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ISEE Upper Level Math : How to find the area of an acute / obtuse triangle

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Example Questions

Example Question #51 : Triangles

Two sides of a scalene triangle measure 4 centimeters and 7 centimeters, and their corresponding angle measures 30 degrees. Find the area of the triangle.

Possible Answers:

$7\ cm^2$

$10\ cm^2$

$8\ cm^2$

$5\ cm^2$

Correct answer:

$7\ cm^2$

Explanation:

$Area=\frac{1}{2}absinC$ ,

where and are the lengths of two sides and is the angle measure.

Plug in our given values:

$Area=\frac{1}{2}absinC=\frac{1}{2}\times 4\times 7\times sin30^{\circ}$

$sin30^{\circ}=\frac{1}{2}\Rightarrow Area=\frac{1}{2}\times 4\times 7\times \frac{1}{2}=7\ cm^2$

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Example Question #53 : Triangles

A scalene triangle has a base length and a corresponding altitude of . Give the area of the triangle in terms of .

Possible Answers:

4t^2

t^2

2t^2

Correct answer:

2t^2

Explanation:

$Area =\frac{bh}2{}$ ,

where is the base and is the altitude.

$Area =\frac{bh}2{}=\frac{t\times 4t}{2}=2t^2$

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Example Question #1 : How To Find The Area Of An Acute / Obtuse Triangle

What is the area of a triangle on the coordinate plane with its vertices on the points (0,-4) ,(0,9), (6,7) ?

Possible Answers:

$40 \frac{1}{2}$

Correct answer:

Explanation:

The base can be seen as the (vertical) line segment connecting (0,-4) and (0,9) , which has length 9 - (-4) = 13 . The height is the pependicular distance from (6,7) to the segment; since the segment is part of the -axis, this altitude is horizontal and has length equal to -coordinate .

The area of this triangle is therefore

$A = \frac{1}{2} bh = \frac{1}{2} \cdot13 \cdot 6 = 39$ .

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Example Question #55 : Triangles

What is the area of a triangle on the coordinate plane with its vertices on the points (-5,0), (10,0), (6,8) ?

Possible Answers:

120

Correct answer:

Explanation:

The base can be seen as the (horizontal) line segment connecting $\left (-5,0 \right )$ and $\left (10,0 \right )$ , the length of which is 10 - (-5) = 15 . The height is the pependicular distance from (6,8) to the segment; since the segment is part of the -axis, this altitude is vertical and has a length equal to -coordinate .

The area of this triangle is therefore

$A = \frac{1}{2} bh = \frac{1}{2} \cdot15 \cdot 8 = 60$ .

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Example Question #1 : How To Find The Area Of An Acute / Obtuse Triangle

Right triangle

Figure NOT drawn to scale.

$\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . What percent of $\bigtriangleup ABC$ has been shaded gray?

Choose the closest answer.

Possible Answers:

$30 \%$

$35 \%$

$25 \%$

$20 \%$

Correct answer:

$25 \%$

Explanation:

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 as well as the large triangle.

The similarity ratio of $\bigtriangleup AXB$ to $\bigtriangleup ABC$ is the ratio of the lengths of their hypotenuses. The hypotenuse of the latter is 18; that of the former, from the Pythagorean Theorem, is

$\sqrt{18 ^{2}+ 30 ^{2}} = \sqrt{324+ 900} = \sqrt{1,224}$

The similarity ratio is therefore $\frac{18}{ \sqrt{1,224}}$ . The ratio of their areas is the square of this, or

$\left (\frac{18}{ \sqrt{1,224}} \right )^{2} = \frac{324}{1,224} = \frac{324 \div 36 }{1,224\div 36} = \frac{9}{34}$

The area of $\bigtriangleup AXB$ is

$\frac{9}{34} \times 100 \% \approx 26.5 \%$

of that of $\bigtriangleup ABC$ , so the choice closest to the correct percent is 25%.

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ISEE Upper Level Math : How to find the area of an acute / obtuse triangle

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #51 : Triangles

Example Question #53 : Triangles

Example Question #1 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #55 : Triangles

Example Question #1 : How To Find The Area Of An Acute / Obtuse Triangle

All ISEE Upper Level Math Resources

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