Question

show that each statement is false by providing a counterexample.
(a) if the area of a rectangle is 36, then the length is 12 and the width is 3.
counterexample: length = \boxed{}, width = \boxed{}
(b) if the measures of $\angle p$, $\angle q$, and $\angle r$ sum to $180^\circ$, then all of the angles must be acute.
counterexample: $m\angle p = \boxed{}^\circ$, $m\angle q = \boxed{}^\circ$, $m\angle r = \boxed{}^\circ$
(c) if the length of $\overline{xz}$ is 20 and point $y$ lies on $\overline{xz}$, then $xy = 10$ and $yz = 10$.
counterexample: $xy = \boxed{}$, $yz = \boxed{}$
(d) if $\angle 1$ and $\angle 2$ are supplementary angles, then one of them must be obtuse.
counterexample: $m\angle 1 = \boxed{}^\circ$, $m\angle 2 = \boxed{}^\circ$

Explanation:

Part (a)

Step1: Recall rectangle area formula

The area of a rectangle is given by \( A = \text{length} \times \text{width} \). We need to find a length and width different from 12 and 3 such that their product is 36.

Step2: Find counterexample values

Let's take length = 6 and width = 6. Then the area is \( 6 \times 6 = 36 \), which satisfies the area condition but length and width are not 12 and 3. (Other possible values: length = 9, width = 4; length = 18, width = 2; etc.)

Step1: Recall angle sum and acute angle definition

An acute angle is less than \( 90^\circ \), a right angle is \( 90^\circ \), and an obtuse angle is greater than \( 90^\circ \) but less than \( 180^\circ \). We need three angles that sum to \( 180^\circ \) where not all are acute (i.e., at least one is right or obtuse).

Step2: Find counterexample angles

Let's take \( m\angle P = 90^\circ \), \( m\angle Q = 45^\circ \), \( m\angle R = 45^\circ \). The sum is \( 90 + 45 + 45 = 180^\circ \), but \( \angle P \) is a right angle (not acute). (Another example: \( 100^\circ \), \( 50^\circ \), \( 30^\circ \) where \( 100^\circ \) is obtuse.)

Step1: Recall segment addition postulate

If \( Y \) lies on \( \overline{XZ} \), then \( XY + YZ = XZ \). We need \( XY \) and \( YZ \) such that \( XY + YZ = 20 \) but \( XY
eq 10 \) and \( YZ
eq 10 \).

Step2: Find counterexample values

Let's take \( XY = 5 \) and \( YZ = 15 \). Then \( 5 + 15 = 20 \), which satisfies the length of \( \overline{XZ} \) but \( XY \) and \( YZ \) are not both 10. (Other possible values: \( XY = 8 \), \( YZ = 12 \); \( XY = 2 \), \( YZ = 18 \), etc.)

Answer:

length = \( 6 \), width = \( 6 \) (or other valid pairs like 9 and 4, 18 and 2, etc.)

Part (b)