Is it right-angled? questions with answers

Questions

1. Question 1Is the triangle right-angled? (Yes or No)

   In triangle $PQR$, $PQ = 40$ cm, $QR = 30$ cm and $PR = 50$ cm. Determine whether triangle $PQR$ is a right-angled triangle.

   Show the worked solution

   1. Step 1: Square each side$PQ^2 = 40^2$$= 1600$$QR^2 = 30^2$$= 900$$PR^2 = 50^2$$= 2500$
   2. Step 2: Which side is the longest?The longest side is $PR$.
   3. Step 3: Add the squares of the other two sides$900 + 1600 = 2500$
   4. Step 4: Compare and conclude$PR^2 = QR^2 + PQ^2$, so triangle $PQR$ is right-angled (converse of Pythagoras' theorem). Yes.
2. Question 2Is the triangle right-angled? (Yes or No)

   In triangle $DEF$, $DE = 17$ cm, $EF = 8$ cm and $DF = 15$ cm. Determine whether triangle $DEF$ is a right-angled triangle.

   Show the worked solution

   1. Step 1: Square each side$DE^2 = 17^2$$= 289$$EF^2 = 8^2$$= 64$$DF^2 = 15^2$$= 225$
   2. Step 2: Which side is the longest?The longest side is $DE$.
   3. Step 3: Add the squares of the other two sides$64 + 225 = 289$
   4. Step 4: Compare and conclude$DE^2 = EF^2 + DF^2$, so triangle $DEF$ is right-angled (converse of Pythagoras' theorem). Yes.
3. Question 3Which angle is the right angle?

   In triangle $LMN$, $LM = 8$ cm, $MN = 6$ cm and $LN = 10$ cm. Use the converse of Pythagoras' theorem to find which angle of triangle $LMN$ is a right angle.

   Show the worked solution

   1. Step 1: Square each side$LM^2 = 8^2$$= 64$$MN^2 = 6^2$$= 36$$LN^2 = 10^2$$= 100$
   2. Step 2: Which side is the longest?The longest side is $LN$.
   3. Step 3: Add the squares of the other two sides$36 + 64 = 100$
   4. Step 4: Compare, then name the right angle$LN^2 = MN^2 + LM^2$, so by the converse of Pythagoras' theorem the angle opposite $LN$ is $90^\circ$: $\angle LMN = 90^\circ$.
4. Question 4Which angle is the right angle?

   In triangle $ABC$, $AB = 3.2$ m, $BC = 2.4$ m and $AC = 4$ m. Use the converse of Pythagoras' theorem to find which angle of triangle $ABC$ is a right angle.

   Show the worked solution

   1. Step 1: Square each side$AB^2 = 3.2^2$$= 10.24$$BC^2 = 2.4^2$$= 5.76$$AC^2 = 4^2$$= 16$
   2. Step 2: Which side is the longest?The longest side is $AC$.
   3. Step 3: Add the squares of the other two sides$5.76 + 10.24 = 16$
   4. Step 4: Compare, then name the right angle$AC^2 = BC^2 + AB^2$, so by the converse of Pythagoras' theorem the angle opposite $AC$ is $90^\circ$: $\angle ABC = 90^\circ$.
5. Question 5Right-angled? From (x + y)(x − y)

   In triangle $XYZ$, $XY = 9$ cm, $YZ = x$ cm and $XZ = y$ cm, where $YZ$ is the longest side. It is given that $(x + y)(x - y) = 81$. Determine whether triangle $XYZ$ is right-angled. Choose the answer with its reason.

   Show the worked solution

   1. Step 1: Expand the brackets: (x + y)(x − y) is a difference of two squares$(x + y)(x - y) = x^2 - y^2$, so $x^2 - y^2 = 81$.
   2. Step 2: Square the third side, XY$9^2 = 81$
   3. Step 3: Converse of Pythagoras' theorem: right-angled exactly when the longest side squared equals the sum of the other two squares$x^2 - y^2 = 81$$\phantom{x^2 - y^2} = 9^2$, so $x^2 = y^2 + 9^2$: triangle $XYZ$ is right-angled, with the right angle opposite $YZ$, at $X$.
6. Question 6Right-angled? From (x + y)(x − y)

   In triangle $PQR$, $PQ = 15$ cm, $QR = x$ cm and $PR = y$ cm, where $QR$ is the longest side. It is given that $(x + y)(x - y) = 225$. Determine whether triangle $PQR$ is right-angled. Choose the answer with its reason.

   Show the worked solution

   1. Step 1: Expand the brackets: (x + y)(x − y) is a difference of two squares$(x + y)(x - y) = x^2 - y^2$, so $x^2 - y^2 = 225$.
   2. Step 2: Square the third side, PQ$15^2 = 225$
   3. Step 3: Converse of Pythagoras' theorem: right-angled exactly when the longest side squared equals the sum of the other two squares$x^2 - y^2 = 225$$\phantom{x^2 - y^2} = 15^2$, so $x^2 = y^2 + 15^2$: triangle $PQR$ is right-angled, with the right angle opposite $QR$, at $P$.