Question

find the area of a rectangle whose breadth is 8 centimetre and length of one of its diagonals is 17 cm ​

Answer 1

Answer:

Step-by-step explanation:

Breadth(b)=8 cm

Diagonal(d)=17 cm

In rectangle ABCD , we have

right angled triangle ∆BCD  where  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

                                   = 120cm²

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Answer 2

Answer:

120cm^2

Step-by-step explanation:

breadth=8cm

diagonal = 17cm

length be x

by using phythagoreas theorem

(17)^2=(8)^2+(x)^2

289= 64 +(x)^2

x^2=289-64

x^2=225

x=15×15

x=15cm

area=length ×breadth

= 15×8

=120cm^2