Question

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Answer 1

The length of ED=2 cm.

• Given that AD || BC. AC and BD intersect at E.
• AE : EC = 1:5 and BD = 12cm.
• Now we check the similarity of triangles,

In ΔAED and ΔCEB,

∠AED = ∠CEB  (Opposite angles)

∠EAD =∠ECB    (Alternate angles)

∠EDA = ∠EBC   (Alternate angles)

• Therefore the triangles are similar by AAA test of similarity.
• ΔEAD ~ ΔECD
• Now, By properties of similar triangles

The ratio of sides are all in same proportion.

$\frac{AE}{CE} = \frac{ED}{EB}= \frac{AD}{CB}$      ......(Equation 1)

• Let, $\frac{AE}{CE} = \frac{ED}{EB}= \frac{AD}{CB}$ = x

But AE:EC is given as 1:5 therefore, x=1/5

• Now we have to find the length of DE,

We take the ratio as,

$\frac{AE}{CE} = \frac{ED}{EB}= \frac{AD}{CB}$ = $\frac{1}{5}$

• $\frac{ED}{EB}$ = $\frac{1}{5}$
• Solving the above equation ,

Taking reciprocals,

$\frac{EB}{ED}$ = $\frac{5}{1}$

• Applying componendo on both sides,

$\frac{EB+ED}{ED}$ = $\frac{5+1}{1}$

$\frac{BD}{ED}$ = $\frac{6}{1}$

But BD=12cm(given), we substitute in the above equation

$\frac{12}{ED}$ = $\frac{6}{1}$

ED = 2 cm.