Question

in the figure ab parallel to CD parallel to EF if ab is equal to 6 cm is equal to 10 cm BD is equal to 4 cm CD is equal to X cm and d is equal to 5 cm then the value of x + Y to the nearest cm is -​

Answer 1

x is 6.67 cm and y is 3.75

Given:

From the data given, AB || CD || EF.

AB = 6cm, EF = 10 cm, BD = 4 cm, CD= x cm and DE = y cm

To find:

The value of X and Y

Solution:

Taking Δ ADB and Δ EDF,

Since they are vertical angles, ∠ ADB = ∠EDF

By alternate angles rule, ∠ ABD = ∠ FED and ∠BAD = ∠ EFD

Hence, by using the method of ASA,

we can say that the two triangles namely Δ ADB and Δ EDF are similar.

So,

$\frac{BD}{DE}$ = $\frac{AB}{FE}$

∴$\frac{4}{y} = \frac{6}{10}$

∴ 6y = 40 (by cross multiplication)

y = $\frac{40}{6} = \frac{20}{3}$

∴ y = 6.67 cm

Now, taking the triangles of ΔADE similar to Δ CDE,

$\frac{DE}{BE} = \frac{DC}{BA}$

∴ $\frac{6.67}{10.67} = \frac{x}{6}$

∴ x = 6 x $\frac{6.67}{10.67}$

∴ x = 3.75 cm