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# Difficult Pythagoras Questions (Year 10, Guided Answers)

## ✩ You must already know

1. The Pythagoras Theorem
2. Area of Triangles
3. Simple Algebraic Equations

## Questions

Try to use a minimum number of hints to reach the solution. All the best!

If you find these challenging, learn to solve the Pythagoras problems here.

### 1 Question

In the right triangle ABC, point D is equidistant from AB and BC and AD = 30, DC = 40. Find lengths of sides AB and BC.

#### Answer

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?

### 2 Question

In above right triangle:

Angle ACB is 90 degrees and sides AC = m, BC = n and AB = p

Altitude CD = h

AD = m1 and DB = n1

Prove that:

1. h2 = m1.n1

2. m2 = n1.p

3. n2 = m1.p

4.

#### Answer

1. h2 = m1.n1

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

### 3 Question

Prove that in a right triangle, the hypotenuse is twice the median drawn to the hypotenuse.

#### Answer

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?

### 4 Question

In a right triangle ABC, length of the medians to the sides AB and BC are and respectively. Find the length of its hypotenuse.

#### Answer

Given:

AD =

CE =

Let AE = EB = x

BD = DC = y

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

## Answers

### 1 Answer

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

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

(Equation 4)

(Equation 4)

Solve for the sides of the right triangle.

Using this value of y in equation 2:

From equation 4:

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

### 2 Answer

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.

Multiplying the results m2 = n1p and n2 = m1p:

m2n2 = (n1p)(m1p)

m2n2 = n1m1p2

Replacing m1n1 by h2:

m2n2 = h2p2

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.

### 3 Answer

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:

Using equation 1, we replace h2 + x2 by y2

(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:

y2 = m2

y = m

We are done!

### 4 Answer

Given:

AD =

CE =

Let AE = EB = x

BD = DC = y

Using triangle ABD:

AB2 + BD2 = AD2

(Equation 1)

Using triangle EBC :

EB2 + BC2 = EC2

(Equation 2)

Adding Equation 1 and 2:

4x2 + y2 + x2 + 4y2 = 601 + 244

5x2 + 5y2 = 845

(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 from Equation 3:

4(169) = AC2

AC = 2 × 13 = 26

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