Sofia wants to know how tall the oak tree behind her school is, and she is not about to climb it. She stands in the sun and notices that her shadow is shorter than she is, while the oak's shadow stretches across the lawn. Is that enough to find the oak's height? It is. The idea behind it is hiding inside every right triangle. Some problems call the two main results of this page the altitude theorem and the leg theorem. Here we use their textbook names.
Similar Right Triangles
Draw a right triangle $ABC$ with the right angle at $C$. The side opposite the right angle is the hypotenuse $\overline{AB}$, with length $c$. The two legs are $b = AC$ and $a = BC$.
Now draw the altitude from $C$ down to the hypotenuse. It meets $\overline{AB}$ at a point $D$ and makes a $90^\circ$ angle there. Call its length $h$. It cuts the hypotenuse into two pieces, $p = AD$ and $q = DB$, so $c = p + q$. The piece $p$ is the projection of leg $b$ onto the hypotenuse, and $q$ is the projection of leg $a$.
Why? Each of the three triangles has a right angle, and each pair shares an acute angle. Two equal angles are enough for AA similarity. Let's see the angles in action.
Scale factor
When two triangles are similar, every side of one is the same number of times as long as the matching side of the other. That number is the scale factor $k$. For right triangles, one equal acute angle is enough to prove similarity, because the right angles are already equal.
Why $k^2$? Area has two dimensions. Stretch the base by $k$ and the height by $k$, and the area grows by $k \cdot k$.
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Two right triangles are similar. The first has legs measuring $3$ in and $4$ in. If the shorter leg of the second triangle measures $9$ in, how long is the hypotenuse of…
Two right triangles are similar. The first has legs measuring $3$ in and $4$ in. If the shorter leg of the second triangle measures $9$ in, how long is the hypotenuse of this second triangle?
Two set squares (which have the shape of a scalene right triangle) are similar. If the hypotenuse of the larger set square is three times that of the smaller set square,…
Two set squares (which have the shape of a scalene right triangle) are similar. If the hypotenuse of the larger set square is three times that of the smaller set square, how many times larger is the area of the large set square compared to the small one?
Use the similarity $\triangle ACD \sim \triangle CBD$ from the last section. In $\triangle ACD$ the legs are $p$ (short) and $h$ (long). In $\triangle CBD$ they are $h$ (short) and $q$ (long). Matching sides are proportional:
$$\frac{p}{h} = \frac{h}{q}$$
Cross-multiplying gives $h^2 = pq$. When $\dfrac{p}{h} = \dfrac{h}{q}$, we say $h$ is the geometric mean of $p$ and $q$.
The theorem also works backward. If you know $h$ and one segment, you can find the other.
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In a right triangle, the altitude drawn to the hypotenuse divides it into two segments measuring $4$ in and $9$ in. Calculate the length of this altitude using the…
In a right triangle, the altitude drawn to the hypotenuse divides it into two segments measuring $4$ in and $9$ in. Calculate the length of this altitude using the corresponding theorem.
The altitude relative to the hypotenuse of a right triangle measures $6$ in. If one of the segments into which it divides the hypotenuse measures $3$ in, how long is the…
The altitude relative to the hypotenuse of a right triangle measures $6$ in. If one of the segments into which it divides the hypotenuse measures $3$ in, how long is the other segment of the hypotenuse?
Now compare $\triangle ABC$ with $\triangle ACD$. They share $\angle A$. In $\triangle ABC$ the hypotenuse is $c$ and the leg next to $A$ is $b$. In $\triangle ACD$ the hypotenuse is $b$ and the leg next to $A$ is $p$. Matching sides are proportional, so
$$\frac{c}{b} = \frac{b}{p} \quad\Longrightarrow\quad b^2 = p \cdot c.$$
Comparing $\triangle ABC$ with $\triangle CBD$ in the same way gives $a^2 = q \cdot c$.
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The hypotenuse of a right triangle measures $25$ in and the projection of one of its legs onto it measures $9$ in. Calculate the length of this leg applying the leg…
The hypotenuse of a right triangle measures $25$ in and the projection of one of its legs onto it measures $9$ in. Calculate the length of this leg applying the leg theorem.
A leg of a right triangle measures $8$ in and its orthogonal projection onto the hypotenuse measures $3,2$ in. Determine the total length of the hypotenuse.
A leg of a right triangle measures $8$ in and its orthogonal projection onto the hypotenuse measures $3,2$ in. Determine the total length of the hypotenuse.
In a right triangle, the projections of the two legs onto the hypotenuse measure $5$ in and $12$ in, respectively. How long is the hypotenuse of the triangle?
In a right triangle, the projections of the two legs onto the hypotenuse measure $5$ in and $12$ in, respectively. How long is the hypotenuse of the triangle?
Right triangles with an altitude to the hypotenuse show up in triangular sails, roof trusses and stone arches. A seam, a post or a beam acts as the altitude and splits a long side into two parts.
You also get a second area formula. Any side can be a base, so with the hypotenuse as the base and the altitude as the height:
$$\text{Area} = \tfrac12 \cdot c \cdot h = \tfrac12 \cdot a \cdot b$$
Arches work the same way. An angle inscribed in a semicircle is a right angle, so any point $C$ on a semicircle with diameter $\overline{AB}$ makes $\triangle ACB$ a right triangle. A vertical post from $C$ to the diameter is the altitude to the hypotenuse.
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In the design of a truss for a pitched roof in the shape of a right triangle, the vertical beam (which acts as the altitude to the hypotenuse) measures $2,4$ ft and…
In the design of a truss for a pitched roof in the shape of a right triangle, the vertical beam (which acts as the altitude to the hypotenuse) measures $2,4$ ft and separates the lower beam (hypotenuse) into two sections, one of which measures $1,8$ ft. Calculate the total perimeter of the outer structure of the truss.
The main sail of a sailboat has the shape of a right triangle. The seam that marks the altitude drawn from the vertex of the right angle to the hypotenuse measures $6$…
The main sail of a sailboat has the shape of a right triangle. The seam that marks the altitude drawn from the vertex of the right angle to the hypotenuse measures $6$ ft. If the projection of the shorter side of the sail onto the hypotenuse is $4,5$ ft, calculate the total area of the sail in square feet.
A stone bridge arch has the shape of a perfect semicircle with a diameter of $20$ ft. At $4$ ft from one of the ends of the bridge, a vertical pillar needs to be placed…
A stone bridge arch has the shape of a perfect semicircle with a diameter of $20$ ft. At $4$ ft from one of the ends of the bridge, a vertical pillar needs to be placed reaching from the ground to the arch. Using the altitude theorem, determine the exact length this pillar must have.
Sometimes the problem gives a relationship between $p$, $q$ and $h$ instead of numbers: "one piece is four times the other," or "the pieces are $x$ and $x + 3$." Write the pieces with a variable, plug them into $h^2 = pq$, and solve.
The leg theorem also relates the legs to the projections. Divide $b^2 = pc$ by $a^2 = qc$ and the $c$ cancels:
Watch the letters. Some problems call the two legs $b$ and $c$ and give the hypotenuse as a number. The rule is the same: the ratio of the squares of the two legs equals the ratio of their projections. For example, if the legs satisfy $\dfrac{b^2}{c^2} = \dfrac{9}{16}$ and the hypotenuse is $50$ in., then the projections are in the ratio $9 : 16$. Write them as $9k$ and $16k$, so $25k = 50$, $k = 2$, and the projections are $18$ in. and $32$ in. Then use the leg theorem with the whole hypotenuse $50$ to get each leg: $\sqrt{18 \cdot 50} = 30$ in. and $\sqrt{32 \cdot 50} = 40$ in.
There is a second way to use a ratio. Divide $b^2 = pc$ by $c^2$ (the hypotenuse squared) and you get
$$\frac{b^2}{c^2} = \frac{p}{c}.$$
So if you are told the ratio of a leg squared to the hypotenuse squared, that ratio is the fraction of the hypotenuse taken up by that leg's projection: $p = \dfrac{b^2}{c^2} \cdot c$. Then $q = c - p$, and the leg theorem finishes the job.
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In a plot of land shaped like a right triangle, an irrigation trench has been dug that coincides with the altitude to the hypotenuse and measures $12$ ft. If the ratio…
In a plot of land shaped like a right triangle, an irrigation trench has been dug that coincides with the altitude to the hypotenuse and measures $12$ ft. If the ratio between the two segments into which this trench divides the hypotenuse is $\frac{9}{16}$, calculate the dimensions of the three sides that bound the land.
In a right triangle, the projection of the shorter leg onto the hypotenuse measures $x$ in, the projection of the longer leg measures $(x+5)$ in, and the altitude…
In a right triangle, the projection of the shorter leg onto the hypotenuse measures $x$ in, the projection of the longer leg measures $(x+5)$ in, and the altitude relative to the hypotenuse measures $(x+2)$ in. Set up the corresponding equation, find the value of $x$, and calculate the total area of the triangle.
In a right triangle, the ratio between the squares of the legs is $\frac{b^2}{c^2} = \frac{9}{16}$. If the hypotenuse measures $50$ in, use the proportionality…
In a right triangle, the ratio between the squares of the legs is $\frac{b^2}{c^2} = \frac{9}{16}$. If the hypotenuse measures $50$ in, use the proportionality relationship deduced from the leg theorems to first find the measure of the projections of both legs onto the hypotenuse, and subsequently the exact length of the legs.
Some problems give you the area instead of the pieces. The area gives the hypotenuse, because $\text{Area} = \tfrac12 c h$ means $c = \dfrac{2 \cdot \text{Area}}{h}$. Then you know two things about $p$ and $q$:
$$p + q = c \qquad\text{and}\qquad p \cdot q = h^2.$$
Substitute $q = c - p$ into the second equation and you get a quadratic.
The Pythagorean Theorem also teams up with these ideas, for example in tangent-line problems. A line tangent to a circle is perpendicular to the radius at the point of tangency, so it creates a right angle.
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The area of a solar panel shaped like a right triangle is $150$ in$^2$ and the altitude relative to its hypotenuse measures $12$ in. By setting up a system of equations…
The area of a solar panel shaped like a right triangle is $150$ in$^2$ and the altitude relative to its hypotenuse measures $12$ in. By setting up a system of equations with the projections of the legs, determine the exact length of the two legs of the solar panel.
If we consider the Earth to be a sphere of radius $R = 6371$ miles, and a person is in a hot air balloon at a height of $h = 2$ miles above the surface, the line of…
If we consider the Earth to be a sphere of radius $R = 6371$ miles, and a person is in a hot air balloon at a height of $h = 2$ miles above the surface, the line of sight tangent to the surface forms a right triangle at the point on the horizon. Use the leg theorem (or the Pythagorean theorem in similar triangles) to find the straight-line distance $d$ from the hot air balloon to the visible horizon, rounding to two decimal places.
Now we come back to Sofia's oak. At the same moment of the day, sun rays are parallel. So the sun hits every vertical object at the same angle. Each object, its shadow and the ray from the top of the object to the tip of the shadow form a right triangle. These triangles share an acute angle, so they are similar by AA.
Back to the start: Sofia's shadow is $\tfrac35$ as long as she is tall, so the oak's shadow is $\tfrac35$ of the oak's height. A $24$-ft shadow means a $40$-ft tree.
A lamp works like the sun, but its light rays spread out from one point. Now the lamp, the person and the ground form two right triangles. The small one is inside the big one.
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A vertical post $2$ ft high casts a shadow of $1,5$ ft on the ground. At the same time and under the same conditions, how long will the shadow cast by a…
A vertical post $2$ ft high casts a shadow of $1,5$ ft on the ground. At the same time and under the same conditions, how long will the shadow cast by a telecommunications tower $40$ ft high be?
A street lamp $4,5$ ft high illuminates a person $1,8$ ft tall who is standing $3$ ft from the base of the lamp. Use the similarity of right triangles to find out the…
A street lamp $4,5$ ft high illuminates a person $1,8$ ft tall who is standing $3$ ft from the base of the lamp. Use the similarity of right triangles to find out the length of the shadow the person casts on the ground.
Light and moving balls both follow the law of reflection: the angle of arrival equals the angle of departure. Put a mirror on the ground, and that means two right triangles with an equal acute angle at the mirror. They are similar by AA.
A billiard ball bouncing off a cushion obeys the same law, so you get the same picture.
A pinhole camera is a closed box with a tiny hole. Light from the top of an object goes straight through the hole and lands on the bottom of the back wall, so the image is upside down. The object and the image make two right triangles that meet at the pinhole. Their angles at the pinhole are vertical angles, so they are equal and the triangles are similar.
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To measure the height of a building, a student places a flat mirror on the ground $15$ ft from the base of the building. Stepping back $1,2$ ft from the mirror, they see…
To measure the height of a building, a student places a flat mirror on the ground $15$ ft from the base of the building. Stepping back $1,2$ ft from the mirror, they see the top of the building reflected in it. If the student's eyes are $1,6$ ft from the ground, what is the height of the building?
On a rectangular billiard table, a ball is hit without spin against the opposite cushion, forming two similar right triangles in its path. The ball starts at $40$ in…
On a rectangular billiard table, a ball is hit without spin against the opposite cushion, forming two similar right triangles in its path. The ball starts at $40$ in from the side cushion and rebounds at $60$ in from the same cushion. If the total longitudinal distance the ball has advanced along the table is $150$ in, determine the longitudinal distances traveled before and after the rebound.
In a homemade pinhole camera $20$ in deep, an inverted image of a tree located $15$ ft from the pinhole is formed. If the image formed at the back of the camera has a…
In a homemade pinhole camera $20$ in deep, an inverted image of a tree located $15$ ft from the pinhole is formed. If the image formed at the back of the camera has a height of $8$ in, what is the real height of the tree in feet?
A line drawn parallel to one side of a triangle cuts off a smaller triangle that shares an angle with the original. By AA it is similar to the original. That is the idea behind ramps, cones, and any "triangle inside a triangle" problem.
A cone is a right triangle spun around. If you slice a cone through its axis, a conical glass shows a triangle with the vertex at the bottom. The water surface cuts off a smaller, similar triangle. Its two legs are the water depth (measured along the axis from the vertex) and the water radius.
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An accessibility ramp has a total length of $12$ ft resting on the ground and overcomes a maximum height difference of $1,5$ ft. If a wheelchair has advanced $4$ ft up…
An accessibility ramp has a total length of $12$ ft resting on the ground and overcomes a maximum height difference of $1,5$ ft. If a wheelchair has advanced $4$ ft up the ramp from its start at the bottom, at what height is it relative to ground level?
A conical glass has a total depth of $15$ in and a radius of $6$ in at its top opening. It is filled with water until the liquid reaches a height of $10$ in measured…
A conical glass has a total depth of $15$ in and a radius of $6$ in at its top opening. It is filled with water until the liquid reaches a height of $10$ in measured from the bottom (lower vertex). Calculate the radius of the circular surface of the water using the similarity of right triangles formed with the central axis of the glass.
Two classic problems show that similar right triangles can give surprising results.
Crossing cables. Two vertical posts have heights $a$ and $b$. A cable runs from the bottom of each post to the top of the other. The cables cross at height $h$. Does $h$ depend on how far apart the posts are?
A square in a right triangle. A square sits in the right-angle corner of a right triangle, with its opposite corner on the hypotenuse.
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Two vertical posts of heights $a = 3$ ft and $b = 6$ ft are separated by an unknown distance. Two taut cables are stretched from the base of each post to the top of the…
Two vertical posts of heights $a = 3$ ft and $b = 6$ ft are separated by an unknown distance. Two taut cables are stretched from the base of each post to the top of the other post. Using the similarity of the right triangles formed, show that the height $h$ at which the cables cross is independent of the distance separating the posts and calculate its value.
A square is inscribed in a right triangle with legs of $8$ in and $6$ in, such that one of its vertices coincides with the right angle of the triangle, and the two sides…
A square is inscribed in a right triangle with legs of $8$ in and $6$ in, such that one of its vertices coincides with the right angle of the triangle, and the two sides of the square forming that vertex rest on the legs. Calculate the measure of the side of the square $x$ using the similarity of the right triangles formed.