Question

answer plz... the area of trapezium ​

Answer 1

Answer is 30*root 5 hope it helps you

Answer 2

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YOUR ANSWER:-

In triangle BCO,

${17}^{2} = {8}^{2} + {a}^{2}$

$289 = 64 + {a}^{2}$

$289 - 64 = {a}^{2}$

${a}^{2} = 225$

$a = \sqrt{225}$

$a = 15 \: cm$

Area of triangle BCO,

$(1 \div 2)bh$

$= (1 \div 2)8 \times 15$

$= 60 \: {cm}^{2}$

Area of rectangle AOCD,

$l \times b$

$= 6 \times 15$

$= 90 \: {cm}^{2}$

Area of trapezium...

Area of triangle + Area of rectangle

$= 60 + 90$

$= 150 {cm}^{2}$

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