Question

find the perimeter and area of rectangle whose breadth is 8cm and it's diagonal is 17cm

Answer 1

Answer:

Breadth=8cm

Diagonal=17 cm

In rectangle ABCD

In right angled triangle ∆BCD

We have,

Hypotenuse=17cm

Perpendicular=8cm

Base= ?

By using Pythagoras theorem

BD^2= BC^2 + CD^2

17^2= 8^2 + CD^2

289= 64 + CD^2

289-64 = CD^2

225= CD^2

CD= 15cm

We have length= 15cm

Perimeter of the rectangle= 2(l + b)

                                           = 2(15 + 8)

                                           = 2(23)

                                           = 46cm

Area of the rectangle = l * b

                                    =15*8

                                    = 120sq.cm

Answer 2

Answer:

P = 46 cm

A = 120 cm²

Breadth=8cm

Diagonal=17 cm

In rectangle ABCD

In right angled triangle ∆BCD

We have,

Hypotenuse=17cm

Perpendicular=8cm

Base= ?

By using Pythagoras theorem

BD^2= BC^2 + CD^2

17^2= 8^2 + CD^2

289= 64 + CD^2

289-64 = CD^2

225= CD^2

CD= 15cm

We have length= 15cm

Perimeter of the rectangle= 2(l + b)

                                           = 2(15 + 8)

                                           = 2(23)

                                           = 46cm

Area of the rectangle = l * b

                                    =15*8

                                    = 120sq.cm

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