Question

In AngleABC,DE//BC and AD/DB =3/5. AC=5.6. .Find AE.​

Answer 1

Answer:

• In triangle ABC
• DE//BC
• we know that
• TRIANGLE ABC = DE//BC =
• AD/DB=AE/EC
• AD/DB=3/5
• I, e AD=3, DB=5
• AE=5.6,EC=X
• AC=AE+EC
• NOW,
• 3/5=5.6/x
• cross multiplication
• $5.6 \times 5 = 3 \times x$
• $56 \div 10 \times 5 = 3x$
• $56 \div 2 = 3x$
• $56 \div 3 \times 2 = x$
• $56 \div 6 = x$
• $x = 9.3$
• EC=X=9.3
• AC=AE+EC
• AC = 5.6+9.3
• AC=14.9

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Answer 2

Answer is in the above attachment.