Pythagoras: Practise Online

Calculator 3 marks

Here is a right-angled triangle.

Calculate the length of the side marked \(x\).
Your answer (just for you: it is never sent anywhere)
Show answer
\(17\ \text{cm}\)
The side marked \(x\) is the hypotenuse, so square the two shorter sides and add.

\(x^2 = 15^2 + 8^2 = 225 + 64 = 289\)

\(x = \sqrt{ 289 } = 17\)

\(x = 17\) cm
- M1squares of both shorter sides \(225\) and \(64\)
- M1adds the squares to get \(289\)
- A1square root, \(17\) cm
Calculator 3 marks

Here is a right-angled triangle.

Work out the length of the side marked \(x\).
Your answer (just for you: it is never sent anywhere)
Show answer
\(18\ \text{cm}\)
Here the hypotenuse is known, so subtract rather than add.

\(x^2 = 30^2 - 24^2 = 900 - 576 = 324\)

\(x = \sqrt{ 324 } = 18\)

\(x = 18\) cm
- M1subtracting the squares \(900\) - \(576\)
- M1\(x^2 = 324\)
- A1\(x = 18\) cm
Calculator 3 marks

\(ABCD\) is a rectangle.

Work out the length of the diagonal \(AC\).
Your answer (just for you: it is never sent anywhere)
Show answer
\(34\ \text{cm}\)
\(ABC\) is a right-angled triangle, because every corner of a rectangle is a right angle. \(AC\) is its hypotenuse.

\(AC^2 = 16^2 + 30^2 = 256 + 900 = 1156\)

\(AC = \sqrt{ 1156 } = 34\)

\(AC = 34\) cm
- M1adding the squares \(256\) + \(900\)
- M1\(AC^2 = 1156\)
- A1\(AC = 34\) cm
Calculator 5 marks

\(ABCD\) is a rectangle. The dashed line is the diagonal \(AC\).

(a)
Calculate the length of \(AC\).

(b)
Calculate the area of triangle \(ABC\).
Your answer (just for you: it is never sent anywhere)
Show answer
(a) \(51\ \text{cm}\)
(b) \(540\ \text{cm}^2\)
The diagonal cuts the rectangle into two right-angled triangles. In triangle \(ABC\) the right angle is at \(B\), so \(AC\) is the hypotenuse.

\(AC^2 = 24^2 + 45^2 = 576 + 2025 = 2601\)

\(AC = \sqrt{ 2601 } = 51\)

\(AC = 51\) cm
The diagonal splits the rectangle into two identical triangles, so triangle \(ABC\) is half the rectangle.

\(\frac{1}{2} \times 24 \times 45 = \frac{ 1080 }{2} = 540\)

The area is \(540\) cm².
- (a) M1\(AC^2 = 24^2 + 45^2\)
- (a) M1\(AC^2 = 2601\) and square root
- (a) A1\(AC = 51\) cm
- (b) M1\(\frac{1}{2} \times 24 \times 45\)
- (b) A1area \(540\) cm²
Calculator 3 marks

The diagram shows a right-angled triangle.

Use Pythagoras' theorem to show that \(x = 0.4\)
Your answer (just for you: it is never sent anywhere)
Show answer
\(x^2 = 0.25 - 0.09 = 0.16 = 0.4^2\), so \(x = 0.4\) cm
The hypotenuse is the longest side and is known, so subtract the squares.

\(x^2 = 0.5^2 - 0.3^2\)

\(x^2 = 0.25 - 0.09 = 0.16\)

Check that this is the square of the stated value: \(0.4^2 = 0.16\)

\(x = \sqrt{ 0.16 } = 0.4\) cm
- M1subtracts the squares: \(x^2 = 0.5^2 - 0.3^2\)
- M1squares evaluated and subtracted: \(x^2 = 0.25 - 0.09 = 0.16\)
- A1shows \(0.16 = 0.4^2\), so \(x = 0.4\)
Calculator 2 marks

Here is a right-angled triangle.

Kai works out that \(y = 13\).

Mel says, "If \(y = 13\), the hypotenuse would be shorter than both of the other sides, so \(y\) cannot be \(13\)."

Is Mel correct?

Tick a box. Yes / No

Give a reason for your answer.
Your answer (just for you: it is never sent anywhere)
Show answer
No. With \(y = 13\) the label \(3y\) cm gives a hypotenuse of \(39\) cm, which is longer than the other sides, \(15\) cm and \(36\) cm.
The hypotenuse is the side opposite the right angle. It is labelled \(3y\) cm.

\((3y)^2 = 15^2 + 36^2 = 225 + 1296 = 1521\)

\(3y = 39\) cm

Divide both sides by 3: \(y = 13\) cm. The hypotenuse is 39 cm and the other two sides are 15 cm and 36 cm.

Kai's value is \(y = 13\) cm. Putting this in gives the correct hypotenuse.

Putting \(y = 13\) into the label \(3y\) gives a hypotenuse of \(39\) cm, which is longer than the other sides, 15 cm and 36 cm.

The hypotenuse would be longer than both other sides, which is possible, so Mel is not correct.

The answer is No.
- M1uses the coefficient in the label \(3y\) cm: either solves \(3y = 39\) to get \(y = 13\) cm, or puts \(y = 13\) into the label to get a hypotenuse of \(39\) cm (treating the label as just \(y\) earns nothing)
- A1No, as a hypotenuse of \(39\) cm is longer than the other sides \(15\) and \(36\)
Calculator 1 mark

The diagram shows a right-angled triangle.

Noah is using Pythagoras' theorem to find the value of \(y\). Here is his working.

\(PQ^2 = RP^2 + QR^2\)

\(2y^2 = 7^2 + 24^2\)

\(2y^2 = 49 + 576 = 625\)

\(y = \sqrt{\frac{625}{2}}\)

Here are four statements about Noah's working.

1. Noah has squared only part of the length of the hypotenuse. 2. Noah has added the two shorter sides instead of square rooting the total of their squares. 3. Noah has treated one of the shorter sides as the hypotenuse. 4. Noah has made a slip when adding two numbers.

What mistake has Noah made? Write down the number of the correct statement.
Your answer (just for you: it is never sent anywhere)

Show answer

Statement 1: Noah has squared only part of the length of the hypotenuse.
The right angle is at R, so the hypotenuse is the side opposite it, PQ. Pythagoras' theorem for this triangle is

\(PQ^2 = RP^2 + QR^2\)

Compare each of Noah's lines with this, one at a time.

The second line has written \((2y)^2\) as \(2y^2\): the 2 has not been squared.

Every addition in the working is correct, so statement 4 is not the mistake. The correct statement is 1: Noah has squared only part of the length of the hypotenuse.
Calculator 3 marks

A triangle has sides of length 0.6 cm, 2.5 cm and 2.4 cm.

Is the triangle right-angled? You must show your working.
Your answer (just for you: it is never sent anywhere)
Show answer
No: \(0.6^2 + 2.4^2 = 6.12\) and \(2.5^2 = 6.25\), and 6.12 is less than 6.25, so the angle opposite the longest side is more than 90° and the triangle is not right-angled
The longest side is 2.5 cm, so it is the only side that could be a hypotenuse. Add the squares of the other two sides and compare with the square of the longest.

\(0.6^2 + 2.4^2 = 0.36 + 5.76 = 6.12\)

\(2.5^2 = 6.25\)

6.12 is less than 6.25.

No: the angle opposite the longest side is more than 90° and the triangle is not right-angled.
- M1sum of the squares of the two shorter sides, \(6.12\), found with \(0.6\) and \(2.4\)
- M1square of the longest side, \(6.25\), found from \(2.5\)
- A1No, comparing \(6.12\) with \(6.25\)
Calculator 5 marks

\(ABC\) is an isosceles triangle with \(AB = AC\).

(a)
Work out the perpendicular height of the triangle from \(A\) to \(BC\).

(b)
Work out the area of triangle \(ABC\).
Your answer (just for you: it is never sent anywhere)
Show answer
(a) \(21\ \text{cm}\)
(b) \(420\ \text{cm}^2\)
The perpendicular from \(A\) cuts \(BC\) exactly in half, giving a right-angled triangle with hypotenuse \(29\) cm and one shorter side \(40 \div 2 = 20\) cm.

\(h^2 = 29^2 - 20^2 = 841 - 400 = 441\)

\(h = \sqrt{ 441 } = 21\)

The perpendicular height is \(21\) cm.
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)

\(= \frac{1}{2} \times 40 \times 21 = 420\)

The area is \(420\) cm².
- (a) M1half the base \(20\) cm used as a side of a right-angled triangle with hypotenuse \(29\) cm
- (a) M1\(h^2 = 841 - 400 = 441\)
- (a) A1height \(21\) cm
- (b) M1\(\frac{1}{2} \times 40 \times 21\)
- (b) A1area \(420\) cm²
Calculator 3 marks

Amara sails her yacht 8 km due west, then turns and sails 6 km due north.

Calculate how far she is from the starting point.
Your answer (just for you: it is never sent anywhere)
Show answer
\(10\ \text{km}\)
The two legs of the journey are at right angles, so they are the shorter sides of a right-angled triangle and the distance home is the hypotenuse.

\(d^2 = 8^2 + 6^2 = 64 + 36 = 100\)

\(d = \sqrt{ 100 } = 10\)

Amara is \(10\) km from the starting point.
- M1adding the squares \(64\) + \(36\)
- M1\(d^2 = 100\)
- A1\(10\) km