Question

find the value of $x$, $y$, and $z$, in the rhombus below.

answer
$x = \square$ $y = \square$ $z = \square$

Explanation:

Step1: Use the property of a rhombus (all sides are equal)

Since all sides of a rhombus are equal, we have \(3x - 5=88\) and \(4y + 8 = 88\) and \(-z + 1=88\).

• For \(3x-5 = 88\):

Add \(5\) to both sides: \(3x=88 + 5\), so \(3x=93\).
Divide both sides by \(3\): \(x=\frac{93}{3}\).

• For \(4y + 8=88\):

Subtract \(8\) from both sides: \(4y=88 - 8\), so \(4y=80\).
Divide both sides by \(4\): \(y=\frac{80}{4}\).

• For \(-z + 1=88\):

Subtract \(1\) from both sides: \(-z=88 - 1\), so \(-z=87\).
Multiply both sides by \(- 1\): \(z=-87\).

Step2: Calculate the values

• Calculate \(x\): \(x = 31\) (because \(93\div3 = 31\)).
• Calculate \(y\): \(y = 20\) (because \(80\div4=20\)).
• Calculate \(z\): \(z=-87\) (because \(-z = 87\) implies \(z=-87\)).

Answer:

\(x = 31\), \(y = 20\), \(z=-87\)