A joker's cap is in the form of a right circular.... -Maths 9th

A joker's cap is in the form of a right circular.... -Maths 9th
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Answer :

Radius of the base of the conical cap (r) = 7 cm  Height of the conical cap (h) = 24 cm  Let 'l' be the slant height of the conical cap. Then l = root under( √r2 + h2) = root under( √72 + 242)  l =  root under( √625) = 25 Area of the sheet required for a cap = curved surface area of a conical cap = 22/7 x 7 x 25 cm2  = 550 cm2  ∴ Sheet required for 10 caps = 550 x 10 cm2  = 5500 cm2
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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.

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.

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.

What is the number of surfaces of a right circular cylinder ? -Maths 9th

Description : What is the number of surfaces of a right circular cylinder ? -Maths 9th

Last Answer : Number of surfaces of right circular cylinder are three.

Find the volume of the right circular... -Maths 9th

Description : Find the volume of the right circular... -Maths 9th

Last Answer : Volume of cone = 1/3 πr2h = 1/3 x 22/7 x (3.5)2 x 12 cm3 = 154 cm3.

The curved surface area of a right circular -Maths 9th

Description : The curved surface area of a right circular -Maths 9th

Last Answer : Curved surface area of cylinder = 2 πrh ⇒ 88 = 2 x 22/7 x r x 14 ⇒ r = 88 x 7/2 x 22 x 14 = 1 ∴ Diameter of the base of cylinder = 2r = 2 x 1 = 2 cm

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

A sphere and a right circular cylinder -Maths 9th

Description : A sphere and a right circular cylinder -Maths 9th

Last Answer : Let the radius of sphere and cylinder be r and h be the height of cylinder. Then according to the question. Volume of sphere = Volume of cylinder ⇒ 4/3πr3 = πr2h ⇒ r = 3/4.h Diameter of the cylinder = ... x 100 = h/2 x 1/h x 100 = 50% Thus, the diameter of the cylinder exceeds its height by 50%.

The volume of a right circular cone is 9856 cmcube. -Maths 9th

Description : The volume of a right circular cone is 9856 cmcube. -Maths 9th

Last Answer : Let the height of the cone be h cm. Radius of the base of the cone (r) = 28/2 cm = 14 cm Volume of the cone = 9856 cm3 ⇒ 1/3πr2h = 9856 ⇒ 1/3 x 22/7 x 14 x 14 x h = 9856 ⇒ h = 9856 x 7 x 3/ ... √196 + 2304) = √2500 ∴ l = 50 cm (iii) Curved surface area of cone = πrl = 22/7 x 14 x 50 = 2200 cm2

Define : Right circular cylinder. -Maths 9th

Description : Define : Right circular cylinder. -Maths 9th

Last Answer : A right circular cylinder is a cylinder that has a closed circular surface having two parallel bases on both the ends and whose elements are perpendicular to its base. It is also called a right ... circular surface is at a fixed distance from a straight line known as the axis of the cylinder.

Define : Right circular cone. -Maths 9th

Description : Define : Right circular cone. -Maths 9th

Last Answer : A right circular cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base. That is, the centre point of the circular base is joined with ... cone is a three-dimensional shape having a circular base and narrowing smoothly to a point above the base.

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

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A right circular cylinder and a right circular cone have equal bases and equal volumes. But the lateral surface area of the right circular cone is -Maths 9th

Description : A right circular cylinder and a right circular cone have equal bases and equal volumes. But the lateral surface area of the right circular cone is -Maths 9th

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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

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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

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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

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A right circular solid cone of maximum possible volume is cut off from a solid metallic right circular cylinder of volume V. -Maths 9th

Description : A right circular solid cone of maximum possible volume is cut off from a solid metallic right circular cylinder of volume V. -Maths 9th

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A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Description : A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Last Answer : Height of cone = 3 times height of cylinder = 3 3 = 9 cm Volume of cylinder = volume of cone r2 = 8 8 r = 8 cm l2 = h2 + r2 = (9)2 + (8)2 l = = 12 cm C.S.A (cone) = = 301.71 cm2

A child consumed an ice-cream of inverted right-circular conical shape from the top and left only 12.5% of the cone for her mother. -Maths 9th

Description : A child consumed an ice-cream of inverted right-circular conical shape from the top and left only 12.5% of the cone for her mother. -Maths 9th

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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

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Is Maria much harder than jokers?

Description : I know jokers but I would like to know Maria, but I find it terribly complicated

Last Answer : It's not terribly complicated, but it's more complicated - and especially more demanding on the ability to combine and track (count) cards. It's definitely better - more fun, more interesting, game.

What is the Joker's back-story?

Description : What is the Joker's back-story?

Last Answer : He was an engineer and bats made him fall down in a certain container of some liquids he was dumped and due to his having some strange clothes he got his skin and hair affectex

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.

An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Description : An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Last Answer : (i) From figure , it is clear that angle between the lines joining stalls C,D and stalls C, B is ∠BCD . Given, ∠DBC = 60° and ∠BAC = 40° CD is a chord of circle. Here, ∠CBD and ... for society. (iii) Other social issues for which such compaigns are required, are old age home, orphanages, etc.

An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Description : An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Last Answer : (i) From figure , it is clear that angle between the lines joining stalls C,D and stalls C, B is ∠BCD . Given, ∠DBC = 60° and ∠BAC = 40° CD is a chord of circle. Here, ∠CBD and ... for society. (iii) Other social issues for which such compaigns are required, are old age home, orphanages, etc.

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

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

30 circular plates, each of radius 14 cm -Maths 9th

Description : 30 circular plates, each of radius 14 cm -Maths 9th

Last Answer : Height of the cylinder formed (h) = 30 x 3 = 90 cm Radius of the base of the cylinder formed (r) = 14 cm (i) Total surface area of the cylinder = 2 πr(r + h) = 2 x 22/7 x 14(14 + 90) = 2 x 22/7 x 14 x 104 = 9152 cm2 (ii) Volume of the cylinder formed = πr2h = 22//7 x 14 x 14 x 90 = 55440 cm3

A semi-circular sheet of metal of diameter 28 cm -Maths 9th

Description : A semi-circular sheet of metal of diameter 28 cm -Maths 9th

Last Answer : When semi-circular sheet is bent to form an open conical cup, the radius of the sheet becomes slant height of the cup and the semi-circular part of the sheet becomes the circumference of the base of the cone. ∴ Slant height of the ... 3 x 22/7 x 7 x 7 x 7√3 = 1078/3.√3 = 359.3 x 1.732 = 622.31 cm3

If a circular sheet of perimeter 2πr touching each side of a given quadrilateral sheet of perimeter 2p -Maths 9th

Description : If a circular sheet of perimeter 2πr touching each side of a given quadrilateral sheet of perimeter 2p -Maths 9th

Last Answer : Let ABCD be the quadrilateral from which a circular sheet is cut off touching each side of the quadrilateral. Also, given AB + BC + CD + DA = 2p ...(i) Circumference of the circle = 2πr ⇒ ... = pr.∴ Required remaining area = Area of quadrilateral - Area of circle = pr - πr2 = r(p - πr).

A goat is tethered to one end of a rope of length 20m, while the other end is fixed at the centre of a large circular field. -Maths 9th

Description : A goat is tethered to one end of a rope of length 20m, while the other end is fixed at the centre of a large circular field. -Maths 9th

Last Answer : (b) 3.5 x 100πThe goat can graze the area CABEODC + quarter circle CDF + quarter circle FEB= \(rac{3}{4}\)x π x (20)2 + \(rac{1}{4}\) x π x (10)2 + \(rac{1}{4}\) x π x (10)2 = 300π + 25π + 25π= 3.5 x 100π.

There are 10 persons who are to be seated around a circular table. Find the probability that two particular persons will always sit together. -Maths 9th

Description : There are 10 persons who are to be seated around a circular table. Find the probability that two particular persons will always sit together. -Maths 9th

Last Answer : Total number of ways in which 10 person can sit around a circular table = 9! (∵ We shall keep one place fixed and the rest of the 9 places will be filled in (9 8 7 6 5 4 3 2 1) ways asthere is ... probability = \(rac{2 imes8!}{9!}\) = \(rac{2 imes8!}{9 imes8!}\) = \(rac{2}{9}.\)

In how many ways can 8 people sit around a circular Table? (a) 5040 (b) 40320 (c) 20160 (d) 2520 -Maths 9th

Description : In how many ways can 8 people sit around a circular Table? (a) 5040 (b) 40320 (c) 20160 (d) 2520 -Maths 9th

Last Answer : As is known in case of circular permutations, we keep one place fixed, so 8 people can sit around a circular table in (8 – 1) ! ways = 7! ways = 5040 ways.

There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

Description : There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

Last Answer : Since the chairs are numbered, so they are distinguishable. Therefore 3 boys can be arranged on 3 alternate chairs in 3! ways. 3 girls can be arrenged in 3! ways Also, the girls can be seated before the boys. Total number of required ways = 3! × 3! + 3! × 3! = 2 × (3!)2

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

Description : 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

Last Answer : answer:

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

What kind of fallacy is the following: "I wore this exact same baseball cap when I got an "A" last week and the week before. If I wear my lucky baseball cap, I will get "A"s on all my tests." a) Circular Reasoning b) False Analogy c) Ad Hominem Argument d) Faulty Cause-Effect Relationship e) None of the above

Description : What kind of fallacy is the following: "I wore this exact same baseball cap when I got an "A" last week and the week before. If I wear my lucky baseball cap, I will get "A"s on all my ... Reasoning b) False Analogy c) Ad Hominem Argument d) Faulty Cause-Effect Relationship e) None of the above

Last Answer : d) Faulty Cause-Effect Relationship

"I wore this exact same baseball cap when I got an "A" last week and the week before. If I wear my lucky baseball cap, I will get "A"s on all my tests." a) Circular Reasoning b) False Analogy c) Ad Hominem Argument d) Faulty Cause-Effect Relationship e) None of the above

Description : "I wore this exact same baseball cap when I got an "A" last week and the week before. If I wear my lucky baseball cap, I will get "A"s on all my tests." a) Circular Reasoning b) False Analogy c) Ad Hominem Argument d) Faulty Cause-Effect Relationship e) None of the above

Last Answer : d) Faulty Cause-Effect Relationship

The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Description : 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : Solution: Given that, Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O. To prove that, The Quadrilateral ABCD is a square. Proof, In ΔAOB and ΔCOD, AO = ... right angle. Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square. Hence Proved.

4. Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Description : 4. Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Last Answer : Solution: Let ABCD be a square and its diagonals AC and BD intersect each other at O. To show that, AC = BD AO = OC and ∠AOB = 90° Proof, In ΔABC and ΔBAD, AB = BA (Common) ∠ABC = ∠BAD = ... = ∠COB ∠AOB+∠COB = 180° (Linear pair) Thus, ∠AOB = ∠COB = 90° , Diagonals bisect each other at right angles

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

Description : 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral whose diagonals bisect each other at right angles. Given that, OA = OC OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90° To show that, if the ... a parallelogram. , ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle. Hence Proved.

In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th

Description : In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th

Last Answer : The sum of angles of a triangle is180° If one aangke is of 90° then the sum of two angles is 90° It means that the angle forming 90° is biggest angle We know , Angle opposite to the longest side is largest. It means hypotenuse is the biggest side of right angled triangle

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

If the diagonals of a quadrilateral bisect each other at right angles , then name the quadrilateral . -Maths 9th

Description : If the diagonals of a quadrilateral bisect each other at right angles , then name the quadrilateral . -Maths 9th

Last Answer : Quadrilateral will be Rhombus .

In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

Description : In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

Last Answer : Given : A parallelogram ABCD such that the bisectors of adjacent angles A and B intersect at P. To prove : ∠APB = 90° Proof : Since ABCD is a | | gm ∴ AD | | BC ⇒ ∠A + ∠B = 180° [sum of consecutive interior ... 90° + ∠APB + ∠2 = 180° [ ∵ ∠1 + ∠2 = 90° from (i)] Hence, ∠APB = 90°