How to find the area of a triangle - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

Triangle B has a height that is twice that of Triangle A and a base that is one-half that of Triangle A. Which is the greater quantity?

(a) The area of Triangle A

(b) The area of Triangle B

Answer

Let and be the base and height of Triangle A. Then the base and height of Triangle B are $\frac{1}{2} b$ and , respectively.

(a) The area of Triangle A is $A = \frac{1}{2} bh$ .

(b) The area of Triangle B is $A = \frac{1}{2}\cdot \frac{1}{2} b \cdot 2h = \frac{1}{2} bh$ .

Therefore, (a) and (b) are equal.

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Question

Two triangles on the coordinate plane have a vertex at the origin and a vertex at , where .

Triangle A has its third vertex at .

Triangle B has its third vertex at .

Which is the greater quantity?

(a) The area of Triangle A

(b) The area of Triangle B

Answer

(a) Triangle A has as its base the horizontal segment connecting and , the length of which is 10. Its (vertical) altitude is the segment from to this horizontal segment, which is part of the -axis; its height is therefore the -coordinate of this point, or .

The area of Triangle A is therefore $A = \frac{1}{2} \cdot 10 \cdot d = 5d$

(b) Triangle B has as its base the vertical segment connecting and , the length of which is 10. Its (horizontal) altitude is the segment from to this vertical segment, which is part of the -axis; its height is therefore the -coordinate of this point, or .

The area of Triangle B is therefore $A = \frac{1}{2} \cdot 10 \cdot c = 5c$

, so . (b), the area of Triangle B, is greater.

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Question

A triangle has sides 30, 40, and 80. Give its area.

Answer

By the Triangle Inequality Theorem, the sum of the lengths of the two shorter sides of a triangle must exceed the length of its longest side. However,

;

Therefore, this triangle cannot exist, and the correct answer is "none of the other responses is correct".

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Question

The above depicts Square ; , and are the midpoints of $\overline{AD}$ , $\overline{BC}$ , and $\overline{AB}$ , respectively. Which is the greater quantity?

(a) The area of $\bigtriangleup BZY$

(b) The area of $\bigtriangleup ZXD$

Answer

For the sake of simplicity, assume that the square has sidelength 2; this reasoning is independent of the actual sidelength.

Since , , and are the midpoints of their respective sides, , as shown in this diagram.

The area of $\bigtriangleup BZY$ , it being a right triangle, is half the product of the lengths of its legs:

$\frac{1}{2} \cdot BZ \cdot BY = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$

The area of $\bigtriangleup ZXD$ is half the product of the length of a base and the height. Using $\overline{XD}$ as the base, and $\overline{AZ}$ as an altitude:

$\frac{1}{2} \cdot XD \cdot AZ= \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$

The two triangles have the same area.

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Question

Construct rectangle , and locate midpoint of side $\overline{RE}$ . Now construct segment $\overline{MC}$ .

Which is the greater quantity?

(a) The area of Quadrilateral

(b) Three times the area of $\Delta MEC$

Answer

$\Delta MEC$ is a right triangle with right angle $\angle E$ , so its legs measure and ; its area is $\frac{1}{2}\left ( ME \right ) \left (EC \right )$ . Since is the midpoint of $\overline{RE}$ , $ME = \frac{1}{2} \left ( RE \right )$ , making the area of the triangle

$\frac{1}{2}\left ( ME \right ) \left (EC \right ) = \frac{1}{2} \cdot \frac{1}{2} \left ( RE \right ) \left (EC \right ) = \frac{1}{4} \left ( RE \right ) \left (EC \right )$

Rectangle has area $\left ( RE \right ) \left (EC \right )$ .

Quadrilateral , which is the portion of not in $\Delta MEC$ , has as its area

$\left ( RE \right ) \left (EC \right ) - \frac{1}{4} \left ( RE \right ) \left (EC \right ) = \frac{3}{4} \left ( RE \right ) \left (EC \right )$

Therefore, the area of Quadrilateral is three times that of $\Delta MEC$ , making (a) and (b) equal.

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Question

The length of a side of a square is two-thirds the length of a leg of an isosceles right triangle. Which is the greater quantity?

(a) The area of the square

(b) The area of the triangle

Answer

Let be the length of a leg of the right triangle. Then the sidelength of the square is $\frac{2}{3}s$ .

(a) The square has area $A = \left ( \frac{2}{3} s \right ) ^{2} = \frac{4}{9} s ^{2}$

(b) The isosceles right triangle has base and height area $A = \frac{1}{2}s^{2}$

$\frac{1}{2} > \frac{4}{9}$ , so (b) is greater.

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Question

Two triangles are on the coordinate plane. Each has a vertex at the origin.

Triangle A has its other two vertices at $\left (18,0 \right )$ and .

Triangle B has its other two vertices at $\left (2b,0 \right )$ and .

Which is the greater quantity?

(a) The area of Triangle A

(b) The area of Triangle B

Answer

Each triangle is a right triangle with legs along the - and -axes, so the area of each can be calculated by taking one-half the product of the two legs.

(a) The horizontal and vertical legs have measures 18 and , respectively, so the triangle has area $A = \frac{1}{2} \cdot 18 \cdot b = 9b$ .

(b) The horizontal and vertical legs have measures and 9, respectively, so the triangle has area $A = \frac{1}{2} \cdot 2b \cdot 9= 9b$ .

The areas are equal.

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Question

Construct rectangle . Let and be the midpoints of $\overline{RE}$ and $\overline{EC}$ , respectively, and draw the segments $\overline{MC}$ and $\overline{RN }$ . Which is the greater quantity?

(a) The area of $\Delta MEC$

(b) The area of $\Delta NER$

Answer

Each triangle is a right triangle, and each has its two legs as its base and height.

(a) is the midpoint of $\overline{RE}$ , so $ME = \frac{1}{2} RE$ .

The area of $\Delta MEC$ is $A = \frac{1}{2}bh= \frac{1}{2} \left ( ME \right ) \left ( EC\right )= \frac{1}{2} \left ( \frac{1}{2} RE \right ) \left ( EC\right ) =\frac{1}{4} \left ( RE \right ) \left ( EC\right )$ .

(b) is the midpoint of $\overline{EC}$ , so $NE = \frac{1}{2} EC$ .

The area of $\Delta NER$ is

$A = \frac{1}{2}bh= \frac{1}{2} \left ( RE \right ) \left ( NE \right )= \frac{1}{2} \left ( \frac{1}{2} RE \right ) \left ( \frac{1}{2} EC\right ) =\frac{1}{4} \left ( RE \right ) \left ( EC\right )$ .

The triangles have equal area.

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Question

The length of a side of a square is one-half the length of the hypotenuse of a $30 ^{\circ }- 60 ^{\circ }-90 ^{\circ }$ triangle. Which is the greater quantity?

(a) The area of the square

(b) The area of the triangle

Answer

(a) Let be the sidelength of the square. Then its area is $A = s^{2}$ .

(b) In a $30 ^{\circ }- 60 ^{\circ }-90 ^{\circ }$ triangle, the shorter leg is one-half as long as the hypotenuse. The hypotenuse has length , so the shorter leg has length . The longer leg is $\sqrt{3}$ times as long as the shorter leg, so the longer leg will have length $s\sqrt{3}$ . The area of the triangle is

$A = \frac{1}{2} bh = \frac{1}{2} s \cdot s\sqrt{ 3} = \frac{\sqrt{ 3}}{2} s^{2}$ .

$\frac{\sqrt{3} }{2}\approx \frac{1.7 }{2} < 1$ , so $\frac{\sqrt{ 3}}{2} s^{2} < s^{2}$ ; the square has the greater area.

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Question

Note: Figures NOT drawn to scale.

Refer to the above figures - a right triangle and a square. The area of the triangle is what percent of the area of the square?

Answer

The area of the triangle is

$A = \frac{1}{2} \cdot 18 \cdot 30 = 270$ square inches.

The sidelength of the square is $2 \times 12 = 24$ inches, so the area of the square is

$A = 24 ^{2} = 24 \times 24 = 576$ .

The question becomes "what percent of 576 is 270", which is answered as follows:

$P = \frac{270}{576} \times 100 = 46 \frac{7}{8}$

The correct answer is $46 \frac{7}{8} %$ .

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Question

Give the area of the above right triangle in terms of .

Answer

The area of a triangle is half the product of its base and its height; for a right triangle, the legs, which are perpendicular, serve as base and height.

$A = \frac{1}{2} (k+2) (k-2)$

$A = \frac{1}{2} (k^{2}-2^{2})$

$A = \frac{1}{2} (k^{2}-4)$

$A = \frac{1}{2} k^{2}-2$

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Question

The lengths of the hypotenuses of ten similar right triangles form an arithmetic sequence. The smallest triangle has legs of lengths 5 and 12 inches; the second-smallest triangle has a hypotenuse of length one and one half feet.

Which of the following responses comes closest to the area of the largest triangle?

Answer

The hypotenuse of the smallest triangle can be calculated using the Pythagorean Theorem:

$c = \sqrt{a^{2}+b^{2}} = \sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169}=13$ inches.

Let $a_{n}$ be the lengths of the hypotenuses of the triangles in inches. $a_{1} = 13$ and $a_{2} = 1 \frac{1}{2} \cdot 12 = 18$ , so their common difference is

$d = a_{2} - a_{1} = 18- 13= 5$

The arithmetic sequence formula is

$a_{n} = a_{1} + (n-1)d$

The length of the hypotenuse of the largest triangle - the tenth triangle - can be found by substituting $a_{1} = 13, n= 10, d = 5$ :

$a_{10} = a_{1} + (10-1)5 = 13+ 9 \cdot 5= 13 + 45 = 58$ inches.

The largest triangle has hypotenuse of length 58 inches. Since the triangles are similar, corresponding sides are in proportion. If we let and be the lengths of the legs of the largest triangle, then

$\frac{a}{5} = \frac{58}{13}$

$\frac{a}{5}\cdot 5= \frac{58}{13} \cdot 5$

$a \approx 22.3$

Similarly,

$\frac{b}{12} = \frac{58}{13}$

$\frac{b}{12} \cdot 12= \frac{58}{13} \cdot 12$

$b \approx 53.5$

The area of a right triangle is half the product of its legs:

$A = \frac{ab}{2}= \frac{22.3 \cdot 53.5}{2} \approx 596.5$ square inches.

Divide this by 144 to convert to square feet:

$596.5 \div 144 \approx 4.1$

Of the given responses, 4 square feet is the closest, and is the correct choice.

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Question

In Square . is the midpoint of $\overline{SE}$ , is the midpoint of $\overline{SA}$ , and is the midpoint of $\overline{QA}$ . Draw the segments $\overline{QX}$ and $\overline{UR}$ .

Which is the greater quantity?

(a) The area of $\bigtriangleup SQX$

(b) The area of $\bigtriangleup UAR$

Answer

The figure referenced is below:

Let be the common sidelength of Square .

Then $SX = SQ = \frac{1}{2}s$ .

The area of right triangle $\bigtriangleup SQX$ is half the product of its legs, so

$A _{\bigtriangleup SQX} = \frac{1}{2} \cdot SQ \cdot SX$

$= \frac{1}{2} \cdot \frac{1}{2}s \cdot \frac{1}{2}s = \frac{1}{8}s ^{2}$

$QA = \frac{1}{2}s$ , so

$UA = \frac{1}{2} \cdot QA = \frac{1}{2} \cdot \frac{1}{2}s = \frac{1}{4}s$

The area of right triangle $\bigtriangleup UAR$ is half the product of its legs, so

$A _{\bigtriangleup UAR } = \frac{1}{2} \cdot UA \cdot AR$

$= \frac{1}{2} \cdot \frac{1}{4}s \cdot s = \frac{1}{8}s ^{2}$

$\bigtriangleup SQX$ and $\bigtriangleup UAR$ have the same area.

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Question

Figure NOT drawn to scale

Refer to the above diagram, in which $\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . Which is the greater quantity?

(a) Four times the area of $\bigtriangleup AXB$

(b) Three times the area of $\bigtriangleup CXB$

Answer

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ - divides the triangle into two triangles similar to each other. The ratio of the hypotenuse of $\bigtriangleup CXB$ to that of $\bigtriangleup AXB$ (which are corresponding sides) is

$\frac{BC}{AB} = \frac{40}{30} =\frac{4}{3}$ ,

making this the similarity ratio. The ratio of the areas of two similar triangles is the square of their similarity ratio, which here is

$\left ( \frac{4}{3} \right )^{2} = \frac{4^{2}}{3^{2}} = \frac{16}{9}$ , or .

Therefore, if is the area of $\bigtriangleup AXB$ and is the area of $\bigtriangleup CXB$ , it follows that

$\frac{Y}{X} = \frac{16}{9}$

$\frac{Y}{X} \cdot X = \frac{16}{9} \cdot X$

$Y = \frac{16}{9} X$

Four times the area of $\bigtriangleup AXB$ is ; three times the area of $\bigtriangleup CXB$ is

$3 Y = 3 \cdot \frac{16}{9} X = \frac{48}{9} X = 5 \frac{4}{9} \cdot X$

$5 \frac{4}{9} > 4$ , so three times the area of $\bigtriangleup CXB$ is the greater quantity.

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Question

Figure NOT drawn to scale.

Refer to the above diagram, in which $\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . Which is the greater quantity?

(a) Twice the area of $\bigtriangleup AXB$

(b) The area of $\bigtriangleup CXB$

Answer

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ - divides the triangle into two triangles similar to each other. Also, since $\angle ABC$ measures 90 degrees and $\angle C$ measure 30 degrees, $\angle A$ measures 60 degrees, making $\bigtriangleup ABC$ a 30-60-90 triangle.

Because of this, the ratio of the measures of the legs of $\bigtriangleup ABC$ is

$\frac{BC}{AB} = \frac{\sqrt{3}}{1}$ ,

Since these legs coincide with the hypotenuses of $\bigtriangleup CXB$ and $\bigtriangleup AXB$ , this is also the similarity ratio of the latter to the former. The ratio of the areas is the square of this, or

$\left ( \frac{\sqrt{3}}{1} \right )^{2} = \frac{3}{1}$

Therefore, the area of $\bigtriangleup CXB$ is three times that of $\bigtriangleup AXB$ . This makes (b) the greater quantity.

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Question

The above figure depicts Trapezoid . Which is the greater quantity?

(a) The area of $\bigtriangleup TRP$

(b) The area of $\bigtriangleup TRA$

Answer

The area of a triangle is one half times the product of its height and the length of its base. As can be seen in the diagram below, both $\bigtriangleup TRP$ and $\bigtriangleup TRA$ have height and base of length :

Since both base length and height are the same between the two triangles, it follows that they have the same area.

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