Question

In the given figure, ABC is an equilateral triangle. DE is parallel to BC such that area of quadrilateral DBCE is equal to one half the area of AABC. If BC = 2 cm, then DE ​

Answer 1

Answer:

DE = √2 cm. Pls mark me as brainliest.

Step-by-step explanation:

∠A is common

∠ABC = ∠ADE  (Corresponding Amgles)

ΔADE is similar to ΔABC  (AA similarity criterion)

Note : The triangles are similar and not congruent

$DE^2 : BC^2 :: A(ADE) : A(ABC)$ (A() denotes area of)

[Side Area relationship of similar triangles]

So, $DE^2 : BC^2 = 1:2$  [Given]

$2DE^2 = BC^2$  (Product of means is equal to product of extremes)

2$DE^2 = 2^2 = 4$

$DE^2 = 2$

DE = √2 cm

Answer 2

Given :  ABC is an equilateral triangle.

DE is parallel to BC such that area of quadrilateral DBCE is equal to one half the area of ΔABC.  BC = 2 cm,

To Find : DE ​

Solution:

area of ΔABC  = area of quadrilateral DBCE  +   area of ΔADE

area of quadrilateral DBCE is equal to one half the area of ΔABC

=>  area of ΔABC  = (1/2) area of ΔABC  +   area of ΔADE

=> area of ΔADE  = (1/2) area of ΔABC  

DE || BC

=> ∠D = ∠B  and ∠E = ∠C  corresponding angles

=> ΔADE ~ ΔABC   Using AA similarity

=> Area of  ΔADE  / Area of  ΔABC   = (DE/BC)²

=> 1/2  =  (DE/2)²

=> DE² = 2

=> DE =  √2

Hence DE is √2

Learn More:

थेल्स प्रमेय(BPT) का कथन लिखकर उसे ...

brainly.in/question/21435493

if a line divides any two sides of a triangle in the same ratio then the ...

brainly.in/question/25977302