A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th

A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th
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A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Description : A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Last Answer : Surface area of sphere = 4πr2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r Radius of cylinder = r CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h) = 4πr2 (iii) Ratio ... sphere)/CSA of Cylinder) = 4r2/4r2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1.

In Fig., a right circular cylinder just encloses a sphere of radius r. Find -Maths 9th

Description : In Fig., a right circular cylinder just encloses a sphere of radius r. Find -Maths 9th

Last Answer : (i) Surface areas S1 of the sphere = 4 πr2 (ii) We have Radius of the cylinder = r Height of the cylinder = h = 2r ∴ Curved surface area S2 of the cylinder ... 2 πrh = 2 πr x 2r = 4 πr2 (iii) S1/S2 = 4 πr2/4 πr2 = 1/1 ∴ S1 : S2 = 1 : 1

Find the ratio of surface area and volume of the sphere of unit radius. -Maths 9th

Description : Find the ratio of surface area and volume of the sphere of unit radius. -Maths 9th

Last Answer : Required ratio = 4πr2 / 4/3.πr3 = 3 x 4 x π x (1)2 / 4 x π x (1)3 = 3/1 (Since, r = 1) i.e., 3 : 1

A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area S. -Maths 9th

Description : A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area S. -Maths 9th

Last Answer : answer:

A sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter ? -Maths 9th

Description : A sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter ? -Maths 9th

Last Answer : answer:

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Last Answer : NEED ANSWER

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Last Answer : According to question find the radius of the sphere

Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Description : Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Last Answer : Formula: Surface area of sphere (SA) = 4πr2 (i) Radius of sphere, r = 10.5 cm SA = 4 (22/7) 10.52 = 1386 Surface area of sphere is 1386 cm2 (ii) Radius of sphere, r = 5.6cm Using formula, SA = 4 (22 ... 75 cm Surface area of sphere = 4πr2 = 4 (22/7) 1.752 = 38.5 Surface area of a sphere is 38.5 cm2

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

If the radius of a sphere is 2r, then its volume will be -Maths 9th

Description : If the radius of a sphere is 2r, then its volume will be -Maths 9th

Last Answer : As, r=2r Volume of sphere = 4​/3π(2r)^3 =32/3​πr^3

If the radius of a sphere is 2r, then its volume will be -Maths 9th

Description : If the radius of a sphere is 2r, then its volume will be -Maths 9th

Last Answer : (d) Given, radius of a sphere = 2r Volume of a sphere =4/3 π(Radius)3 = 4/3 π(2r)3 = 4/3 π 8r3 = (32 πr3)/3 cu units Hence the volume of a sphere is (32 πr3)/3 cu units.

The radius of sphere is 2r, then find its volume. -Maths 9th

Description : The radius of sphere is 2r, then find its volume. -Maths 9th

Last Answer : Volume of the sphere = 4/3.π.(2r)3 = 32/3πr3

If the radius of a sphere is doubled... -Maths 9th

Description : If the radius of a sphere is doubled... -Maths 9th

Last Answer : Surface area of sphere = 4πr2 When radius is doubled then new surface area = 4π(2r)2 = 4π x 4r2 = 4(4πr2 ) = 4 x original surface area. ∴​ Surface area becomes 4 tim es.

Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

Description : Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

Last Answer : Let 'r' be the radius of sphere Surface area of sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 154 = 4 x 22/7 x r2 ⇒ r 2 = 154 x 7/4 x 22 = 49/4 ⇒ r = 7/2 cm = 3.5 cm

A cylinder, a cone and a sphere are of the same radius -Maths 9th

Description : A cylinder, a cone and a sphere are of the same radius -Maths 9th

Last Answer : Let r be the common radius of a cylinder, cone and a sphere. Then, height of the cylinder = Height of the cone = Height of the sphere = 2r Let 'I' be the slant height of the cone. Then l = root under( √r2 + h2) = root under( ... , S1 : S2 :S3 = 4 πr2 : √5 πr2 : 4 πr2 ∴ S1 : S2 : S3 = 4 : √5 : 4

The surface area of a sphere of radius 5 cm -Maths 9th

Description : The surface area of a sphere of radius 5 cm -Maths 9th

Last Answer : Radius of the sphere (r1) = 5 cm Radius of the base of cone (r2) = 4 cm Let r сm be the height of the cone. Surface area of sphere = 4 πr2 ⇒ 4 π(5)2 = 100 π cm2 Curved surface area of cone = πrl = 4 πl ... ∴ Volume of cone = 1/3 πr2h = 1/3 x 22/7 x 42 x 3 = 352/7 cm3 = 50.29 cm3 (Approximately)

In a sphere of radius 2 cm a cone of height 3 cm is inscribed. What is the ratio of volumes of the cone and sphere ? -Maths 9th

Description : In a sphere of radius 2 cm a cone of height 3 cm is inscribed. What is the ratio of volumes of the cone and sphere ? -Maths 9th

Last Answer : answer:

A sphere, a cylinder and a cone respectively are of the same radius and same height. Find the ratio of their curved surfaces. -Maths 9th

Description : A sphere, a cylinder and a cone respectively are of the same radius and same height. Find the ratio of their curved surfaces. -Maths 9th

Last Answer : answer:

A cube of side 4cm is painted with 3 colors red, blue and green in such a way that opposite sides are painted in the same color. This cube is now cut into 64 cubes of equal size. ➢ How many have at least two sides painted in different colors. ➢ How many cubes have only one side painted. ➢ How many cubes have no side painted. ➢ How many have exactly one side not painted.

Description : A cube of side 4cm is painted with 3 colors red, blue and green in such a way that opposite sides are painted in the same color. This cube is now cut into 64 cubes of equal size. ➢ How many ... one side painted. ➢ How many cubes have no side painted. ➢ How many have exactly one side not painted.

Last Answer : Here are the answers. Cubes that have at least two sides painted in different colours are 24 + 8 = 32. Cubes that have only one side painted are 24. Cubes that have no side painted = 8. Cubes that have exactly one side not painted = 0.

A cube of side 4cm is painted with 3 colors red, blue and green in such a way that opposite sides are painted in the same color. This cube is now cut into 64 cubes of equal size. 1. How many have at least two sides painted in different colors. 2. How many cubes have only one side painted. 3. How many cubes have no side painted. 4. How many have exactly one side not painted.

Description : A cube of side 4cm is painted with 3 colors red, blue and green in such a way that opposite sides are painted in the same color. This cube is now cut into 64 cubes of equal size. 1. How many ... side painted. 3. How many cubes have no side painted. 4. How many have exactly one side not painted.

Last Answer : Here are the answers. 1. Cubes that have at least two sides painted in different colours are 24 + 8 = 32. 2. Cubes that have only one side painted are 24. 3. Cubes that have no side painted = 8. 4. Cubes that have exactly one side not painted = 0.

There are two identical cubes. Out of one cube, a sphere of maximum volume (VS) is cut off. Out of the second cube, a cone of maximum volume -Maths 9th

Description : There are two identical cubes. Out of one cube, a sphere of maximum volume (VS) is cut off. Out of the second cube, a cone of maximum volume -Maths 9th

Last Answer : answer:

A sphere of radius r is cut from larger sphere of radius R. The distance between their centre is a. The centroid of the remaining valoume lies on the line of centres and the sistanc from the centre of the larger sphere is a.ar2 / (R2 - r2) b.107 dynes c.aR/(R2-r2) d.aR/(R - r) e.ar3 / (R3 - r3)

Description : A sphere of radius r is cut from larger sphere of radius R. The distance between their centre is a. The centroid of the remaining valoume lies on the line of centres and the sistanc from the centre of the larger sphere is a.ar2 / (R2 - r2) b.107 dynes c.aR/(R2-r2) d.aR/(R - r) e.ar3 / (R3 - r3)

Last Answer : e. ar3 / (R3 - r3)

The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Description : The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Last Answer : Total surface area of one brick = 2(lb +bh+lb) = [2(22.5 10+10 7.5+22.5 7.5)] cm2 = 2(225+75+168.75) cm2 = (2 468.75) cm2 = 937.5 cm2 Let n bricks can be painted out by the ... 93750 cm2 So, we have, 93750 = 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.

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

If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -Maths 9th

Description : If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -Maths 9th

Last Answer : Semi-perimeter of triangle (s) = \(rac{3+4+5}{2}\)cm = 6 cm∴ Area of triangle A = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{6 imes3 imes2 imes1}\) cm2 = 6 cm2∴ Radius of circum-circle = \(rac{abc}{4( ext{Area of}\,\Delta)}\) = \(rac{3+4+5}{4 imes60}\) cm = 2.5 cm

If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Description : If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Last Answer : (b) 7.5 cm.Area of a triangle = \(rac{1}{2}\)x base x height= In-radius x semi-perimeter of the Δ \(\big[ ext{Using r =}rac{\Delta}{s}\big]\)Let the sides of triangle be 4x, 5x and 6x respectively. Given: In-radius = 3 ... x \(rac{15x}{2}\) = \(rac{6x}{2}\) x h ⇒ h = \(rac{45}{6}\) = 7.5 cm.

If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Description : If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Last Answer : Let r be the radius of the sphere. and Volume of a sphere = surface area of the sphere ⇒ 4 / 3πr3 = 4πr2 ⇒ r = 3 cm ∴ Diameter of the sphere = 2r = 2 × 3 = 6 cm

If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Description : If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Last Answer : Let r be the radius of the sphere. and Volume of a sphere = surface area of the sphere ⇒ 4 / 3πr3 = 4πr2 ⇒ r = 3 cm ∴ Diameter of the sphere = 2r = 2 × 3 = 6 cm

From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. -Maths 9th

Description : From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. -Maths 9th

Last Answer : answer:

The diameter of a solid mettalic right circular cylinder is equal to its height. After culting out the largest possible solid sphere -Maths 9th

Description : The diameter of a solid mettalic right circular cylinder is equal to its height. After culting out the largest possible solid sphere -Maths 9th

Last Answer : answer:

A sphere and a cone have equal bases. If their heights are also equal, the ratio of their curved surface will be : -Maths 9th

Description : A sphere and a cone have equal bases. If their heights are also equal, the ratio of their curved surface will be : -Maths 9th

Last Answer : answer:

The radius of gyration of a solid sphere of radius r is equal to a.0.8 r2 b.0.2 r2 c.0.3 r2 d.0.4 r2 e.0.5 r2

Description : The radius of gyration of a solid sphere of radius r is equal to a.0.8 r2 b.0.2 r2 c.0.3 r2 d.0.4 r2 e.0.5 r2

Last Answer : d. 0.4 r2

What is The surface area of a sphere can be approximated as follows Surface area 4πr2 where r is the radius of the sphere π is a constant that is roughly equal to 3. Using the simple approximat?

Description : What is The surface area of a sphere can be approximated as follows Surface area 4πr2 where r is the radius of the sphere π is a constant that is roughly equal to 3. Using the simple approximat?

Last Answer : 300 m^2

When a bar magnet is cut into two equal halves, the pole strength of each piece (1) becomes double (2) becomes half (3) becomes zero (4) remains the same

Description : When a bar magnet is cut into two equal halves, the pole strength of each piece (1) becomes double (2) becomes half (3) becomes zero (4) remains the same

Last Answer : (4) remains the same Explanation: The poles remains same whether the magnet is cut into two equal half or more and also pole strength remains same but magnetic moment reduces due to decrease in the length of ... is called north pole (N-poled of the magnet and the other end the south pole (S-pole).

When a bar magnet is cut into two equal halves, the pole strength of each piece – (1) becomes double (2) becomes half (3) becomes zero (4) remains the same

Description : When a bar magnet is cut into two equal halves, the pole strength of each piece – (1) becomes double (2) becomes half (3) becomes zero (4) remains the same

Last Answer : remains the same

If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2