Пікірлер

• 
  @jks7856Күн бұрын

  You are doing a very good job. Keep it up. Subscribed your channel. 👍

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  @ronney6121Күн бұрын

  I have a passion for geometry. I solved it in 2 minutes. 😎

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  @ashokgupta310Күн бұрын

  Very beautiful question

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  @hklrao2 күн бұрын

  Please listen to the wonderful and motivating address of Prime Minister Modiji on Olympied Mathematics day and his conversation with high rank Maths students and get motivated to learn and promote Maths Dr.LakshmanRao PhD Krishnapuŕi Chennai 600 028

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  @starlordplays12062 күн бұрын

  Add the area of quarter and semi circle

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  @Shri_123563 күн бұрын

  Semi circle and quarter circle ke area ko ek dusre me add karna tha minus nahi 👎🏻

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  @anandharamang32894 күн бұрын

  BE /AB = tan 30 =1/√3 BE= AB/√3= 10/√3 Red Area = 1/2 *AB*(AD+(BC-BE)) = 1/2 * 10*(10+(10-10/√3)) =5*20-50/√3 = 100-50√3/3 =71.132 sq.u

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  @anandharamang32894 күн бұрын

  Draw reflection below and make it a semi circle. Then we will get three points on the circle. Join to make a triangle with sides √12^2+16^2 , 9+9, √(12+9+9)^2+16^2 Sides calculated are 20,18, 34 Radius= abc/4∆, ∆ = 1/2*(9+9)*16 =144 Radius= 20*18*34/(4*144 )= 85/4=21.5 u

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  @pas62954 күн бұрын

  21 units

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  @DR-kz9li4 күн бұрын

  15=(9-r)+(12-4r) r=2

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  @ronney61213 күн бұрын

  r = 6/5 aa rha hai

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  @prossvay874411 күн бұрын

  Answer is 26.44

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  @erwinkurniadi185012 күн бұрын

  This is an example of a difficult way to solve it, even though there are other easy ways

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  @markandeysingh135513 күн бұрын

  How to solve by Geometry, not Trignometry ?

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  @prossvay874413 күн бұрын

  Red area=1/2(12)(6√5)=36√5=80.50

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  @prossvay874413 күн бұрын

  Quarter area=1/4(π)(√10/2)^2=5π/8=1.96 square units

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  @bpark1000113 күн бұрын

  What are you doing?!!!! You only need to translate the segments to form a right triangle 6/8/? AB = √(6² + 8²) = √(36 + 64) = √(100) = 10.

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  @ashwinbapat14 күн бұрын

  sir +class 10 coordinate geometry ke questions bata digiay

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  @JangiLalPatel-wd4do15 күн бұрын

  It is not easy way to solve

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  @sorourhashemi324916 күн бұрын

  Thanks. You did great. My problem is I don't know anything about cos. is it possible to be solved the other way?

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  @amitavadasgupta698516 күн бұрын

  ar of tr.=4×3/2=6 unit=6r when r is radius of inscribed circle.=>r=1 unit make ar of circle=pai.r^1=pai.1=pai when pai=22/7.proved

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  @amitavadasgupta698516 күн бұрын

  AB=(6^2+8^2)^1/2=(36+64)^1/2=100^1/2=10 Unit.ans

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  @salimahmad741416 күн бұрын

  AB = 10 units of length.

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  @erwinkurniadi185018 күн бұрын

  15 x 15 = 36 x BD BD = 6.25 diameter = 36 + 6.25 = 42.25 radius = 21.125 area of the circle = π 21.125² area = 1401.274 cm²

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  @lasalleman679220 күн бұрын

  Find inradius first. Inradius is = to the area of the triangle /semi-perimeter. Here, area of triangle 54/semi-perimeter of 18 = inradius of 3 . Area of circle is thus PI * 3^2 = Area of circle 28.2743.

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  @ManojkantSamal21 күн бұрын

  Given l=13 I found out b=60/13 Area =l×b=13×(60/13)=60 Let BE=x So,AE=13-x In triangle ADE AD^2=12^2-(13-x)^2 =144-(169+x^2-26x) =144-169-x^2+26x In triangle BCE BC^2=5^2-x^2 =25-x^2 We know BC=AD SO, 144-169-x^2+26x=25-x^2 -25+26x=25 26x=50 X=50/26=25/13 In triangle BCE BC^2=25-x^2 =25-(25/13)^2 =25-(625/169) ={(169×25)-625}/169 ={25(169-25)}/169 =(25×144)/169 BC=(5×12)/13 =60/13 Area of rectangle =CD×BC =13×(60/13)=60.... Respected Sir,pranam... I prefer the way your answering

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  @krishnaswamyv-nb5zh21 күн бұрын

  S

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  @krishnaswamyv-nb5zh21 күн бұрын

  IS ANGLE DEC IS 90.’

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  @MeenakshiSinha-dc6uu24 күн бұрын

  Sir 2.77-1 =1.77 not 2.76. So sorry

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  @sreenathreddy917324 күн бұрын

  Triangle EDC and Triangle BEC are similar triangles by Angle Angle Angle. that means BC/12 = 5/13 BC = 60/13... Area of rectanlge = BC * AB = 60 S.Units

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  @amitavadasgupta698524 күн бұрын

  ar of rt triangle=12×5/2=30 sq unit.

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  @amitavadasgupta698524 күн бұрын

  ar. of rectangle=2×ar.of triangle with same base and same parallel.=2×30=60 sq unit.ans

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  @santokhsidhuatla704524 күн бұрын

  Let h is height of Rectangle 12^2-h^2+5^-h^2=13 144+25-13=h^+h^2 156 =2h^2 h=156/2=78^1/2 Area of Rectangle =13 * 78^1/2 =114.813 square units

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  @mariocohen581625 күн бұрын

  This triangle is rectangular because 13^ 2=12^2+5^2 thus Area (5×12):2=30square unit

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  @mariocohen581625 күн бұрын

  We also have if h is the height of triangle relative to the base 13 this height is also the wish of there rectangle since the triangle is rectangular we have the relation 13×h=12×5 h=4,615.Thus Area Rectangle= 4,615×13=60

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  @sorourhashemi324925 күн бұрын

  Area of Triangle=30 square units by Heroin F. Height or width= 4.66 area ofrectangle=60 square units

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  @AbdulJaleel-rd5ul25 күн бұрын

  Area of triangle 30 square unit. So area of rectangle is double double of that.

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  @plamenpenchev26226 күн бұрын

  Square side = 2a, chord theorem: 2×(2a) = a^2, a = 0, a = 4 Then 2r = 2×4 +2, r = 5

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  @AtharvaAgarwal27 күн бұрын

  This is olympiad? Its easier than my school

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  @joysebastianpoovakulam710228 күн бұрын

  wrong, i feel answer is 25.12

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  @sorourhashemi324928 күн бұрын

  Heron result is radical 720; so the area is 26.832. And area of red semi circle 17.242

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  @amitavadasgupta698529 күн бұрын

  ar=4.5 pai.(pai=22/7)

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  @amitavadasgupta698529 күн бұрын

  BD=10-6=4

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  @santokhsidhuatla7045Ай бұрын

  Side of square =a= Ur2xa=10xur2 a=10 Area of square =10x10=100 su Srea of circle=pi/4x100=78.5 su Net area=100-78.5=21.5su Read shaded area=21.5/4=5.375sq unit

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  @PrithwirajSen-nj6qqАй бұрын

  Red area and other three each of the same shaped areas are equal -- it has been said without any logic. This may be discussed also to make the solution a clear one

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  @peterotto712Ай бұрын

  It would be interesting to learn what language you are talking

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  @AbhayPratapSinghRajawat-ch4uzАй бұрын

  but it is is not ,GIVEN THAT DIAGONAL OF SQUARE PASSES THROUGH CENTER OF CIRCLE or RADII JOINING VERITICE OF SMALL AND BIG SQUARE FORM A LINE

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  @deviprasad8993Ай бұрын

  Why centre of circle and small square are collinear.

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  @ashwinbapatАй бұрын

  bhohat sahi sir maja aa gaya .

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  @OverclockingCowboyАй бұрын

  Another problem wherein the red area is not dependent on the radius of the circles with radius R, (R+r)/2, and r.

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  @user-zr1gk7ho6yАй бұрын

  +CircleQuarter(r) = п/4 * 8² = 16п +CircleQuarter(r/2) = 16п/4 = 4п - CircleQuarter(r/4)*2 = 2*4п/4 = -2п RedArea = 16п + 4п - 2п = 18п

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  @plamenpenchev262Ай бұрын

  I solved the same way. But a minute later: As inspected, the area does not depend on three radii. This means that you can make upper semi-circle with r = 0 and obtained a semi-circle (red) with diameter 12. Then S = pi*6^2/2 = 18×pi.

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  @OverclockingCowboyАй бұрын

  Yeah, the red area is constant regardless of the value of r ( r = 6-x ).