7,244 views
10 votes
10 votes
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

The perimeter of parallelogram ABCD is 92 cm. AD is 1 cm more than twice AB. Find-example-1
User Capoeira
by
2.8k points

1 Answer

19 votes
19 votes

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

Step-by-step 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} \\ (90)/(6)=(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

The perimeter of parallelogram ABCD is 92 cm. AD is 1 cm more than twice AB. Find-example-1
User Soheila
by
2.9k points