the perimeter of a rectangle is 44 cm. if its length is increased by 4 cm and breadth is increased by 2 cm its area is increased by 72 sqcm. find the dimensions of the rectangle

11 cm, 11 cm

Step-by-step explanation:

Let the length and width of the rectangle be x cm and y cm respectively.

Therefore,

Perimeter of rectangle = 44 cm

2(x + y) = 44

x + y = 44/2

x + y = 22 (1)

Area of rectangle = xy

When length is increased by 4 cm and breadth is increased by 2 cm.

New length = (x + 4) cm

New width = (y + 2) cm

New area of rectangle = original area + 72.. (given)

Therefore,

(x + 4) (y + 2) = xy + 72

xy + 2x + 4y + 8 = xy + 72

2x + 4y = 72 - 8

2x + 4y = 66

2(x + 2y) = 66

x + 2y = 66/2

x + 2y = 33 (2)

Subtracting equation (1) from equation (2)

x + 2y - (x + y) = 33 - 22

x + 2y - x - y = 11

y = 11

Substituting y = 11 in equation (1)

x + 11 = 22

x = 22 - 11

x = 11

Thus, the dimensions of the rectangle are 11 cm and 11 cm.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

is it 35

step-by-step explanation: