Question

6. The perimeter of a rectangle is 28m
and its diagonal is 10 m.. what are the

the lengths of its side?​

Answer 1

Answer:

l=8m

or

6m

d Diagonal

10

m

P Perimeter

28

m

Using the formulas

P=2(l+w)

d=w2+l2

There are 2 solutions forl

l=P

4+1

48d2﹣P2=28

4+1

4·8·102﹣282=8m

l=P

4﹣1

48d2﹣P2=28

4﹣1

4·8·102﹣282=6m

Answer 2

Answer:

Length is 8 m and breadth is 6 m.

Step-by-step explanation:

Let's assume that, the length of the rectangle is x, and it's breadth is y.

Therefore, it's perimeter = 2(x + y) = 28

∴ x + y = 14 . . . . . . . . . . . .  (i)

Also, $x^2 + y^2 = 10^2 = 100$

$=>(x + y)^2 - 2xy = 100$  [using the identity $(a + b)^2 = a^2 + 2ab + b^2$]

$=>14^2 - 2xy = 100$  [using equation (i)]

$=>196 - 2xy = 100$

$=>98 - xy = 50$  [dividing by 2]

$=>xy = 98 - 50 = 48$

$=>x(14 - x) = 48$  [using equation (i)]

$=>x^2 - 14x + 48 = 0$

$=>x^2 - 6x - 8x + 48 = 0$

$=>x(x - 6) - 8(x - 6) = 0$

$=>(x - 6)(x - 8) = 0$

∴ x = 6 or x = 8

∴ y = 8 or y = 6.

Length is 8 m and breadth is 6 m.