Input two rectangle dimensions (length, width, diagonal, area and perimeter), and the calculator will calculate the unknown property. The calculator accepts all types of input values, including fractions and square roots, and provides step-by-step explanation.
solution
$$ P = 12 $$explanation
To find perimeter $ P $ use formula:
$$ P = 2 a + 2 b $$After substituting $ a = \frac{ 7 }{ 2 } $ and $ b = \frac{ 5 }{ 2 } $ we have:
$$ P = 2 \cdot \frac{ 7 }{ 2 } + 2 \cdot \frac{ 5 }{ 2 } $$ $$ P = 7 + 5 $$ $$ P = 12 $$$$ A = ab $$ |
area |
$$ P = 2a + 2b $$ |
perimeter |
$$ d^2 = a^2 + b^2 $$ |
diagonal |
This calculator uses the following formulas to find the missing values of a rectangle:
Area: | $$ A = a \cdot b $$ | |
Perimeter: | $$ P = 2a + 2b $$ | |
Diagonal: | $$ d^2 = a^2 + b^2 $$ |
What is the area of a rectangle with a base of 12 cm and a height of 3/2 cm?
base $ a = 6 $
height $ b = \dfrac{9}{2} $
$$ \color{blue}{A = a \cdot b} = 6 \cdot \frac{9}{2} = \frac{54}{2} = 27 $$What is the perimeter of a rectangle with a length of 7/2cm and a width of 5/2cm?
length $ a = \dfrac{7}{2} \, cm $
width $ b = \dfrac{5}{2} \, cm $
$$ \color{blue}{P = 2a + 2b} = 2 \cdot \frac{7}{2} + 2 \cdot \dfrac{5}{2} = 7 + 5 = 12 $$The area of a rectangle is 42 cm2. Find its perimeter if the width is 7cm.
We'll need two steps to solve this one:
Step 1: find length ( b ):
width $ a = 7 cm $
area: $ A = 42 cm $
$$ \begin{aligned} A & = a \cdot b \\[ 1 em] 42 & = 7 \cdot b \\[ 1 em] b & = \frac{42}{7}\\[ 1 em] b & = 6 \\[ 1 em] \end{aligned} $$Step 2: find perimeter ( P )
width $ a = 7 cm $
length $ b = 6 cm $
$$ \begin{aligned} P & = 2a + 2b \\[ 1 em] P & = 2 \cdot 7 + 2 \cdot 6 \\[ 1 em] P & = 14 + 12 \\[ 1 em] P & = 28 \, cm^2 \\[ 1 em] \end{aligned} $$What is the diagonal of a rectangle if the perimeter is P = 11/2 cm and a width is a = 3/2 cm ?
Step 1: find length ( b ):
width $ a = \dfrac{3}{2} cm $
perimeter: $ P = \dfrac{11}{2} cm $
$$ \begin{aligned} P & = 2a + 2b \\[ 1 em] \frac{11}{2} & = 2 \cdot \frac{3}{2} + 2b \\[ 1 em] \frac{11}{2} & = 3 + 2b \\[ 1 em] 2b &= \frac{11}{2} - 3 \\[1 em] 2b &= \frac{5}{2} \\[1 em] b &= \frac{5}{4} \end{aligned} $$Step 2: find diagonal ( d )
width $ a = \dfrac{3}{2} cm $
length $ b = \dfrac{5}{4} cm $
$$ \begin{aligned} d^2 & = a^2 + b^2 \\[ 1 em] d^2 & = \left( \frac{3}{2} \right)^2 + \left( \frac{5}{4} \right)^2 \\[ 1 em] d^2 & = \frac{9}{4} + \frac{25}{16} \\[ 1 em] d^2 & = \frac{61}{16} \\[ 1 em] d & = \frac{\sqrt{61}}{4} \end{aligned} $$