Quadrilaterals — Hard Practice Quiz
A Geometry cheat sheet for Quadrilaterals — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Square
Where: \(s\) = side, \(d\) = diagonal
Rectangle
Where: \(l\) = length, \(w\) = width
Parallelogram Area
Where: \(b\) = base, \(h\) = height, \(\theta\) = angle
Rhombus Area
Where: \(d_1, d_2\) = diagonals
Trapezoid Area
Where: \(b_1, b_2\) = bases, \(h\) = height
Kite Area (alternate form)
Where: \(\theta\) = angle between diagonals
Cyclic Quadrilateral (Brahmagupta)
Where: \(s\) = semi-perimeter
Practice quiz
A square has a perimeter $P_S$. A rectangle has a perimeter $P_R$ such that $P_R = P_S$. If the length of the rectangle is three times its width, what is the ratio of the area of the square to the area of the rectangle, $A_S / A_R$?
- $4/3$
- $3/4$
- $1$
- $2/3$
Answer: $4/3$
A rhombus has diagonals $d_1$ and $d_2$. A square has a diagonal equal to $d_1$. If the area of the rhombus is equal to the area of the square, what is the relationship between $d_1$ and $d_2$?
- $d_2 = d_1$
- $d_2 = 2d_1$
- $d_2 = d_1 / 2$
- $d_2 = d_1 \sqrt{2}$
Answer: $d_2 = d_1$
A parallelogram has sides $a$ and $b$ and an angle $\theta$ between them. A square has a perimeter equal to the perimeter of this parallelogram. If the area of the parallelogram is $A_P = ab\sin\theta$, what is the area of the square, $A_S$, in terms of $a$, $b$, and $\theta$?
- $\frac{(a+b)^2}{4}$
- $ab\sin\theta$
- $\frac{a^2+b^2}{4}$
- $\frac{a^2+b^2+2ab\sin\theta}{4}$
Answer: $\frac{(a+b)^2}{4}$
A trapezoid has bases $b_1$ and $b_2$ and height $h$. A rectangle is constructed such that its length is the average of the trapezoid's bases, and its width is the trapezoid's height. How does the area of this rectangle, $A_R$, compare to the area of the trapezoid, $A_T$?
- $A_R = A_T$
- $A_R = 2A_T$
- $A_R = A_T / 2$
- $A_R = A_T \sqrt{2}$
Answer: $A_R = A_T$
A kite has diagonals $d_1$ and $d_2$, and the angle between them is $\theta$. If this kite is also a rhombus, what is the relationship between its area formula $A = \frac{1}{2}d_1 d_2 \sin\theta$ and the general rhombus area formula $A = \frac{1}{2}d_1 d_2$?
- The kite formula simplifies to the rhombus formula because $\sin\theta = 1$ for a rhombus
- The formulas are always different
- The kite formula is only for non-rhombic kites
- The rhombus formula is a special case of the kite formula when $d_1 = d_2$
Answer: The kite formula simplifies to the rhombus formula because $\sin\theta = 1$ for a rhombus
A cyclic quadrilateral has sides $a, b, c, d$. If this quadrilateral is a rectangle, what does Brahmagupta's formula $A = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ simplify to?
- $lw$
- $l^2+w^2$
- $\frac{1}{2}lw$
- $\sqrt{l^2+w^2}$
Answer: $lw$
If the diagonal of a square is increased by a factor of $k$, by what factor does its area increase?
- $k^2$
- $k$
- $\sqrt{k}$
- $2k$
Answer: $k^2$
A rectangle has a fixed perimeter $P$. What is the maximum possible area it can have, and what shape is the rectangle when its area is maximized?
- $A = \frac{P^2}{16}$, a square
- $A = \frac{P^2}{4}$, a square
- $A = \frac{P^2}{16}$, a rectangle with $l=2w$
- $A = \frac{P^2}{8}$, a square
Answer: $A = \frac{P^2}{16}$, a square
A trapezoid has bases $b_1$ and $b_2$ and height $h$. If $b_1 = b_2 = b$, what does the trapezoid area formula $A = \frac{1}{2}(b_1 + b_2)h$ simplify to, and what shape does the trapezoid become? How does this simplified formula relate to the parallelogram area formula $A = bh$?
- It simplifies to $A = bh$, becoming a parallelogram, which is the parallelogram area formula
- It simplifies to $A = bh$, becoming a rectangle, which is the parallelogram area formula
- It simplifies to $A = \frac{1}{2}bh$, becoming a triangle
- It remains $A = \frac{1}{2}(b_1 + b_2)h$ as it's still a trapezoid
Answer: It simplifies to $A = bh$, becoming a parallelogram, which is the parallelogram area formula
A rhombus has a perimeter $P$. If one of its diagonals is $d_1$, what is its area in terms of $P$ and $d_1$?
- $\frac{d_1}{4}\sqrt{P^2 - 4d_1^2}$
- $\frac{d_1}{2}\sqrt{P^2 - d_1^2}$
- $\frac{P}{4}\sqrt{d_1^2 - P^2}$
- $\frac{P d_1}{4}$
Answer: $\frac{d_1}{4}\sqrt{P^2 - 4d_1^2}$
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