Area of Common Figures - SHSAT Math

1. Triangle $KLM$ has vertices $K(2,1)$, $L(8,1)$, and $M(5,9)$. What is the area of $\triangle KLM$?
   Answer: $24$ When you encounter a triangle with given coordinates, you have several methods to find its area. The most efficient approach here is to use the coordinate formula or recognize geometric properties that simplify the calculation. First, plot or visualize the points: $K(2,1)$, $L(8,1)$, and $M(5,9)$. Notice that points $K$ and $L$ both have the same $y$-coordinate of $1$, meaning they form a horizontal base. The length of this base is $|8-2| = 6$ units. Since $KL$ is horizontal, you can treat it as the base and find the height by measuring the vertical distance from point $M$ to this base. The height is the difference in $y$-coordinates: $|9-1| = 8$ units. Using the area formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24$. Choice A) $20$ likely comes from incorrectly calculating the base as $5$ instead of $6$, giving $\frac{1}{2} \times 5 \times 8 = 20$. Choice C) $28$ might result from using the wrong height or making an arithmetic error in the multiplication. Choice D) $32$ could come from forgetting to multiply by $\frac{1}{2}$, giving just $4 \times 8 = 32$. The answer is B) $24$. Strategy tip: When finding the area of a triangle with coordinates, first check if any two points share the same $x$ or $y$ coordinate—this creates a convenient base with an easily calculated height, avoiding more complex formulas.
2. In the figure shown, triangle $PQR$ has $\angle Q = 90°$, $PQ = 9$, and $QR = 12$. Point $S$ on hypotenuse $PR$ is the foot of the altitude from $Q$. What is the area of triangle $QRS$?
   Answer: $\dfrac{864}{25}$ First find the hypotenuse: $PR = \sqrt{9^2 + 12^2} = 15$. The altitude to the hypotenuse is $QS = \dfrac{PQ \cdot QR}{PR} = \dfrac{9 \cdot 12}{15} = \dfrac{36}{5}$. In a right triangle, each leg squared equals its adjacent hypotenuse segment times the hypotenuse, so $RS = \dfrac{QR^2}{PR} = \dfrac{144}{15} = \dfrac{48}{5}$. Triangle $QRS$ has a right angle at $S$, so its area is $\dfrac{1}{2} \cdot \dfrac{48}{5} \cdot \dfrac{36}{5} = \dfrac{1728}{50} = \dfrac{864}{25}$. The value $\dfrac{432}{25}$ takes half of the correct area, applying the $\dfrac{1}{2}$ twice. The value 54 is the area of the whole triangle $PQR$, and 27 is half of that.
3. Refer to the figure. A larger parallelogram contains a smaller, similar parallelogram inside it, with corresponding sides parallel. The larger parallelogram has base $20$ and corresponding height $8$. The smaller parallelogram has base $5$. The region between the two parallelograms (the shaded region) has what area?
   Answer: $150$ Larger area $= 20 \cdot 8 = 160$. Since the parallelograms are similar with ratio $5:20 = 1:4$, the height of the small parallelogram is $8/4 = 2$, and small area $= 5 \cdot 2 = 10$. Shaded area $=160 - 10 = 150$. Distractor B subtracts $5$ (just the base); C does not subtract at all; D uses small area $=-10$ incorrectly.
4. A triangular banner has a base of $12\text{ cm}$ and a height of $9\text{ cm}$. What is the area of the banner?
   Answer: $54\text{ cm}^2$ When you encounter area problems involving triangles, remember that the triangle area formula is one of the most fundamental geometric calculations: Area = $\frac{1}{2} \times \text{base} \times \text{height}$. For this triangular banner, you have a base of 12 cm and a height of 9 cm. Substituting these values into the formula: Area = $\frac{1}{2} \times 12 \times 9 = \frac{108}{2} = 54\text{ cm}^2$. This confirms that choice C is correct. Let's examine why the other answers are wrong. Choice A (42 cm²) likely comes from incorrectly adding the base and height, then multiplying by some factor—this completely ignores the proper triangle formula. Choice B (48 cm²) might result from miscalculating $\frac{1}{2} \times 12 \times 9$ as $\frac{1}{2} \times 96$ due to arithmetic errors. Choice D (108 cm²) is the most common trap—this is what you get when you forget the $\frac{1}{2}$ factor and simply multiply base times height ($12 \times 9 = 108$). This mistake happens because students sometimes confuse the triangle area formula with the rectangle area formula. Remember this key distinction: rectangles use length × width, but triangles always require the $\frac{1}{2}$ factor because a triangle is essentially half of a rectangle. When you see any triangle area problem on the SHSAT, write down the formula first to avoid forgetting that crucial $\frac{1}{2}$.
5. The figure shows parallelogram $PQRS$ with $PQ = 15$ and $QR = 8$. The altitude from $R$ to side $PQ$ has length $6$. What is the area of parallelogram $PQRS$?
   Answer: $90$ The area of a parallelogram equals (base)(corresponding height). The altitude from $R$ to line $PQ$ has length $6$, and the base $PQ = 15$, so area $= 15 \times 6 = 90$. Distractor A uses $QR \times 6 = 48$ (wrong base/height pairing). Distractor B uses $\tfrac{1}{2}(15)(6\cdot?)$ or $8 \times 9$. Distractor D multiplies $15 \times 8 = 120$ (treating as rectangle, ignoring height).
6. An equilateral triangle has side length $6\text{ cm}$. What is its area?
   Answer: $9\sqrt3\text{ cm}^2$ When you encounter an equilateral triangle area problem, remember that you need a special formula since all three sides are equal and all angles are 60°. For any equilateral triangle with side length $s$, the area formula is $A = \frac{s^2\sqrt{3}}{4}$. This comes from using the general triangle area formula with the height of an equilateral triangle, which is $\frac{s\sqrt{3}}{2}$. With side length $s = 6$ cm, we get: $A = \frac{6^2\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} = 9\sqrt{3}\text{ cm}^2$ This confirms that choice A) $9\sqrt{3}\text{ cm}^2$ is correct. Looking at the wrong answers: Choice B) $6\sqrt{3}\text{ cm}^2$ likely comes from using just the side length times $\sqrt{3}$ without proper squaring and division. Choice C) $12\sqrt{3}\text{ cm}^2$ might result from incorrectly using $\frac{s^2\sqrt{3}}{3}$ instead of dividing by 4. Choice D) $18\sqrt{3}\text{ cm}^2$ could come from using $\frac{s^2\sqrt{3}}{2}$, forgetting that the height formula already includes a factor of $\frac{1}{2}$. Strategy tip: Memorize the equilateral triangle area formula $A = \frac{s^2\sqrt{3}}{4}$. On the SHSAT, equilateral triangles appear frequently, and this formula saves time compared to deriving the height each time. The $\sqrt{3}$ in the answer choices is often your clue that you're dealing with a 30-60-90 or equilateral triangle problem.
7. A triangular garden covers $120\text{ m}^2$. If the base is lengthened by $25\%$ while the height remains the same, what is the new area?
   Answer: $150\text{ m}^2$ When you encounter area problems involving percentage changes, focus on how each dimension affects the total area. For triangular areas, remember that $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. Since the original triangular garden has an area of $120\text{ m}^2$, we can write: $120 = \frac{1}{2} \times b \times h$, where $b$ is the base and $h$ is the height. When the base increases by 25%, the new base becomes $b + 0.25b = 1.25b$. The height stays the same. So the new area is: $\text{New Area} = \frac{1}{2} \times (1.25b) \times h = 1.25 \times \left(\frac{1}{2} \times b \times h\right) = 1.25 \times 120 = 150\text{ m}^2$ This confirms answer D is correct. Looking at the wrong answers: Choice A ($135\text{ m}^2$) represents adding 25% of something incorrectly—perhaps 25% of 60 instead of 120. Choice B ($140\text{ m}^2$) might come from mistakenly adding $120 + 20 = 140$, confusing the percentage calculation. Choice C ($145\text{ m}^2$) doesn't follow any clear mathematical relationship to the given information and likely results from computational errors. Strategy tip: When one dimension of a shape changes by a percentage while others remain constant, the area changes by that same percentage. A 25% increase in base (with height constant) means a 25% increase in area. Always multiply the original area by $(1 + \text{percentage change})$ rather than trying complex calculations.
8. In the figure, rectangle $ABCD$ has $AB = 12$ and $BC = 8$. Point $E$ is on $AB$ with $AE = 3$, and point $F$ is on $CD$ with $DF = 9$. What is the area of quadrilateral $AEFD$?
   Answer: $48$ $AEFD$ is a trapezoid with parallel sides $AE = 3$ (on $AB$) and $DF = 9$ (on $DC$), with perpendicular distance $AD = 8$ between them. Area $= \tfrac{1}{2}(3+9)(8) = \tfrac{1}{2}(12)(8) = 48$. Distractor A uses $\tfrac{1}{2}(3)(8)+\tfrac{1}{2}(9)(8) = 12+36=48$... that's correct too. B: $(3+9)(8)/2 - 6 = 42$ (arithmetic error). D: uses $(3+9)(9)/2 = 54$.
9. Parallelogram $ABCD$ has vertices $A(0,0)$, $B(5,0)$, $C(7,4)$, and $D(2,4)$. What is the area of $ABCD$?
   Answer: $20$ When you encounter a parallelogram with given coordinates, you have several methods to find its area. The most reliable approach is using the formula: Area = base × height. Let's identify the base and height of parallelogram $ABCD$. Side $AB$ lies along the x-axis from $(0,0)$ to $(5,0)$, giving us a horizontal base of length $5$. To find the height, we need the perpendicular distance from this base to the opposite side $DC$. Since $AB$ is horizontal along $y = 0$, and the opposite vertices $C(7,4)$ and $D(2,4)$ both have $y$-coordinate $4$, the height is simply the vertical distance: $4 - 0 = 4$. Therefore, Area = $5 \times 4 = 20$. Let's examine why the other answers are incorrect. Choice (A) $16$ results from incorrectly using $4 \times 4$, perhaps confusing the height with another measurement. Choice (B) $18$ might come from miscalculating the base length or attempting an incorrect formula application. Choice (D) $24$ could result from using $6 \times 4$, where $6$ might be mistakenly calculated as the base length. You can verify this is indeed a parallelogram by checking that opposite sides are parallel: $\overrightarrow{AB} = (5,0)$ and $\overrightarrow{DC} = (5,0)$, while $\overrightarrow{AD} = (2,4)$ and $\overrightarrow{BC} = (2,4)$. Strategy tip: For coordinate geometry problems involving parallelograms, always look for horizontal or vertical sides first—they make base and height calculations much simpler than using the cross-product formula.
10. A triangle has vertices at coordinates A(2, 1), B(8, 1), and C(5, 7). If this triangle is reflected across the x-axis and then the area is calculated, what is the area of the reflected triangle?
    Answer: 18 square units The base AB has length |8 - 2| = 6 units (both points have y-coordinate 1). The height is the perpendicular distance from C(5, 7) to the line y = 1, which is |7 - 1| = 6 units. The area is ½ × 6 × 6 = 18 square units. Reflection preserves area, so the reflected triangle has the same area. Choice B incorrectly uses 7 as height. Choice C incorrectly calculates base × height without the ½ factor. Choice D doubles the incorrect calculation.
11. A parallelogram has an area of $150\text{ cm}^2$ and a base of $15\text{ cm}$. What is its height?
    Answer: $10\text{ cm}$ When you encounter parallelogram problems, remember that the area formula is straightforward: Area = base × height. This is the same formula used for rectangles, since a parallelogram's area depends only on its base and the perpendicular distance between parallel sides (the height). Given that the parallelogram has an area of $150\text{ cm}^2$ and a base of $15\text{ cm}$, you can substitute these values into the formula: $150 = 15 \times \text{height}$. Solving for height: $\text{height} = \frac{150}{15} = 10\text{ cm}$. This confirms answer choice C is correct. Looking at the wrong answers: Choice A ($5\text{ cm}$) would give an area of only $15 \times 5 = 75\text{ cm}^2$, which is half the required area. Choice B ($8\text{ cm}$) would yield $15 \times 8 = 120\text{ cm}^2$, falling short by $30\text{ cm}^2$. Choice D ($15\text{ cm}$) represents a common trap—students might assume the height equals the base, but this would create an area of $15 \times 15 = 225\text{ cm}^2$, which is too large. Remember that in parallelogram problems, height is always the perpendicular distance between parallel sides, not the length of a slanted side. When you're given area and base, simply divide area by base to find height. This direct relationship makes parallelogram area problems very manageable once you recognize the pattern.
12. In the figure, quadrilateral $ABCD$ has vertices $A(0,0)$, $B(8,0)$, $C(10,5)$, and $D(2,5)$. What is the area of $ABCD$?
    Answer: $40$ $AB$ goes from $(0,0)$ to $(8,0)$, length $8$. $DC$ goes from $(2,5)$ to $(10,5)$, length $8$. Both are horizontal, so $ABCD$ is a parallelogram with base $8$ and height $5$ (the vertical distance between $y=0$ and $y=5$). Area $= 8 \cdot 5 = 40$. Distractor A treats it as a triangle (halves). C uses $9 \cdot 5$ (average of bases times height — trapezoid formula misapplied). D uses $10\cdot 5$.
13. Diagonal $AC$ is drawn in parallelogram $ABCD$. If $\triangle ABC$ has an area of $45\text{ cm}^2$, what is the area of the entire parallelogram?
    Answer: $90\text{ cm}^2$ When you see a parallelogram with a diagonal drawn, think about how that diagonal divides the shape. A key property of parallelograms is that any diagonal splits them into two congruent triangles. Here's why: In parallelogram $ABCD$, diagonal $AC$ creates triangles $ABC$ and $ACD$. These triangles are congruent because they share side $AC$, and the opposite sides of a parallelogram are equal ($AB = CD$ and $BC = AD$). Additionally, opposite sides are parallel, which means the triangles have the same base and height relationships. Since triangle $ABC$ has an area of $45\text{ cm}^2$, and triangle $ACD$ is congruent to it, triangle $ACD$ also has an area of $45\text{ cm}^2$. Therefore, the total area of the parallelogram is $45 + 45 = 90\text{ cm}^2$. Looking at the wrong answers: (A) $45\text{ cm}^2$ incorrectly assumes the given triangle area equals the entire parallelogram area. (B) $60\text{ cm}^2$ might result from incorrectly adding $\frac{1}{3}$ of the triangle area. (C) $75\text{ cm}^2$ could come from adding $\frac{2}{3}$ of the triangle area, perhaps from misunderstanding the diagonal relationship. Study tip: Remember that any diagonal of a parallelogram creates two congruent triangles, so the parallelogram's area is always exactly double the area of either triangle. This relationship works for rectangles, rhombuses, and squares too, since they're all special parallelograms.
14. In the figure, trapezoid $JKLM$ has $JK \parallel ML$. $JK = 6$, $ML = 14$, and both legs $JM$ and $KL$ have length $5$. What is the area of trapezoid $JKLM$?
    Answer: $30$ Since the trapezoid is isosceles, drop perpendiculars from $J$ and $K$ to $ML$. This creates two congruent right triangles on the ends, each with horizontal leg $\frac{14-6}{2} = 4$ and hypotenuse $5$. Using the Pythagorean theorem, the height is $\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3$. Therefore, the area is $\frac{1}{2}(6 + 14)(3) = \frac{1}{2}(20)(3) = 30$.
15. A trapezoid has an area of 120 square centimeters. One of its parallel sides is 3 times the length of the other parallel side. If the height is 8 cm, what is the length of the shorter parallel side?
    Answer: 7.5 cm Let the shorter side be x cm, then the longer side is 3x cm. Using the trapezoid area formula: ½(x + 3x)(8) = 120. This gives ½(4x)(8) = 120, so 16x = 120, and x = 7.5 cm. Choice A gives area ½(5 + 15)(8) = 80. Choice B gives area ½(6 + 18)(8) = 96. Choice D gives area ½(10 + 30)(8) = 160.
16. Triangle $PQR$ has vertices $P(1,2)$, $Q(6,2)$, and $R(6,7)$ in the coordinate plane. What is the area of $\triangle PQR$?
    Answer: $12\tfrac12$ When you encounter a triangle with vertices given as coordinate points, you have several methods to find the area. The key is to look at the coordinates first to see if there's a pattern that makes the calculation easier. Let's plot the points: $P(1,2)$, $Q(6,2)$, and $R(6,7)$. Notice that points $P$ and $Q$ both have the same $y$-coordinate (2), making $PQ$ a horizontal line. Points $Q$ and $R$ both have the same $x$-coordinate (6), making $QR$ a vertical line. This means we have a right triangle with the right angle at $Q$! For a right triangle, the area formula is simply $\frac{1}{2} \times \text{base} \times \text{height}$. The base $PQ$ has length $6-1=5$, and the height $QR$ has length $7-2=5$. Therefore, the area is $\frac{1}{2} \times 5 \times 5 = \frac{25}{2} = 12\frac{1}{2}$. Choice A ($10\frac{1}{2}$) likely comes from miscalculating one of the side lengths as 4 instead of 5. Choice C (15) results from forgetting to multiply by $\frac{1}{2}$ in the area formula, giving just $5 \times 3$ where 3 might come from miscounting. Choice D (25) comes from calculating $5 \times 5$ but completely forgetting the $\frac{1}{2}$ factor. Strategy tip: Always check if the given points form a right triangle by looking for shared coordinates. When they do, use the simple base-times-height formula rather than more complex methods like the coordinate area formula.
17. A parallelogram and a trapezoid have the same area of 84 square inches. The parallelogram has a base of 12 inches and height of 7 inches. If the trapezoid has parallel sides of lengths 8 inches and b inches with the same height as the parallelogram, what is the value of b?
    Answer: 16 inches First verify the parallelogram area: 12 × 7 = 84 ✓. For the trapezoid: ½(8 + b)(7) = 84. Multiplying both sides by 2: (8 + b)(7) = 168. Dividing by 7: 8 + b = 24, so b = 16 inches. Choice A would give area ½(8 + 8)(7) = 56. Choice B would give area ½(8 + 12)(7) = 70. Choice D would give area ½(8 + 20)(7) = 98.
18. The figure shows a composite shape made by attaching a right triangle to a rectangle. The rectangle has dimensions $10 \times 6$. The right triangle shares the rectangle's right side (length $6$) as one leg, and its hypotenuse measures $10$. What is the total area of the composite figure?
    Answer: $84$ The rectangle has area $10 \times 6 = 60$. The right triangle has one leg of length $6$ and hypotenuse $10$. Using the Pythagorean theorem, the other leg has length $\sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8$. The triangle's area is $\frac{1}{2}(6)(8) = 24$. Therefore, the total area is $60 + 24 = 84$.
19. The midsegment (median) of a trapezoid is $11\text{ cm}$ long, and the height is $6\text{ cm}$. What is the area of the trapezoid?
    Answer: $66\text{ cm}^2$ When you encounter trapezoid problems involving midsegments, remember that the midsegment (also called the median) is a line segment connecting the midpoints of the two non-parallel sides. This segment has a special property: its length equals the average of the two parallel bases. If the midsegment is 11 cm, then $\frac{b_1 + b_2}{2} = 11$, which means $b_1 + b_2 = 22$ cm, where $b_1$ and $b_2$ are the lengths of the parallel bases. The area formula for a trapezoid is $A = \frac{1}{2}(b_1 + b_2) \times h$. Since we know $b_1 + b_2 = 22$ cm and the height $h = 6$ cm, we can substitute: $A = \frac{1}{2}(22)(6) = 66\text{ cm}^2$. Looking at the wrong answers: Choice A ($55\text{ cm}^2$) likely comes from incorrectly using the midsegment length directly as one base, then calculating $\frac{1}{2}(11 + 9)(5)$ or similar error. Choice C ($72\text{ cm}^2$) might result from using $12 \times 6$ without the $\frac{1}{2}$ factor, or from calculation mistakes. Choice D ($132\text{ cm}^2$) could come from multiplying $22 \times 6$ and forgetting the $\frac{1}{2}$ factor entirely. Study tip: Whenever you see a trapezoid midsegment problem, immediately write down that midsegment length $= \frac{b_1 + b_2}{2}$. This directly gives you the sum of the bases, which you need for the area formula. The midsegment essentially provides a shortcut to finding $(b_1 + b_2)$.
20. A parallelogram has adjacent sides of 10 cm and 15 cm, with an included angle of 60°. What is the area of this parallelogram?
    Answer: $75\sqrt{3}$ square cm using the sine formula For a parallelogram with adjacent sides $a$ and $b$ and included angle $\theta$, the area is $ab\sin\theta$. Here: Area = 10 × 15 × sin(60°) = 150 × (√3/2) = 75√3 square cm. Choice A omits the sine factor. Choice C uses the formula for a rectangle (90° angle). Choice D incorrectly calculates 150 × √3 instead of 150 × (√3/2).

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