Find the area of a rectangle with length 4 and diagonal 5?

Answer 1

you have to find the width

a^2+b^2=c^2
4^2 +b^2 = 5^2
16 + b^2 =25
b^2=9
b= 3

A=wl
A= 4*3
A= 12 unit^2

Answer 2

3*4 = 12

Answer 3

Find the width

Pythagorean Theorem

c² = a² - b²

c² - a² = b²

√(5)² - (4)² = √b²

√25 - 16 = √b²

√9 = √b²

3 = b

The width is 3

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Area of a rectangle formula

A + L x W

A = 4 x 3

A = 12

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Answer 4

12 is right.

let's say:
length = y = 4
hight = x = ?
diagonal = z = 5
angle between y and z = alpha


solution 1 with pythagoras:

triangle: x^2+y^2=z^2 so x^2+16=25 -> x=Squareroot(9)=3
the area is calculated: area=height*length=x*y=3*4=12



solution 2 with sinus/cosinus:

cos(alpha)=y/z=4/5
alpha=36,87
sin(alpha)=sin(36,87)=3/5 and also sin(alpha)=x/z=x/5
therefore: x=5*sin(alpha)=5*sin(36,87)=3
so now we have the height and the length of the rectangle.
the area is calculated: area=height*length=x*y=3*4=12

Answer 5

rectangle is two triangles.
1st triangle
a^2 + b^2 = c^2
(4^2) + b^2 = (5^2)
16 + b^2 = 25
b^2 = 9
b= 3 = length of triangle = length of rectangle

A= l * w
A = 3 * 4
A = 12

Also this triangle is a prefect and well know 3,4,5 triangle

Answer 6

Here's a hint.
For a triangle: c^2 = a^2 + b^2

c would be the diagonal.
a would be the length.
Solve for b.

Then use the formula Area = length x width

Answer 7

3,4,5 are the sides and diagonal 3x4=12

Answer 8

12 unit squared.

Answer 9

I'd say 12

Answer 10

wouldn't it be 12? you gotta use Pythagoras' Theorum to find how long the other side would be :-)