Pythagoras' theorem: problems questions with answers

Questions

1. Question 1Ladders and cables

   A ladder $3$ m long leans against an upright wall. Its foot is on level ground, $0.8$ m from the wall. Find the height, $h$ m, that the ladder reaches up the wall, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Name the hypotenuse: opposite the right angle between the upright and the groundThe hypotenuse is the ladder, opposite the right angle at the foot of the wall.
   2. Step 2: Use Pythagoras' theorem, then square-root at the end$h^2 = 3^2 - 0.8^2$$= 8.36$$h = \sqrt{8.36}$$= 2.8913\ldots$$= 2.89$ m (3 s.f.)
2. Question 2Diagonals of rectangles

   The diagram shows a rectangular field $ABCD$. It is $60$ m long and its diagonal $AC$ is $75$ m. Find its width, $x$ m.

   Show the worked solution

   1. Step 1: The diagonal cuts the rectangle into two right-angled triangles: name the hypotenuseIn triangle $ABC$, $\angle ABC = 90^\circ$, so the hypotenuse is $AC$.
   2. Step 2: Use Pythagoras' theorem, then square-root at the end$x^2 = 75^2 - 60^2$$= 2025$$x = \sqrt{2025}$$= 45$ m
3. Question 3Due north, then due east

   A ship travels $5$ km due north from $P$ to $Q$, then $11$ km due east from $Q$ to $R$. Find the distance $PR$, $x$ km, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: North–south and east–west are at right angles: where is the right angle?$\angle PQR = 90^\circ$, so $PR$ is the hypotenuse.
   2. Step 2: Use Pythagoras' theorem, then square-root at the end$x^2 = 5^2 + 11^2$$= 146$$x = \sqrt{146}$$= 12.0830\ldots$$= 12.1$ km (3 s.f.)
4. Question 4Heights of isosceles triangles

   The cross-section of a tent is an isosceles triangle $ABC$ with $AB = AC = 5$ m and base $BC = 6$ m. $D$ is the midpoint of $BC$. Find the height $AD$, $h$ m.

   Show the worked solution

   1. Step 1: Find BD$BD = \dfrac{6}{2}$$= 3$
   2. Step 2: In right-angled triangle ABD, find h$h^2 = 5^2 - 3^2$$= 16$$h = \sqrt{16}$$= 4$ m
5. Question 5Two triangles, two steps

   In the diagram, $\angle PQR = 90^\circ$ and $S$ lies on $QR$. $PS = 20$ cm, $QS = 12$ cm and $SR = 6$ cm. Find $x$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Triangle PQS: find PQ$PQ^2 = 20^2 - 12^2$$= 256$$PQ = \sqrt{256}$$= 16$
   2. Step 2: QR is the two parts added$QR = 12 + 6$$= 18$
   3. Step 3: Triangle PQR: find x (use the unrounded PQ)$x^2 = 16^2 + 18^2$$= 580$$x = \sqrt{580}$$= 24.0831\ldots$$= 24.1$ cm (3 s.f.)
6. Question 6Shortest distance to the hypotenuse (area two ways)

   In the diagram, $\angle DEF = 90^\circ$, $DE = 14$ cm and $EF = 48$ cm. $G$ is the point on $DF$ such that $EG$ is perpendicular to $DF$. Find the shortest distance from $E$ to $DF$, $h$ cm, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: First find the hypotenuse DF: ADD the squares, then square-root$DF^2 = 14^2 + 48^2$$= 2500$$DF = \sqrt{2500}$$= 50$
   2. Step 2: The area of triangle DEF, using the two shorter sides (they are at right angles)Area $= \dfrac{1}{2} \times 14 \times 48$$= 336$ cm²
   3. Step 3: The same area, with DF as the base: the height is the perpendicular h$\dfrac{1}{2} \times 50 \times h = 336$$h = \dfrac{2 \times 336}{50}$$= 13.4400\ldots$$= 13.4$ cm (3 s.f.)