The diagonals divide each figure into four triangles.
Which of the figures are always divided into four congruent triangles by the diagonals?
7 answers
Square and rhombus are always divided into four congruent triangles by the diagonals.
a quadrilateral with four right angles
Which of the following classifications fits the description above?
I. parallelogram
II. rectangle
III. rhombus
IV. square
V. trapezoid
Which of the following classifications fits the description above?
I. parallelogram
II. rectangle
III. rhombus
IV. square
V. trapezoid
II. rectangle
III. rhombus
IV. square
III. rhombus
IV. square
Kenny drew a rhombus and a trapezoid on a piece of paper. He is asked to explain to another student how to tell the two quadrilaterals apart. Which is a reason that Kenny can give?
A. A rhombus has exactly two congruent sides, and a trapezoid has at least two congruent sides.
B. A rhombus has only has one pair of parallel sides, and a trapezoid is a parallelogram.
C. A rhombus is a parallelogram, and a trapezoid only has one pair of parallel sides.
D. A rhombus has four congruent sides, and a trapezoid always has at least two congruent sides.
A. A rhombus has exactly two congruent sides, and a trapezoid has at least two congruent sides.
B. A rhombus has only has one pair of parallel sides, and a trapezoid is a parallelogram.
C. A rhombus is a parallelogram, and a trapezoid only has one pair of parallel sides.
D. A rhombus has four congruent sides, and a trapezoid always has at least two congruent sides.
A. A rhombus has exactly two congruent sides, and a trapezoid has at least two congruent sides.
Solve the following inequality.
x/3 + 8 <_ 2/7
x/3 + 8 <_ 2/7
To solve the inequality \( \frac{x}{3} + 8 \leq \frac{2}{7} \), we can first simplify the inequality:
\( \frac{x}{3} + 8 \leq \frac{2}{7} \)
Subtract 8 from both sides:
\( \frac{x}{3} \leq \frac{2}{7} - 8 \)
\( \frac{x}{3} \leq -\frac{54}{7} \)
Multiply both sides by 3 to isolate x:
\( x \leq -\frac{54}{7} \times 3 \)
\( x \leq -\frac{162}{7} \) or approximately \( x \leq -23.14 \)
\( \frac{x}{3} + 8 \leq \frac{2}{7} \)
Subtract 8 from both sides:
\( \frac{x}{3} \leq \frac{2}{7} - 8 \)
\( \frac{x}{3} \leq -\frac{54}{7} \)
Multiply both sides by 3 to isolate x:
\( x \leq -\frac{54}{7} \times 3 \)
\( x \leq -\frac{162}{7} \) or approximately \( x \leq -23.14 \)