Question

The perimeter of parallelogram ABCD is 92 cm. AD is 1 cm more than twice AB. Find the lengths of all four sides of ABCD. (1 point) AB = 15 cm, AD = 15 cm BC = 31 cm, CD = 31 cm O AB = 22.5 cm, AD = 23.5 cm, BC = 23.5 cm, CD = 22.5 cm O AB = 16 cm, AD = 33 cmn BC = 33 cm CD = 16 cm O AB = 15 cm, AD = 31 cm BC = 31 cm CD = 15 cm о

Answer 1

AB=15 AD=31 BC=31 CD=15

Explanation

Step 1

the perimeter of the parallelogram ABCD is

$\begin{gathered} \text{Perimeter}=AB+BC+CD+AD \\ \text{Also} \\ \text{Perimeter}=92\text{ cm} \\ so \\ 92=AB+BC+CD+AD\text{ Equation(1)} \end{gathered}$

on a parallelogram, 2 length are similear

$\begin{gathered} BC=AD(\text{blue lines)} \\ \text{and} \\ AB=CD(\text{black lines)} \end{gathered}$

now, replace in equation (1)

$\begin{gathered} 92=AB+BC+CD+AD\text{ Equation(1)} \\ 92=AB+AD+AB+AD \\ 92=2AB+2AD \\ 92=2(AB+AD)\text{ Equation (2)} \end{gathered}$

Step 2

now,AD is 1 cm more than twice AB, in other words, you have to add 1 to twice AB to ger AD

$AD=1+2AB\text{ Equation (3)}$

Step 3

solve equation (2) and (3)

replace equation (3) in equation(2)

$\begin{gathered} 92=2(AB+AD)\text{ Equation (2)} \\ 92=2(AB+(1+2AB))\text{ } \\ 92=2(AB+1+2AB) \\ 92=2(1+3AB) \\ 92=2+6AB \\ \text{subtract 2 in both sides} \\ 92-2=2+6AB-2 \\ 90=6AB \\ \text{divide both sides by 6} \\ \frac{90}{6}=\frac{6AB}{6} \\ 15=AB \end{gathered}$

Step 4

replace the value of AB in equation (3) to find AD

$\begin{gathered} AD=1+2AB\text{ Equation (3)} \\ AD=1+2(15) \\ AD=1+30 \\ AD=31 \end{gathered}$

I hope this helps you