Question

In the given figure, AB = 8cm, BC = 3cm and BE is parallel to CD.

(i) FIND—
a) BE/CD ,
b) area of ∆ABC/area of quadrilateral BCDE

(ii) What is the special name given to the quadrilateral BCDE?.​

please answer correctly!​

Answer 1

Answer:

i) a) BE/CD is 8/11.

  b) area of ∆ABC/area of quadrilateral BCDE is 64/57.

ii)  The special name given to the quadrilateral BCDE is Trapezium.

Step-by-step explanation:

i) a)

In ΔACD, AB = 8cm, BC= 3cm and BE ║ CD. (Given)

By Thales Theorem,

$\frac{AB}{BC}$ = $\frac{AE}{ED}$ = $\frac{8}{3}$

∠B = ∠C (Corresponding Angles)

∠E = ∠D (Corresponding Angles)

∠A = ∠A (Common Angle)

By AAA Similarity, ΔABE ~ ΔACD.

∴ $\frac{AB}{AC} = \frac{BE}{CD} = \frac{AE}{AD}$

⇒ $\frac{AB}{AC}= \frac{AB}{AB+BC}= \frac{8}{8+3}=\frac{8}{11}$

∴ $\frac{BE}{CD}=\frac{8}{11}$

b)

We know that the ratios of areas of two similar triangles is the ratio of squares of corresponding sides.

$\frac{ar(\triangle ABE)}{ar(\triangle ACD)} =(\frac{AB}{AC})^2=(\frac{8}{11})^2=\frac{64}{121}$

and area (BCDE)= area(ΔACD) - area(ΔABE) = 121 k²-64 k² = 57 k²

Now, $\frac{ar(\triangle ABE)}{ar(BCDE)}=\frac{64k^{2} }{57k^2} =\frac{64}{57}$

ii)

In BCDE,

BE ║ CD,

∴ one pair of opposite sides are parallel to each other; hence it is a Trapezium.

To learn more about Similarity:

https://brainly.in/question/20663980

To learn more about Trapezium:

https://brainly.in/question/53780850