Question

Calculate the area of as shown in figure​

Answer 1

Answer:

Length of CO :

17² = 8²+x²

289 = 64 + x²

289-64 = x²

225 = x²

√225 = x

15 = x

Area of AOCD = 6×15 = 90cm²

Area of OBC = 1/2×8×15 = 60cm²

Total area = (90+60)cm² = 150cm²

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Answer 2

Step-by-step explanation:

$\bf\underline{\underline{\red{Given:-}}}$

• ABCD is a trapezium.

• AB // CD

• CO ⊥ AB

• AO = 6cm, OB = 8cm , CB = 17cm and CD = 6cm

$\bf\underline{\underline{\blue{To\:Find:-}}}$

• The area of the trapezium ABCD

$\bf\underline{\underline{\green{Solution:-}}}$

$\rm CO\: is\: drawn\: perpendicular\: to\: AB$

$\rm In \: ∆BOC\: is \: right\: angled \: at \:O$

$\rm By \: applying \: Pythagoras \: theorem\: in \triangle BOC$

$\rm \implies {BC}^{2} = {OB}^{2} + {OC}^{2}$

$\rm \implies {17}^{2} = {8}^{2} + {OC}^{2}$

$\rm \implies {OC}^{2} = {17}^{2} - {8}^{2}$

$\rm \implies {OC}^{2} = 289 - 64$

$\rm \implies {OC}^{2} = 225$

$\rm \implies OC = \sqrt{225}$

$\rm \implies OC = 15cm$

Now:-

• CD = 6cm

• AB = 6 + 8 = 14cm

$\rm The\: two\: parallel\: sides \: are\: 6cm \: and \: 8cm$

$\rm Height \: (OC) = 15cm$

$\boxed{ \rm \leadsto Area \: of \: trapezium = \frac{1}{2} \times (Sum \: of \: parallel \: sides) \times Height }$

$\rm = \dfrac{1}{2} \times ( CD + AO ) \times OC$

$\rm = \dfrac{1}{2} \times ( 6 + 14 ) \times 15$

$\rm = \dfrac{1}{2}\times 20 \times$

$\rm = 10 \times 15$

$\rm = 150cm^{2}$

$\underline{\boxed{\purple {\rm \therefore Area \: of \: trapezium = 150cm^{2}}}}$