴ Lateral surface area = 2h(l + b) = 2 x 20(80 + 40) cm² = 40 x 120 = 4800 cm² (ii) Total surface area = 2(lb ... x 40 + 40 x 20 + 20 x 80) cm² = 2(3200 + 800 + 1600) cm² = 5600 x 2 = 11200 cm²

For a given Reynold number as d/D for an orifice increases, Cd will (where, d & D are orifice & pipe diameters respectively). (A) Increase (B) Decrease (C) Remain constant (D) Either (A) or (B); depends on other factors

Description : For a given Reynold number as d/D for an orifice increases, Cd will (where, d & D are orifice & pipe diameters respectively). (A) Increase (B) Decrease (C) Remain constant (D) Either (A) or (B); depends on other factors

Last Answer : (A) Increase

The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.

Find the resistance of a hollow cylindrical conductor of length 1.0m and inner and outer radii 1.0mm and 2.0mm respectively. The resistivity of the ma

Description : Find the resistance of a hollow cylindrical conductor of length 1.0m and inner and outer radii 1.0mm and 2.0mm respectively. The resistivity of the ma

Last Answer : Find the resistance of a hollow cylindrical conductor of length 1.0m and inner and outer radii 1.0mm and 2 ... the material is `2.0xx10^(-8)(Omega)m`.

The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Description : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Last Answer : s= 2 a+b+c​ = 2 35+54+61​ =75 Area, A= s(s−a)(s−b)(s−c)​ = 75(75−35)(75−54)(75−61)​ =420 5​ cm 2 Now, Area of the triangle is also given as A= 2 1​ ×a×h Where, h is the longest altitude. Therefore, 2 1​ ×a×h=420 5​ Hence, h=24 5​ cm

The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Description : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Last Answer : The length of its longest altitude

What is the shear stress acting on the outer surface of a hollow shaft subjected to a torque of 100 Nm?(The inner and outer diameter of the shaft is 15 mm and 30 mm respectively.) a. 50.26 N/mm2 b. 40.24 N/mm2 c. 20.120 N/mm2 d. 8.74 N/mm2

Description : What is the shear stress acting on the outer surface of a hollow shaft subjected to a torque of 100 Nm?(The inner and outer diameter of the shaft is 15 mm and 30 mm respectively.) a. 50.26 N/mm2 b. 40.24 N/mm2 c. 20.120 N/mm2 d. 8.74 N/mm2

Last Answer : c. 20.120 N/mm2

A hollow circular pipe with non-reflective inner surface is cut into 5 pieces and arranged as shown in the figure below. If it is illuminated from below using a torch as shown in the figure, what would be the profile of the light falling on the sheet kept above?

Description : A hollow circular pipe with non-reflective inner surface is cut into 5 pieces and arranged as shown in the figure below. If it is illuminated from below using a torch as shown in the figure, what would be the profile of the light falling on the sheet kept above?

Last Answer : A

Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. -Maths 9th

Description : Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. -Maths 9th

Last Answer : The lines are perpendicular to a common line AB. Hence, the angle between these lines will be 90+90=180∘ Since, the angle between them is 180∘, the lines are parallel to each other.

Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. -Maths 9th

Description : Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. -Maths 9th

Last Answer : To draw a line perpendicular to AB through A and B, respectively. Use the following steps of construction. 1.Draw a line segment AB = 4 cm. 2.Taking 4 as centre and radius more than ½ AB (i.e ... [since, it sum of interior angle on same side of transversal is 180°, then the two lines are parallel]

In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Description : In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Last Answer : Solution :- Jion AB ∠ABD = 90° (Angle in a semicircle) Similarly, ∠ABC = 90° So, ∠ABD + ∠ABC = 90° + 90° = 180° Therefore,DBC is a line i.e.,B lies on the segment DC.

If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Description : If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Last Answer : 20.8 cm Let the edge of the single cube be ‘a’ cm. Then, total volume melted = Volume of cube formed ⇒ (6)3 + (8)3 + (10)3 = a3 ⇒ a3 = 216 + 512 + 1000 = 1728 ... Diagonal of the new cube = 3–√a=(3–√×12)3a=(3×12) cm = 20.8 cm (approx.)

If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π cm3

If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π cm3

50 circular plates each of radius .... Find its total surface area -Maths 9th

Description : 50 circular plates each of radius .... Find its total surface area -Maths 9th

Last Answer : 50 circular plate diameter of plate =14cm Radius of plate = d/2=214​=7cm thickness =0.5cm → Height of the cylinder = No. of plates × thickness of plate =50×0.5 =25cm → total surface Area = curved surface + 2 × area of base =2πr.h+2×πr2 =2πr(h+r) =2π×7(25+7) =2×722​×7×32 =1408cm2

The magnitude of the volume of a closed right circular cylinder of unit height divided by the magnitude of the total surface area of the -Maths 9th

Description : The magnitude of the volume of a closed right circular cylinder of unit height divided by the magnitude of the total surface area of the -Maths 9th

Last Answer : answer:

Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Description : Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Last Answer : When two cubes are joined end to end, then Length of the cuboid = 6 + 6 = 12 cm Breadth of the cuboid = 6 cm Height of the cuboid = 6 cm Total surface area of the cuboid = 2 (lb + bh + hi) = 2(12 x6 + 6×6 + 6×12) = 2(72 + 36 + 72) = 2(180) = 360 cm2

Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Description : Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Last Answer : When two cubes are joined end to end, then Length of the cuboid = 6 + 6 = 12 cm Breadth of the cuboid = 6 cm Height of the cuboid = 6 cm Total surface area of the cuboid = 2 (lb + bh + hi) = 2(12 x6 + 6×6 + 6×12) = 2(72 + 36 + 72) = 2(180) = 360 cm2

Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Last Answer : Edge of cube (a) = 10 cm (i) ∴ Lateral surface area = 4a² = 4 x (10)² = 4 x 100 cm²= 400 cm² (ii) Total surface area = 6a² = 6 x(10)² cm² = 6 x 100 = 600 cm²

Find the total surface area of a hemisphere of radius 10 cm -Maths 9th

Description : Find the total surface area of a hemisphere of radius 10 cm -Maths 9th

Last Answer : Radius of hemisphere, r = 10cm Formula: Total surface area of hemisphere = 3πr2 = 3×3.14×102 = 942 The total surface area of given hemisphere is 942 cm2.

A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Description : If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Last Answer : Let one of the remaining sides be x cm.Then, other side = \(\sqrt{5^2-x^2}\) cm∴ Area = \(rac{1}{2} imes{x} imes\sqrt{25-x^2}\) = 6⇒ \(x\sqrt{25-x^2}\) = 12 ⇒ x2(25 - x2) = 144⇒ 25x2 - x4 = 144 ⇒ x4 - 25x2 ... (x2 - 16) (x2 - 9) = 0 ⇒ x2 = 16 or x2 = 9 ⇒ x = 4 or 3∴ The two sides are 4 cm and 3 cm.

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral,prove that it is a rectangle. -Maths 9th

Description : If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral,prove that it is a rectangle. -Maths 9th

Last Answer : Solution :- Let, ABCD be a cyclic quadrilateral such that its diagonal AC and BD are the diameters of the circle though the vertices A,B,C and D. As angle in a semi-circle is 900 ∴ ∠ABC = 900 and ∠ADC = 900 ∠DAB = 900 ... Hence, ABCD is a rectangle.

AC and BD are chords of a circle that bisect each other. Prove that AC and BD are diameters and ABCD is a rectangle. -Maths 9th

Description : AC and BD are chords of a circle that bisect each other. Prove that AC and BD are diameters and ABCD is a rectangle. -Maths 9th

Last Answer : Solution :- Let AC and BD bisect each other at point 0. Then, OA = OC and OB = OD In triangles AOB and COD, we have OA = OC OB = OD and ∠ AOB = ∠ COD (Vertically opposite angles) ∴ △ AOB ... ∠ADC Also, ∠BAD = 90° = ∠BCD Also, AB = CD and BC = DA (Proved above) Hence, ABCD is a rectangle.

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm

Nominal Pipe Size (NPS) of a pipe less than 12 inches in diameter indicates its(A) Inner diameter(B) Outer diameter(C) Thickness(D) Neither inner nor outer diameter

Description : Nominal Pipe Size (NPS) of a pipe less than 12 inches in diameter indicates its (A) Inner diameter (B) Outer diameter (C) Thickness (D) Neither inner nor outer diameter

Last Answer : (D) Neither inner nor outer diameter

Nominal size of a pipe is an indication of its __________ diameter. (A) Inner (B) Outer (C) Approximate (D) None of these

Description : Nominal size of a pipe is an indication of its __________ diameter. (A) Inner (B) Outer (C) Approximate (D) None of these

Last Answer : (C) Approximate

The base and the corresponding altitude of a parallelogram are 10 cm and 7 cm, respectively. Find its area. -Maths 9th

Description : The base and the corresponding altitude of a parallelogram are 10 cm and 7 cm, respectively. Find its area. -Maths 9th

Last Answer : Area of parallelogram = Base × Corresponding altitude = 10 × 7 = 70 cm2.

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC. We have, ABC is a right angled triangle at B. Now, AC2=AB2+BC2 [by Pythagoras theorem]Now, AC2=AB2+BC2 ... =24(1+6-√)cm2=24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6-√−−−−−−√cm2241+6cm2 .

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC.

A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Description : A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Last Answer : Answer We have, r=0.7m, h=8m ∴ Total surface area = 2πr2+2πrh=2πr(r+h)=2×722​×0.7×8.7m2 Required cost = Rs. {2×722​×0.7×8.7×10}=Rs.382.80

If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th

Description : If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th

Last Answer : answer:

The figure below shows the structure of a mitochondrion with its four parts labelled A, B, C and D. Select the part correctly matched with its function. D B A C (a) D (outer membrane) – gives rise to inner membrane by splitting (b) B (inner membrane) – forms infoldings called cristae (c) C (crista) – possesses single circular DNA molecule and ribosomes (d) A (matrix) – major site for respiratory chain enzyme

Description : The figure below shows the structure of a mitochondrion with its four parts labelled A, B, C and D. Select the part correctly matched with its function. D B A C (a) D (outer ... possesses single circular DNA molecule and ribosomes (d) A (matrix) - major site for respiratory chain enzyme

Last Answer : (b) B (inner membrane) – forms infoldings called cristae