Question

ABC is an isosceles triangle right angled at B. two equilateral triangles are constructed with side BC and AC. prove that ar(BCD)= 1/2 ar(ACE)

Answer 1

Answer:

Step-by-step explanation:

Hope this helps ...you could also use similarity for this

Ps. Area of an equilateral triangle is root 3,into a sq. Upon 4

Answer 2

Answer:

area (BCD)= 1/2 area (ACE)

Step-by-step explanation:

Given :

The isosceles triangle ABC has a right angle at B. Two equilateral triangles are constructed with side BC and AC respectively.

AB = BC

This implies, $AB^2=BC^2$

According to the Pythagoras theorem,

$AC^2=AB^2+BC^2$

$AC^2$  $=2BC^2$

$\frac{AC^2}{BC^2} = \frac{2}{1}$

We know that triangle ACE is right angled at side AC and the  triangle BCD is right angled at side BC.

Area of triangle ACE / Area of triangle BCD =   $\frac{\sqrt3}{4}a_1^2$ / $\frac{\sqrt3}{4}a_2^2$

                                                                        =  $\frac{\sqrt3}{4}AB^2$ / $\frac{\sqrt3}{4}BC^2$

                                                                        = 2 / 1

                                                                        = 2

∴ Area of triangle ACE / Area of triangle BCD = 2

⇒ Area of triangle BCD / Area of triangle ACE = 1 / 2

⇒ area (BCD)= 1/2 area (ACE)

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