Similar Right Triangles
Aaliyah wants to know how tall the flagpole in front of her school is. She can't climb it, and her 25-ft tape measure is far too short. But she can measure distances on the ground, and she has a small mirror in her backpack. Is that enough? Yes, because similar right triangles turn short, easy measurements into tall, impossible ones. By the end of this page you will see exactly how.
Are two right triangles similar?
Two right triangles already have one pair of equal angles: the $90^\circ$ angles. So the AA test only needs one more pair of equal acute angles.
You can also test with sides. Two right triangles are similar when:
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their two legs are in proportion, or
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the hypotenuse and one leg are in proportion.
Why does the second test work? If the hypotenuse and one leg are scaled by the same factor $k$, the Pythagorean Theorem forces the other leg to be scaled by $k$ too.
A carpenter cuts two wooden wedges in the shape of a right triangle. The first has legs of $15\text{ cm}$ and $20\text{ cm}$. The second has a hypotenuse of…
A carpenter cuts two wooden wedges in the shape of a right triangle. The first has legs of $15\text{ cm}$ and $20\text{ cm}$. The second has a hypotenuse of $50\text{ cm}$ and a leg of $30\text{ cm}$. Mathematically check if both wedges are similar.
Scale factor, area and volume
If two figures are similar with scale factor $k$, then:
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lengths are multiplied by $k$,
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areas are multiplied by $k^2$,
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volumes are multiplied by $k^3$.
Why? Area has two dimensions, and each one is stretched by $k$. Volume has three dimensions.
When a plane parallel to the base cuts a pyramid, the small pyramid on top is similar to the whole one. Its scale factor is $k = \dfrac{\text{distance from the vertex}}{\text{total height}}$.
Two traffic signs shaped like right triangles are similar. The shorter leg of the small sign measures $30\text{ cm}$ and its area is $600\text{ cm}^2$. If the shorter…
Two traffic signs shaped like right triangles are similar. The shorter leg of the small sign measures $30\text{ cm}$ and its area is $600\text{ cm}^2$. If the shorter leg of the large sign measures $60\text{ cm}$, what is the area of the large sign?
A right square-based pyramid is cut by a plane parallel to the base at one third of its height (measuring from the vertex). If the area of the cross-section obtained is…
A right square-based pyramid is cut by a plane parallel to the base at one third of its height (measuring from the vertex). If the area of the cross-section obtained is $15\text{ cm}^2$, what is the volume of the original pyramid if its total height is $18\text{ cm}$?
Shadows and light sources
The sun's rays are essentially parallel. So at the same time and place, every vertical object and its shadow form a right triangle with the same acute angle. These triangles are similar by AA, and $\dfrac{\text{height}}{\text{shadow}}$ is the same for all of them.
A lamp or spotlight works the same way, except the rays now fan out from a point.
A volleyball post casts a shadow of $3,5\text{ m}$ on the horizontal ground. At the same time and in the same place, a water bottle $0,25\text{ m}$ high casts a shadow…
A volleyball post casts a shadow of $3,5\text{ m}$ on the horizontal ground. At the same time and in the same place, a water bottle $0,25\text{ m}$ high casts a shadow of $0,35\text{ m}$. Calculate the height of the post using similarity of triangles.
An opaque fence $2\text{ m}$ high is located in front of the vertical wall of a building. A spotlight on the ground $1,5\text{ m}$ from the fence casts its shadow onto…
An opaque fence $2\text{ m}$ high is located in front of the vertical wall of a building. A spotlight on the ground $1,5\text{ m}$ from the fence casts its shadow onto the building, reaching a height of $6\text{ m}$. If we move the spotlight $1\text{ m}$ further from the fence (placing it at $2,5\text{ m}$), how much will the height of the shadow cast on the building decrease?
The mirror method
On a cloudy day there are no shadows, but a flat mirror on the ground still works. Light bounces off a mirror so that the angle of incidence equals the angle of reflection. Both the observer and the object form right triangles with the ground, and these triangles have equal acute angles at the mirror. That makes them similar by AA.
Back to the flagpole: with nothing but a mirror and a tape measure on the ground, Aaliyah found that the pole is $40$ ft tall.
To measure the height of a wind turbine, an engineer places a flat mirror on the ground $12\text{ m}$ from the base of the tower. She steps back $1,5\text{ m}$ from the…
To measure the height of a wind turbine, an engineer places a flat mirror on the ground $12\text{ m}$ from the base of the tower. She steps back $1,5\text{ m}$ from the mirror until she sees the highest point of the wind turbine reflected exactly in the center of the mirror. If the engineer's eyes are $1,6\text{ m}$ from the ground, calculate the height of the wind turbine.
The altitude to the hypotenuse
Draw the altitude from the right angle to the hypotenuse of a right triangle. It cuts the triangle into two smaller right triangles. Each of them shares an acute angle with the big one, so all three triangles are similar.
Here the altitude splits the hypotenuse $c$ into two pieces called projections of the legs onto the hypotenuse: $x$ is the piece next to leg $a$, and $y$ is the piece next to leg $b$. The geometric mean of two positive numbers $p$ and $q$ is $\sqrt{pq}$.
Why? Compare the small triangles with matching sides. The two small triangles give $\dfrac{x}{h} = \dfrac{h}{y}$, so $h^2 = xy$. The big triangle and the one containing $a$ give $\dfrac{c}{a} = \dfrac{a}{x}$, so $a^2 = xc$. The area rule is just the area of the big triangle counted two ways: $\tfrac{1}{2}ab = \tfrac{1}{2}ch$.
A nice check: $a^2 + b^2 = xc + yc = (x + y)c = c^2$, which is the Pythagorean Theorem again. Some problems call these results the altitude theorem and the leg theorem.
In a pitched roof that forms a right angle at its top vertex, the altitude drawn from this vertex divides the base (hypotenuse) into two segments of $2\text{ m}$ and…
In a pitched roof that forms a right angle at its top vertex, the altitude drawn from this vertex divides the base (hypotenuse) into two segments of $2\text{ m}$ and $8\text{ m}$. Apply the altitude theorem to calculate the maximum height of the roof.
A skate ramp has the profile of a right triangle resting on its hypotenuse. The horizontal base (hypotenuse) measures $10\text{ m}$ and the projection of the upward…
A skate ramp has the profile of a right triangle resting on its hypotenuse. The horizontal base (hypotenuse) measures $10\text{ m}$ and the projection of the upward section onto the base is $3,6\text{ m}$. Use the leg theorem to calculate the length of the upward section.
In a metal structure shaped like a right triangle, a vertical pillar (altitude to the hypotenuse) divides the lower beam (hypotenuse) into two parts. If an inclined side…
In a metal structure shaped like a right triangle, a vertical pillar (altitude to the hypotenuse) divides the lower beam (hypotenuse) into two parts. If an inclined side (leg) measures $5\text{ m}$ and the total lower beam measures $10\text{ m}$, how long is the projection of that side onto the lower beam?
Finding every length
A right triangle with its altitude has six lengths: $a, b, c, h, x, y$. Knowing any two that are independent usually lets you find all the rest. Plan:
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Mark which projection touches which leg.
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Pick the theorem that links what you know with one unknown.
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Use $c = x + y$ whenever you need a missing piece.
From $a^2 = xc$ you get a handy formula: $$x = \dfrac{a^2}{c}$$ The projection of a leg is the leg squared over the hypotenuse.
The central mast of a tent forms a right angle with the two main ropes that secure it from the top to the ground. The mast is $4\text{ m}$ high and one of the ropes is…
The central mast of a tent forms a right angle with the two main ropes that secure it from the top to the ground. The mast is $4\text{ m}$ high and one of the ropes is anchored at $3,2\text{ m}$ from the base of the mast. At what distance from the base is the other rope anchored?
In a technical drawing class, a student draws a right triangle with a hypotenuse of $13\text{ cm}$ and legs of $5\text{ cm}$ and $12\text{ cm}$. Then, they draw the…
In a technical drawing class, a student draws a right triangle with a hypotenuse of $13\text{ cm}$ and legs of $5\text{ cm}$ and $12\text{ cm}$. Then, they draw the altitude relative to the hypotenuse. Use the leg theorem to calculate the lengths of the two projections of the legs onto the hypotenuse.
The facade of a modern building has a large window shaped like a right triangle resting on its hypotenuse. The height of the window with respect to the base is…
The facade of a modern building has a large window shaped like a right triangle resting on its hypotenuse. The height of the window with respect to the base is $4,8\text{ m}$ and the projection of the shorter side onto the base is $3,6\text{ m}$. Calculate the length of the two legs and the hypotenuse.
Setting up quadratic equations
Sometimes the problem gives you a relationship between two lengths rather than their values. Then you call one length $x$, write the other in terms of $x$, and let a theorem build an equation.
Another useful setup is to use the Pythagorean Theorem when the hypotenuse and a relationship between the legs are given.
An agricultural plot of land has the shape of a right triangle. The projection of one of the legs onto the hypotenuse is $4\text{ dam}$ greater than the projection of…
An agricultural plot of land has the shape of a right triangle. The projection of one of the legs onto the hypotenuse is $4\text{ dam}$ greater than the projection of the other leg. If the altitude relative to the hypotenuse measures $4,8\text{ dam}$, set up a quadratic equation and calculate the total length of the hypotenuse.
The design of a windsurf sail has the shape of a right triangle. It is known that the lower edge (hypotenuse) measures $5\text{ m}$ and that the trailing edge (one of…
The design of a windsurf sail has the shape of a right triangle. It is known that the lower edge (hypotenuse) measures $5\text{ m}$ and that the trailing edge (one of the legs) measures $1,2\text{ m}$ more than its projection onto the hypotenuse. Determine the two possible lengths the trailing edge can have using the leg theorem.
A cable-stayed bridge has a main pillar that forms a right angle with the two longest stay cables. The total length of the deck between the anchorages of these cables…
A cable-stayed bridge has a main pillar that forms a right angle with the two longest stay cables. The total length of the deck between the anchorages of these cables (hypotenuse) is $125\text{ m}$. If one of the cables measures $25\text{ m}$ more than the other, determine the height of the main pillar (altitude to the hypotenuse).
Ratios and rectangles
When the two projections are given as a ratio, write them as $x = pt$ and $y = qt$ for the same unknown $t$. Then $h^2 = xy$ gives you one equation for $t$.
The altitude and a projection can also be the sides of a rectangle inside the triangle. Such a rectangle has sides $h$ and $x$ (or $h$ and $y$), so its area is $h \cdot x$.
In a museum, a rectangular painting is framed by a right triangle resting on its hypotenuse, such that the altitude to the hypotenuse coincides with one of the sides of…
In a museum, a rectangular painting is framed by a right triangle resting on its hypotenuse, such that the altitude to the hypotenuse coincides with one of the sides of the painting. If the projections of the legs onto the base measure $16\text{ cm}$ and $9\text{ cm}$, and the painting occupies exactly the rectangle formed by the altitude and the shorter projection, what is the area of the painting?
In an architectural design, the profile of a roof is a right triangle resting on its hypotenuse. The altitude drawn from the vertex of the right angle divides the…
In an architectural design, the profile of a roof is a right triangle resting on its hypotenuse. The altitude drawn from the vertex of the right angle divides the hypotenuse into two segments that are in the ratio $9:16$. If the altitude measures $6\text{ m}$, calculate the total perimeter of the triangle forming the roof.
Using the altitude twice, and a proof
The tool $c \cdot h = a \cdot b$ lets you find the altitude without projections. And the small triangles cut by the altitude are right triangles too, so you can draw another altitude inside them and apply the theorems again.
Now an invitation to prove something. Take the right triangle with legs $9$ and $12$ (hypotenuse $15$). The altitude is $h = \dfrac{9 \cdot 12}{15} = 7.2$, so $\dfrac{1}{h^2} = \dfrac{1}{51.84}$. Now add the reciprocals of the squares of the legs: $$\dfrac{1}{81} + \dfrac{1}{144} = \dfrac{16 + 9}{1296} = \dfrac{25}{1296} = \dfrac{1}{51.84}$$ They match! This is not a coincidence, and it works for every right triangle: $\dfrac{1}{h^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}$.
An industrial designer creates a flat part in the shape of a right triangle whose legs measure $15\text{ in}$ and $20\text{ in}$. To reinforce the part, they place a…
An industrial designer creates a flat part in the shape of a right triangle whose legs measure $15\text{ in}$ and $20\text{ in}$. To reinforce the part, they place a straight rod perpendicular to the hypotenuse passing through the vertex of the right angle (that is, the altitude). In turn, from the foot of this altitude, they place another rod perpendicular to the longer leg. How long is this last rod?
Prove that in any right triangle, the reciprocal of the square of the altitude to the hypotenuse is equal to the sum of the reciprocals of the squares of the legs…
Prove that in any right triangle, the reciprocal of the square of the altitude to the hypotenuse is equal to the sum of the reciprocals of the squares of the legs ($\frac{1}{h^2} = \frac{1}{b^2} + \frac{1}{c^2}$). Then, verify this relationship by calculating the altitude in both ways in a triangle with legs $10\text{ cm}$ and $24\text{ cm}$.
Overlapping triangles
Some figures hide two right triangles that share an angle. Both have a right angle, so they are similar by AA. Then you only have to match the sides carefully and solve a proportion.
On a billiard table, a ball must hit the cushion and rebound towards a pocket, forming two similar right triangles. The ball is $20\text{ cm}$ from the cushion and the…
On a billiard table, a ball must hit the cushion and rebound towards a pocket, forming two similar right triangles. The ball is $20\text{ cm}$ from the cushion and the pocket is $60\text{ cm}$ from the same cushion. If the horizontal distance parallel to the cushion between the ball and the pocket is $100\text{ cm}$, at what horizontal distance from the ball's initial position must the impact on the cushion occur?
In a piece of jewelry, a square of side $x$ is inscribed within a right triangle with legs of $6\text{ mm}$ and $8\text{ mm}$, such that the vertex of the right angle of…
In a piece of jewelry, a square of side $x$ is inscribed within a right triangle with legs of $6\text{ mm}$ and $8\text{ mm}$, such that the vertex of the right angle of the triangle coincides with a vertex of the square and the opposite vertex of the square rests on the hypotenuse. Calculate the value of $x$ using similarity of triangles.
A rectangular sheet of paper $8\text{ cm}$ wide and $15\text{ cm}$ long is folded so that two opposite vertices coincide. The fold formed is a straight line. Calculate…
A rectangular sheet of paper $8\text{ cm}$ wide and $15\text{ cm}$ long is folded so that two opposite vertices coincide. The fold formed is a straight line. Calculate the length of this fold based on the similarity of the right triangles formed.
Hourglass figures: rivers and crossing cables
When two lines cross, the vertical angles are equal. If two triangles are placed tip to tip like an hourglass and each has a right angle (or the sides are parallel), AA gives similar triangles. Surveyors use this to measure a river's width without crossing it.
The same hourglass shape appears with two vertical posts and two crossing cables.
A surveyor wants to calculate the width of a river. To do this, they stand on one bank and observe a tree on the opposite bank. They walk $30\text{ m}$ along the bank…
A surveyor wants to calculate the width of a river. To do this, they stand on one bank and observe a tree on the opposite bank. They walk $30\text{ m}$ along the bank (perpendicular to the initial line of sight) and place a ranging pole. Then they walk another $6\text{ m}$ in the same direction, turn $90^\circ$ away from the river and walk $4\text{ m}$ until they see the pole aligned with the tree. What is the width of the river?
In a statics study, two vertical posts of heights $a = 120\text{ cm}$ and $b = 80\text{ cm}$ are separated by a distance $d$. From the base of each post, a taut cable is…
In a statics study, two vertical posts of heights $a = 120\text{ cm}$ and $b = 80\text{ cm}$ are separated by a distance $d$. From the base of each post, a taut cable is pulled to the top of the other post. The cables cross at a height $h$. Show using similarity of triangles that the height $h$ at which they cross does not depend on the distance $d$ and calculate its value.