Similar Right Triangles and Indirect Measurement
How tall is your school's flagpole? You can't climb it with a tape measure, but you can find out with a yardstick and a sunny afternoon. The secret is that right triangles hide similar triangles, and similar triangles turn easy measurements into hard ones. In this section you'll learn the tools, then use them on flagpoles, rivers, pool tables, and more.
The Geometric Mean Theorems
Draw a right triangle and drop the altitude from the right angle straight down to the hypotenuse. It splits the hypotenuse into two pieces, called segments (or projections of the legs onto the hypotenuse). It also splits the big triangle into two smaller right triangles.
Both small triangles are right triangles, and each shares an acute angle with the big triangle. By AA similarity, all three triangles are similar. Matching up their sides gives three proportions:
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The left small triangle and the big triangle: $\dfrac{m}{a} = \dfrac{a}{c}$
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The right small triangle and the big triangle: $\dfrac{n}{b} = \dfrac{b}{c}$
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The two small triangles: $\dfrac{h}{m} = \dfrac{n}{h}$
Cross-multiplying each one gives the formulas below.
Some problems call these the altitude theorem and the leg theorem. They're the same formulas.
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.
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?
Finding Every Side
If you know any two of the lengths $m, n, h, a, b, c$ (and they're enough to pin down the triangle), you can find all the rest. Pick the formula that contains what you know and what you want.
| You know | You want | Use |
|---|---|---|
| $m$ and $n$ | $h$ | $h^2 = m \cdot n$ |
| $h$ and $m$ | $n$ | $n = \dfrac{h^2}{m}$ |
| $c$ and $m$ | leg $a$ | $a^2 = m \cdot c$ |
| $a$ and $c$ | $m$ | $m = \dfrac{a^2}{c}$ |
| $m$ and $n$ | $c$ | $c = m + n$ |
There's one more handy fact. The area of the triangle is $\tfrac{1}{2}ab$ (using the legs as base and height) and also $\tfrac{1}{2}ch$ (using the hypotenuse as the base). So $c \cdot h = a \cdot b$. The Pythagorean Theorem, $a^2 + b^2 = c^2$, is always there as a check.
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.
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?
When the Theorems Lead to a Quadratic
Sometimes the problem gives you a relationship between lengths ("one segment is 5 more than the other") instead of the lengths themselves. Then:
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Let $x$ be one unknown length and write the others in terms of $x$.
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Apply the right theorem to get an equation. It will usually be quadratic.
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Solve it (factor or use the quadratic formula).
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Throw out any negative or impossible solution, then check.
The graph is $y = x^2 + 5x - 36$. Its x-intercepts, $-9$ and $4$, are the solutions of the equation in the first example below.
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, Reciprocals, and Altitudes Inside Altitudes
Three more ideas make the hardest problems manageable.
Segments given as a ratio. If the segments are in the ratio $1:4$, call them $t$ and $4t$. The altitude theorem turns into an equation in $t$.
The reciprocal identity. The area equation $c \cdot h = a \cdot b$ plus the Pythagorean Theorem gives a neat relationship between the altitude and the legs.
An altitude inside an altitude. Once you draw the altitude, you get new right triangles. A new altitude drawn in one of them can be handled with the same theorems.
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 $24\text{ cm}$.
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.
Are Two Right Triangles Similar?
Two right triangles already share one pair of equal angles (the right angles). So you only need one more piece of information to prove similarity:
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both pairs of legs are proportional, or
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the hypotenuses and one pair of legs are proportional.
To check, first match sides by size: shorter leg with shorter leg, longer leg with longer leg, hypotenuse with hypotenuse. Then compare the ratios.
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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every length is multiplied by $k$,
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every area is multiplied by $k^2$,
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every volume is multiplied by $k^3$.
Why? An area is length times length, so both lengths get multiplied by $k$: $k \cdot k = k^2$. A volume has three lengths, so you get $k^3$.
A plane slice parallel to the base of a pyramid makes a smaller copy of the base. The small pyramid on top is similar to the whole pyramid, and its scale factor is (distance from the apex) $\div$ (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, Mirrors, and Spotlights
Shadows. The sun is so far away that its rays are parallel. At the same moment, every vertical object makes the same angle with the sunlight. So an object, its shadow, and the ray from its tip form right triangles with equal acute angles. They're similar by AA, and:
$$\frac{\text{height}}{\text{shadow}} = \frac{\text{height}}{\text{shadow}}$$Mirrors. Put a flat mirror on the ground. Light bounces off it so that the angle going in equals the angle coming out. When you see the top of a tree in the mirror, the triangle (your eyes, you, the mirror) and the triangle (the treetop, the tree, the mirror) both have a right angle and equal angles at the mirror. They're similar.
Spotlights. A light on the ground, a fence, and a wall make nested triangles. They share the angle at the light and both have a right angle at the ground. Because the triangles are nested, compare the heights with the distances measured from the light.
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.
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.
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?
Rivers, Crossing Cables, and Folded Paper
These three problems are the same idea in disguise: find two similar right triangles, match the sides, and solve a proportion.
Measuring a river. You can measure a river's width from one bank. Look at a tree on the far bank, walk along your bank at a right angle to your line of sight, place a stake, walk farther, turn $90^\circ$ away from the river, and stop when the stake lines up with the tree. You get a big triangle (with the river as a leg) and a small triangle you can measure completely.
Crossing cables. Two vertical posts stand a distance $d$ apart, with heights $a$ and $b$. A cable runs from the base of each post to the top of the other. How high above the ground do the cables cross? Surprisingly, the answer doesn't depend on $d$.
Folded paper. When you fold a rectangle so that two opposite corners meet, the crease is the perpendicular bisector of the diagonal between those corners. The crease and the diagonal make right triangles that are similar to the right triangle cut by the diagonal.
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.
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.
Back to the flagpole: a $3\text{ ft}$ yardstick with a $4\text{ ft}$ shadow, next to a $28\text{ ft}$ shadow, gives $\dfrac{3}{4} = \dfrac{h}{28}$, so the flagpole is $21\text{ ft}$ tall, measured without leaving the ground.