Question

plz plz..... tell fast this answer 3,4 sums

Answer 1

question no.(3) , Given Angles are

A = 2y + 4° , B = 6x - 4°

C = 4y - 4° , angle D = 7x + 2°

sum of opposite angles of given cyclic

quadrilateral will be equal = 180°

Angle A + angle C = 180°

2y + 4° + 4y - 4° = 180°

6y = 180° => y = 30°

Angle A = 2y + 4° = 2(30) + 4 = 64°

amgleC = 4y - 4° = 4(30) - 4° = 116°

AngleB + AngleD = 180°

6x - 4° + 7x + 2° = 180°

13x = 182 => x = 14

thus

AngleB = 6x - 4 = 6(14) - 4

AngleB= 80°

and AngleD = 7x + 2° = 7(14) + 2= 100°

angle D = 100°

Answer:all the angles of cyclic quadrilateral ABCD are A = 64°,B = 80°,
C = 116°, D = 100°

question.no (4):
-------------------------

given AB is parallel to CD

AB = 10cm , CD = 24cm

in quadrilateral ABCD ,

radius of circle = 13cm = OC = OA

CD = 24cm,then CE = 24 / 2 = 12cm=ED

in triangle COE,

(CO)^2 = (CE)^2 + (OE)^2

(13)^2 = (12)^2 + (OE)^2

OE^2 = 169 - 144 = 25

=> OE = √25 = 5cm

so thus , OE = 5cm

now in triangleAOF

since, AB = 10cm ,AF = 10/ 2 = 5cm = BF

(AO)^2 = (OF)^2 + (AF)^2

(13)^2 = (OF)^2 + (5)^2

(OF)^2 = 169 - 25 = 144

OF = √144 = 12cm

therefore,

the shortest distance between AB and CD

EF = OE + OF

EF = 5cm + 12cm = 17cm

Answer: shortest distance = 17cm