Math, asked by victorimoduoba, 3 months ago

if the area of the trapezium is 30cm square. find the value of y.

Answers

Answered by igibrahim07

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is =

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12)

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6)

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 =

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 =

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108 =6

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108 =6 3

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108 =6 3 So area of figure =6

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108 =6 3 So area of figure =6 3

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108 =6 3 So area of figure =6 3 (

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108 =6 3 So area of figure =6 3 ( 2

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108 =6 3 So area of figure =6 3 ( 230+18

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108 =6 3 So area of figure =6 3 ( 230+18 )=6

Given side AB=30 cm ,DC=18 cm and AD=CB=12 cm Then diff of side AB and DC=30-18=12Then height of figure is = (12) 2 −(6) 2 = 144−36 = 108 =6 3 So area of figure =6 3 ( 230+18 )=6 3

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