Finding the hypotenuse questions with answers

Questions

1. Question 1Whole-number answers

   In the diagram, $\angle ABC = 90^\circ$. Find $x$.

   Show the worked solution

   1. Step 1: Name the hypotenuse: the side opposite the right angleThe hypotenuse is $AC$, opposite the right angle at $B$.
   2. Step 2: Use Pythagoras' theorem$AC^2 = AB^2 + BC^2$$x^2 = 24^2 + 32^2$$= 576 + 1024$$= 1600$
   3. Step 3: Square-root only at the end$x = \sqrt{1600}$$= 40$ m
2. Question 2Whole-number answers

   In the diagram, $\angle PQR = 90^\circ$. Find $x$.

   Show the worked solution

   1. Step 1: Name the hypotenuse: the side opposite the right angleThe hypotenuse is $PR$, opposite the right angle at $Q$.
   2. Step 2: Use Pythagoras' theorem$PR^2 = PQ^2 + QR^2$$x^2 = 27^2 + 36^2$$= 729 + 1296$$= 2025$
   3. Step 3: Square-root only at the end$x = \sqrt{2025}$$= 45$ m
3. Question 3Answers to 3 significant figures

   In the diagram, $\angle XYZ = 90^\circ$. Find $x$.

   Show the worked solution

   1. Step 1: Name the hypotenuse: the side opposite the right angleThe hypotenuse is $XZ$, opposite the right angle at $Y$.
   2. Step 2: Use Pythagoras' theorem$XZ^2 = XY^2 + YZ^2$$x^2 = 10^2 + 15^2$$= 100 + 225$$= 325$
   3. Step 3: Square-root only at the end$x = \sqrt{325}$$= 18.0277\ldots$$= 18.0$ cm (3 s.f.)
4. Question 4Answers to 3 significant figures

   In the diagram, $\angle KLM = 90^\circ$. Find $x$.

   Show the worked solution

   1. Step 1: Name the hypotenuse: the side opposite the right angleThe hypotenuse is $KM$, opposite the right angle at $L$.
   2. Step 2: Use Pythagoras' theorem$KM^2 = KL^2 + LM^2$$x^2 = 17^2 + 6^2$$= 289 + 36$$= 325$
   3. Step 3: Square-root only at the end$x = \sqrt{325}$$= 18.0277\ldots$$= 18.0$ cm (3 s.f.)
5. Question 5Decimal lengths: name the side

   In triangle $DEF$, $\angle DEF = 90^\circ$, $DE = 10.8$ cm and $EF = 3.2$ cm. Find the length of $DF$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Name the hypotenuse: the side opposite the right angleThe hypotenuse is $DF$, opposite the right angle at $E$.
   2. Step 2: Use Pythagoras' theorem$DF^2 = DE^2 + EF^2$$DF^2 = 10.8^2 + 3.2^2$$= 116.64 + 10.24$$= 126.88$
   3. Step 3: Square-root only at the end$DF = \sqrt{126.88}$$= 11.2641\ldots$$= 11.3$ cm (3 s.f.)
6. Question 6Decimal lengths: name the side

   In triangle $KLM$, $\angle KLM = 90^\circ$, $KL = 9.5$ cm and $LM = 8.4$ cm. Find the length of $KM$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Name the hypotenuse: the side opposite the right angleThe hypotenuse is $KM$, opposite the right angle at $L$.
   2. Step 2: Use Pythagoras' theorem$KM^2 = KL^2 + LM^2$$KM^2 = 9.5^2 + 8.4^2$$= 90.25 + 70.56$$= 160.81$
   3. Step 3: Square-root only at the end$KM = \sqrt{160.81}$$= 12.6810\ldots$$= 12.7$ cm (3 s.f.)