Difficult Pythagoras Questions (Year 10, Guided Answers)

Diagram and Variables

Let the distance of D from AB be x

∴ ED = DF = x (D is equidistant from AB BC)

also BF = ED = x and EB = DF = x

Let AE = z ∴ AB = AE + EB = z + x

Let FC = y ∴ BC = BF + FC = x + y

Equating the area of triangle with its parts can you find the relation between x, y and z?

Ans.  AB = 42 and BC = 56

1. h² = m₁.n₁

Can you find relation between h, m₁, m using △ ACD?

Let ABC be the right triangle.

Side AB = a, BC = b.

BD = Median to hypotenuse, so AD = DC = m

Construct altitude BE. Let ED = x

Let BE = h, BD = y

So we have to prove y = m

Can you now form three equations using triangles BDE, BEA and BCE and link x, h, y with a, b and m?

Given:

AD = $\sqrt{601}$

CE = $2\sqrt{61}$

Let AE = EB = x

BD = DC = y

Can you use the Pythagorean Theorem to form an equation between x, y, and AD?

Ans.  AC = 26

Diagram and Variables

Let the distance of D from AB be x

∴ ED = DF = x (D is equidistant from AB BC)

also BF = ED = x and EB = DF = x

Let AE = z ∴ AB = AE + EB = z + x

Let FC = y ∴ BC = BF + FC = x + y

Equations

Area of △ AED + Area of square EBFD + Area of △ DFC = Area of △ ABC

$\frac{zx}{2}+x^2+\frac{xy}{2}=\frac{(z+x)(x+y)}{2}$

zx + 2x² + xy = (z + x)(x + y)

zx + 2x² + xy = zx + zy + x² + xy

2x² = zy + x²

x² = zy (Equation 1)

Applying the Pythagoras theorem on △ AED :

z² + x² = AD² = 30² = 900

z² + x² = 900

Replacing x², using equation 1:

z² + zy = 900

z(z + y) = 900 (Equation 2)

Applying the Pythagoras theorem on △ DFC :

y² + x² = DC² = 40² = 1600

y² + x² = 1600

Replacing x² by zx, using equation 1:

y² + zy = 1600

y(y + z) = 1600 (Equation 3)

Dividing equation 3 by 2

$\frac{y(y+z)}{z(y+z)}=\frac{1600}{900}$

$\frac{y}{z}=\frac{16}{9}$

$y÷z=16÷9$

$y/z=16/9$

$y=\frac{16z}{9}$ (Equation 4)

$y=16z/9$ (Equation 4)

Solve for the sides of the right triangle.

Using this value of y in equation 2:

$z(z+\frac{16z}{9})=900$

$z^2+\frac{16z^2}{9}=900$

$\frac{25z^2}{9}=900$

$25z^2=900\times 9$

$z^2=\frac{900\times 9}{25}$

$z=\frac{30\times 3}{5}=6\times 3=18$

From equation 4:

$y=\frac{16\times 18}{9}=16\times 2=32$

Using equation 1:

x² = zy = 18 × 32 = 9 × 64

x = 3 × 8 = 24

Therefore side AB = z + x = 18 + 24 = 42

and BC = x + y = 24 + 32 = 56

1. h² = m₁.n₁

From △ ACD :

h² + m₁² = m²

h² = m² − m₁² Equation 1

From △ CDB :

h² + n₁² = n²

h² = n² − n₁² (Equation 2)

Adding Equation 1 and 2:

h² + h² = m² − m₁² + n² − n₁²

h² + h² = m² + n² − m₁² − n₁²

2h² = m² + n² − m₁² − n₁² (Equation 3)

From △ ABC:

m² + n² = AB² = (m₁ + n₁)² as AB = m₁ + n₁

Using this result to replace m² + n² in Equation 3

2h² = (m₁ + n₁)² − m₁² − n₁²

2h² = m₁² + n₁² + 2m₁n₁ − m₁² − n₁²

2h² = 2m₁n₁

h² = m₁n₁

Proved!

2. m² = n₁.p

From △ ACD:

m² = h² + m₁²

Using ( h² = m₁.n₁), replace h²:

m² = m₁n₁ + m₁²

m² = m₁(n₁ + m₁)

Now m_1 + n_1 = AB = p so

m² = m₁(p)

h² = m₁.p

Proved!

3. n² = m₁.p

If you have done the previous proof, you don’t need hints for this one! The proof is very similar.

4. $h=\frac{mn}{p}$

Multiplying the results m² = n₁p and n² = m₁p:

m²n² = (n₁p)(m₁p)

m²n² = n₁m₁p²

Replacing m₁n₁ by h²:

m²n² = h²p²

$\frac{m^2{n}^2}{p^2}=h^2$

$\frac{mn}{p}=h$

Proved!

Another very simple way to prove this is by using the area of the triangle. I leave it to you to try it out.

Let ABC be the right triangle.

Side AB = a, BC = b.

BD = Median to hypotenuse, so AD = DC = m

Construct altitude BE. Let ED = x

Let BE = h, BD = y

So we have to prove y = m

Equations

From triangle BDE

BD² = BE² + ED²

y² = h² + x² (Equation 1)

From triangle BEA:

AB² = BE² + AE²

a² = h² + (m − x)²

h² = a² − (m − x)² (Equation 2)

Similarly form equation from triangle BCE:

BC² = BE² + EC²

b² = h² + (m + x)²

h² = b² − (m + x)² (Equation 3)

Solve for y

Adding equation 2 and 3:

2h² = a² − (m − x)² + b² − (m + x)²

2h² = a² + b² − [(m − x)² + (m + x)²]

2h² = a² + b² − [m² + x² − 2mx + m² + x² + 2mx]

2h² = a² + b² − [2m² + 2x²]

2h² = a² + b² − 2m² − 2x²

2h² + 2x² = a² + b² − 2m²

Dividing both side by 2:

$h^2+x^2=\frac{a^2+b^2-2m^2}{2}$

Using equation 1, we replace h² + x² by y²

$y^2=\frac{a^2+b^2-2m^2}{2}$ (Equation 4)

Now from right triangle ABC, we know that:

a² + b² = AC² = (2m)² (as AC =AD + DC = m+m = 2m)

a² + b² = 4m² (Equation 5)

Putting value of a² + b² in equation 4:

$y^2=\frac{4m^2-2m^2}{2}$

$y^2=\frac{2m^2}{2}$

y² = m²

y = m

We are done!

Given:

AD = $\sqrt{601}$

CE = $2\sqrt{61}$

Let AE = EB = x

BD = DC = y

Using triangle ABD:

AB² + BD² = AD²

$(2x{)}^2+y^2=(\sqrt{601}{)}^2$

$4x^2+y^2=601$ (Equation 1)

Using triangle EBC :

EB² + BC² = EC²

$x^2+(2y{)}^2=(2\sqrt{61}{)}^2$

$x^2+4y^2=(2\sqrt{61}{)}^2$

$x^2+4y^2=244$ (Equation 2)

Adding Equation 1 and 2:

4x² + y² + x² + 4y² = 601 + 244

5x² + 5y² = 845

$x^2+y^2=169$ (Equation 3)

Let us solve for the hypotenuse using the triangle ABC:

AB² + BC² = AC²

(2x)² + (2y)² = AC²

4x² + 4y² = AC²

4(x² + y²) = AC²

Substituting the value of $x^2+y^2$ from Equation 3:

4(169) = AC²

$AC=\sqrt{4(169)}$

AC = 2 × 13 = 26