In the figure, ABCDE is a pentagon with BE||CD and BC||DE. BC is perpendicular to CD. AB= 5cm, AE=5cm, BE= 7cm, BC= x-y and CD= x+y. If the perimeter of ABCDE is 27cm. find the value of x and y, given x, y ≠0.

OR

Solve the following system of equations:

x = 6 & y = 1

Given – in pentagon ABCDE, BE||CD and BC||DE & BC ⊥ CD

AB = 5cm, AE = 5cm, BE = 7cm, BC = x-y, CD = x+y

Perimeter of ABCDE = 27 cm

To Find – values of x & y

Property – A quadrilateral with opposite sides parallel and four right angles is a rectangle.

Answer –

In quadrilateral BCDE, BE||CD and BC||DE & BC ⊥ CD,

∴ ∠BCD = ∠CDE = ∠DEB = ∠EBC = 90°

As in quadrilateral BCDE, opposite sides are parallel and all angles are right angles, hence BCDE is a rectangle.

We know that, opposite sides of rectangle are equal.

∴ BE = CD = 7cm

∴ x + y = 7 ………(1)

And BC = DE

∴ BC = DE = x – y ………(2)

Now, perimeter of pentagon ABCDE is

P = AB + BC + CD + DE + EA

∴ 27 = 5 + (x – y) + 7 + (x – y) + 5

∴ 27 = 17 + 2x – 2y

∴ 10 = 2x – 2y

∴ x – y = 5 ………(3)

Adding equations (1) & (3),

∴ 2x = 12

∴ x = 6

Substituting x in eq (1),

∴ 6 + y = 7

∴ y = 1

Therefore, values of x & y are 6 & 1 respectively.

OR

x & y = 1

Given equations –

To Find : values of x & y

Answer –

In given equations,

Substituting and , we get,

∴ 21m + 47n = 110 ………(1)

∴ 47m + 21n = 162 ………(2)

Multiplying eq(1) by 47 and eq(2) by 21, we get,

∴ 987m + 2209n = 5170 ………(3)

∴ 987m + 441n = 3402 ………(4)

Subtracting eq(4) from eq(3),

∴ 1768n = 1768

∴ n = 1

∴ y = 1

Substituting n in eq(1),

∴ 21m + 47(1) = 110

∴ 21m = 110 – 47

∴ 21m = 63

∴ m = 3

Therefore, solution for given equations is .