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256cmsquare

Bhai sidha sa (9 - r) + (12 - r) = 15 rakho simple

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ED=EO=c→ c=√5→ OD= Radio =r=√5√2=√10→ EB=√[(√5)²+(√10)²] =√15 → Potencia de E respecto a la circunferencia =(√5)²=5=CE*√15→ CE=5/√15=(√15)/3 → CB=EB+EC =√15+(√15)/3 =(4√15)/3. Gracias y un saludo cordial.

9 - r + 12 - r = 15 21 - 2r = 15 2r = 6 r = 3

54=18R=>R=3=>pai.R^2=9.pai sq.unit is ar.of circle when pai=22/7

r=9+12-21=6/2=3 So that Arean=πr2=9π✓ Aise short cut sikhiye sir tb class lijiye

good

3cm

Good

the radius of incircle = area/ semi parameter

Why you use killer Hasina's picture?

12*20=240=BD(14+FH) = 10*24=FH(14+BD)---> 14BD=14FH ---> BD=FH ---> 240=BD(14+BD)---> BD²+14BD-240=0---> BD=10=FH---> BH=10+14+10=34. Gracias y saludos

9п un.²

15-(12-r)=9-r ; 15-12+r=9-r ; 2r=6; r=3

How do you know AB is the radius of arc BC?

196/5=39.20...... May be....... ^=read as to the power *=read as square root #=read as triangle As per question In #ABC, angle B=90 AD=7, CD=14,DEII BC So, AE/BE=AD/CD=7/14=(1/2) Let AE=R So, R/BE=(1/2) BE=2R....... EQN1 Again DFll AB CF/BF=CD/AD=14/2=2 CF/2R=2 CF=2×2R=4R... EQN2 In #ADE, angle AFD=90 So, AE^2+DE^2=7^2=49 R^2+(2R)^2=49 R^2+4R^2=49 5R^2=49 R^2=49/5 R=*(49/5)=7/*5 2R=2×(7/*5)=14/*5 Area of the square =(2R)^2=4R^2 So, 4R^2=4×(7/*5)^2 =4×(49/5) =196/5=39.20.....

1×1=1×3.14159268

2.4×3=7.2÷2.4142135624=2.9823×2.9823=8.89433×3.14169268=27.95

Good. நன்று

Aapka solution galat hai

🫡

radius r = (P+B-H)/2, now area of circle = TT x r^2

Thanks easy. CV connect o to b . So 3r/2+4r/2=6 ==>,r=12/7 and area =4.53

A=5 B= 1 Next 1.5

ar. triangle=18R=>18R=54=>R=3.=>ar circle=9.pai when pai=22/7.ans

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In Base , two parts : r and (4-r) In Perpendicular , two parts : r. and (3-r) In Hypotenuse ,two parts : (3-r) and {5-(3-r)} {5-(3-r)} = (4-r) r= 1 Area= Π

In Base , two parts : r and 12-r , In Perpendicular , two parts : r and 9-r In Hypotenuses , two parts , (9-r ) and {15-(9-r)} Now {15-(9-r)} =12-r r= 3 Area = 9Π

^= read as to the power *=read as square root #= read as area of the triangle @= read as triangle # of the right angle @ =(1/2)× 9×12=54 S=(9+12+15)/2=36/2=18 Radious of the circle (r)= (#@)/S =54/18 =3 Aera of the circle =π(r^2) =π(3^2)=π×9=9π

El área roja equivale a la de una corona circular creada al desplazar A hasta B---> Área roja =(8/2)²π =16π. Gracias y saludos

9π

For one fourth of area: Ar+2Ab=25; Ar+Ab=(25/4)π; Ab=25(1-π/4); Ar=(25/4)π-25(1-π/4)=25(π/2-1)=14.2699... square units; Total red area is four times this amount: 57.0796... square units.

Area roja =2*[(5²π)-(10²/2)] =2*(25π-50) =50(π-2) ≈57,08 ud². Gracias y saludos.

Antah vritt ki trijya, r,3,hi hoga

1/x^2=1/39^2+1/52^2 So x=31.2.❤

✓(24^2 + 32^2) = 40 (24)(32)/2 = 384 384 - 384 (3/4)^2 = 384 (7/16) = 168

❤

60-4=56sqwarooth 7.4833

8×8-2×2=60sqrtoth =7.7459666

ar.of circle=9.pai where pai=22/7 as radius of circle=3 unit.

Trazamos el rectángulo ABCG y líneas horizontal y vertical por D y obtenemos una matriz de cuatro celdas que denominaremos por su vértice de esquina---> Áreas: B=2*14=28 ; C=(2*32-28)=36 ; A=2a ; ABCG=2(14+32+a)=92+2a ; G=(ABCG)-B-C-A=28 ---> G/A=C/B---> Área roja =a =28*28/2*36=98/9 =10,8888... Gracias y saludos

8*3=24 ; 8*4=32---> 8*5=40---> Razón de semejanza entre DBE y ABC =s=(40-10)/40=3/4---> s²=9/16---> Área sombreada ADEC =(24*32/2)*[1-(9/16)] =168. Gracias y saludos

168