Question

In triangle ABC, angleB = 90°, BE is the perpendicular bisector of AC then (ar BEC)/(ar ABC) =
(a) 1/2
(b) 2/1
(c) 4/1
(d) 1/4

Answer 1

Answer:

option (a) is correct.

Step-by-step explanation:

Given triangle ABC, ∠B = 90°, BE is the perpendicular bisector of AC. we have to find the value of $\frac{ar(BEC)}{ar(ABC)}$

In ΔBEA and ΔBEC

AE=EC        (∵given)

∠AEB=∠CEB     (∵given)

BE=BE           (∵common)

By SAS rule, ΔBEA≅ΔBEC

As congruent triangles have equal area ∴ ar(BEA)=ar(BEC)

Now, ar(ABC)=ar(BEA)+ar(BEC)

                     =ar(BEA)+ar(BEA)

     ar(ABC)=2ar(BEA)

⇒  $\frac{ar(BEC)}{ar(ABC)}=\frac{1}{2}$

Hence, option (a) is correct.

Answer 2

Answer:

The correct option is (a).

Step-by-step explanation:

It is given that BE is the perpendicular bisector of AC. It means BE divides the line AC is 2 equal parts.

Area of a triangle is

$A=\frac{1}{2}\times base \times height$

The area of triangle BEC is

$Area(BEC)=\frac{1}{2}\times BE\times CE$

The area of triangle ABC is

$Area(ABC)=\frac{1}{2}\times BE\times AC$

$Area(ABC)=\frac{1}{2}\times BE\times (2CE)$

$Area(ABC)=2[\frac{1}{2}\times BE\times (2CE)]$

$Area(ABC)=2Area(BEC)$

$\frac{Area(BEC)}{Area(ABC)}=\frac{Area(BEC)}{2Area(BEC)}=\frac{1}{2}$

Therefore option (a) is correct.