Question

In the figure, ABCDE is a pentagon with
BE|| CD O and BC || DE. BL is I to co JF
the perimeter a ABCDE is 21 cm
and the value of xandy.
guys plz do help I have exam on Monday..... ​

Answer 1

Answer:

The value of y = 0 and value of x = 5

Step-by-step explanation:

It is given that,

AB = 3cm = AE = 3cm ,

BC = (x - y) ,

CD = (x + y) ,

BE = 5cm

In the given the figure BE || CD , BC || DE , and , BC ⊥ CD,

So,

BCDE is rectangle,

Hence,

CD = BE = 5cm , BC = DE = (x - y)

=> (x + y) = 5.........(Equation)1

Now,

Perimeter of ABCDE

=> AB + BC + CD + DE + AE

Now,

Put the values of them.

We will get,

21 = 3 + (x - y) + (x + y) + (x - y) + 3

{Perimeter = 21 (Given also)}

=> 21 = 6 + 2(x - y) + (5) From Eq 1

=> 21 - 11 = 2(x - y)

=> 10/2 = (x - y)

=> (x - y) = 5 ........(Equation)2

From Eq(1) and Eq (2)

x + y = 5

x - y = 5

-------------

2x = 10

-------------

x = 5 {Put the value in Eq (1)

We will get,

=> x + y = 5

=> 5 + y = 5

=> y = 5 - 5

=> y = 0 , x = 5

Hence,

The value of x = 5 and y = 0.

NOTE:

Refer to the Attached file for the figure.

Answer 2

$\huge\bf\red{ANSWER :-}$

$\green{AB=3cm}$

$\blue{AE=3cm}$

$\purple{BC=(x-y)}$

$\pink{CD=(x+y)}$

$\orange{BE =5cm}$

In the given figure BE II CD, BC II CD ,and BC perpendicular CD,

So,

(x+y)=5...........(1)

perimeter of ABCDE

21=3+(x-y)+(x+y)+(x-y)+3

21-11=2(x-y)

10=2(x-y)

10/2=(x-y)

(x-y)=5............(2)

From 1 and 2 equation

x+y=5

x-y=5

__________________________

2x=10

x=5

Put this value in 1 equation

x+y=5

5+y=5

y=0

Then,

$<font color="yellow">y=0$

$<font color="brown">x=5$