GCSE Area of a triangle Practice Questions

Free GCSE Area of a triangle practice questions with full step-by-step worked solutions. Covers area of a triangle, half ab sin C, identifying the included angle, exact surd answers. Practise exam-style problems and check your method.

area of a trianglehalf ab sin Cidentifying the included angleexact surd answersequilateral triangleisosceles triangle
GCSE Higher70 questionsStep-by-step solutions
Question 1
2 markseasy
In triangle ABCABC, AB=8AB = 8 cm, AC=5AC = 5 cm and angle BAC=30∘BAC = 30^\circ. Work out the area of triangle ABCABC.
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Worked solution

  1. State the formula for the area of a triangle.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    For any triangle the area is 12absin⁡C\frac{1}{2}ab\sin C, where aa and bb are two sides of the triangle and CC is the angle between them.

  2. Identify the angle between the two given sides.

    a=AB,b=AC,C=∠BACa = AB, \quad b = AC, \quad C = \angle BAC

    The sides ABAB and ACAC both start at AA, so the angle between them is the angle at AA, which is ∠BAC\angle BAC. This is the included angle, and it is the only angle that may be used with these two sides.

  3. State the area of triangle ABC.

    Area=12×8×5×sin⁡30∘=10 cm2\text{Area} = \frac{1}{2} \times 8 \times 5 \times \sin 30^\circ = 10\text{ cm}^2

    The area of triangle ABCABC is 1010 cm2^2.

Answer
Area=10 cm2\text{Area} = 10\text{ cm}^2
Question 2
1 markeasy
In triangle PQRPQR, QP=9QP = 9 cm and QR=11QR = 11 cm. Which angle must be used as CC in the formula area=12absin⁡C\text{area} = \frac{1}{2}ab\sin C with these two sides?
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Worked solution

  1. State the formula for the area of a triangle.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    For any triangle the area is 12absin⁡C\frac{1}{2}ab\sin C, where aa and bb are two sides of the triangle and CC is the angle between them.

  2. Find the vertex the two sides have in common.

    QP∩QR=QQP \cap QR = Q

    Both QPQP and QRQR contain the letter QQ, so both sides start at the vertex QQ. The angle between them therefore has its vertex at QQ.

  3. State the angle that must be used.

    C=∠PQRC = \angle PQR

    The included angle is ∠PQR\angle PQR, the angle at QQ.

Answer
∠PQR\angle PQR
Question 3
2 marksintermediate
In triangle ABCABC, AB=8AB = 8 cm and AC=5AC = 5 cm. The area of triangle ABCABC is 1010 cm2^2. Work out the two possible sizes of angle BACBAC.
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Worked solution

  1. State the formula for the area of a triangle.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    For any triangle the area is 12absin⁡C\frac{1}{2}ab\sin C, where aa and bb are two sides of the triangle and CC is the angle between them.

  2. Identify the angle between the two given sides.

    a=AB,b=AC,C=∠BACa = AB, \quad b = AC, \quad C = \angle BAC

    The sides ABAB and ACAC both start at AA, so the angle between them is the angle at AA, which is ∠BAC\angle BAC. This is the included angle, and it is the only angle that may be used with these two sides.

  3. Rearrange the area formula and substitute to find the sine of the angle.

    sin⁡∠BAC=2×Areaab=2×108×5=12\sin \angle BAC = \frac{2 \times \text{Area}}{ab} = \frac{2 \times 10}{8 \times 5} = \frac{1}{2}

    Doubling the area and dividing by the two sides gives sin⁡∠BAC=12\sin \angle BAC = \frac{1}{2}.

  4. Take the inverse sine to get the acute solution.

    sin⁡−1(12)=30∘\sin^{-1}(\frac{1}{2}) = 30^\circ

    The inverse sine key on a calculator always returns the acute angle, so the first solution is 30∘30^\circ.

  5. Note that the sine gives two possible angles, not one.

    sin⁡30∘=sin⁡150∘⇒∠BAC=30∘ or 150∘\sin 30^\circ = \sin 150^\circ \Rightarrow \angle BAC = 30^\circ \text{ or } 150^\circ

    Between 0∘0^\circ and 180∘180^\circ there are always TWO angles with the same sine: θ\theta and 180∘−θ180^\circ - \theta. Both 30∘30^\circ and 150∘150^\circ are angles of a genuine triangle with this area, so the sine alone cannot decide between them.

  6. State the two possible angles.

    ∠BAC=30∘ or 150∘\angle BAC = 30^\circ \text{ or } 150^\circ

    Angle BACBAC is either 30∘30^\circ or 150∘150^\circ.

Answer
30∘ and 150∘30^\circ \text{ and } 150^\circ
Question 4
4 markshard
In triangle RSTRST, RS=10RS = 10 cm, ST=6ST = 6 cm and TR=14TR = 14 cm. Work out the area of triangle RSTRST. Give your answer in its simplest surd form.
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Worked solution

  1. Note that no angle is given, so one must be found first.

    Area=12absin⁡C needs an angle\text{Area} = \frac{1}{2}ab\sin C \text{ needs an angle}

    The area formula needs two sides and the angle between them. Sides RSRS and STST meet at SS, so the angle needed is ∠RST\angle RST — and the cosine rule can find it from the three sides.

  2. State the cosine rule.

    a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A

    The cosine rule links all three sides of a triangle to one angle: a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A, where AA is the angle opposite the side aa. Rearranged, cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}.

  3. Write the cosine rule for the angle between the two chosen sides.

    cos⁡∠RST=RS2+ST2−TR22×RS×ST\cos \angle RST = \frac{RS^2 + ST^2 - TR^2}{2 \times RS \times ST}

    ∠RST\angle RST sits between RSRS and STST, and the side opposite it is TRTR, so that is the side that gets subtracted.

  4. Substitute the three side lengths.

    cos⁡∠RST=102+62−1422×10×6=−60120=−0.5\cos \angle RST = \frac{10^2 + 6^2 - 14^2}{2 \times 10 \times 6} = \frac{-60}{120} = -0.5

    The three sides give cos⁡∠RST=−0.5\cos \angle RST = -0.5 exactly, as a fraction — there is no need to round it.

  5. Find the sine of that angle from its cosine.

    sin⁡2∠RST=1−(−0.5)2=0.75⇒sin⁡∠RST=0.53\sin^2 \angle RST = 1 - (-0.5)^2 = 0.75 \Rightarrow \sin \angle RST = 0.5\sqrt{3}

    sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, and the angle of a triangle always has a POSITIVE sine, so take the positive root: sin⁡∠RST=0.53\sin \angle RST = 0.5\sqrt{3}.

  6. Substitute the two sides and this sine into the area formula.

    Area=12×10×6×0.53\text{Area} = \frac{1}{2} \times 10 \times 6 \times 0.5\sqrt{3}

    The two sides that meet at SS are 1010 cm and 66 cm, and the sine of the angle between them has just been found.

  7. Work the area out.

    Area=153\text{Area} = 15\sqrt{3}

    Multiplying out gives an area of 15315\sqrt{3} cm2^2.

  8. Check the area a second way with Heron formula.

    s=10+6+142=15,Area=15(5)(9)(1)=675=153s = \frac{10 + 6 + 14}{2} = 15, \quad \text{Area} = \sqrt{15(5)(9)(1)} = \sqrt{675} = 15\sqrt{3}

    Heron formula uses only the three sides, so it is a completely independent check. It gives the same area, 15315\sqrt{3} cm2^2.

  9. Check the units of the answer.

    153 cm215\sqrt{3}\text{ cm}^2

    Both sides are measured in cm, and an area is a length times a length, so the area is in cm squared.

  10. State the area of triangle RST.

    Area=12×10×6×0.53=153 cm2\text{Area} = \frac{1}{2} \times 10 \times 6 \times 0.5\sqrt{3} = 15\sqrt{3}\text{ cm}^2

    The area of triangle RSTRST is 15315\sqrt{3} cm2^2.

Answer
Area=153 cm2\text{Area} = 15\sqrt{3}\text{ cm}^2
Question 5
5 markschallenging
Triangle PP has sides 66 cm and 55 cm with an included angle of 30∘30^\circ. Triangle QQ has sides 88 cm and 44 cm with an included angle of 60∘60^\circ. Triangle RR has sides 99 cm and 55 cm with an included angle of 45∘45^\circ. Put the three triangles in order of area, smallest first.
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Worked solution

  1. State the formula for the area of a triangle.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    For any triangle the area is 12absin⁡C\frac{1}{2}ab\sin C, where aa and bb are two sides of the triangle and CC is the angle between them.

  2. Note that all three triangles are given in the same form.

    two sides and the angle between them\text{two sides and the angle between them}

    Each triangle gives two sides and the included angle, so all three areas can be worked out with the same formula and then compared.

  3. Work out the area of triangle P.

    12×6×5×sin⁡30∘=7.5≈7.50\frac{1}{2} \times 6 \times 5 \times \sin 30^\circ = 7.5 \approx 7.50

    Triangle PP has area 7.57.5 cm2^2, about 7.507.50 cm2^2.

  4. Work out the area of triangle Q.

    12×8×4×sin⁡60∘=83≈13.86\frac{1}{2} \times 8 \times 4 \times \sin 60^\circ = 8\sqrt{3} \approx 13.86

    Triangle QQ has area 838\sqrt{3} cm2^2, about 13.8613.86 cm2^2.

  5. Work out the area of triangle R.

    12×9×5×sin⁡45∘=11.252≈15.91\frac{1}{2} \times 9 \times 5 \times \sin 45^\circ = 11.25\sqrt{2} \approx 15.91

    Triangle RR has area 11.25211.25\sqrt{2} cm2^2, about 15.9115.91 cm2^2.

  6. Compare the three areas as decimals.

    7.50<13.86<15.917.50 < 13.86 < 15.91

    As decimals the three areas are 7.507.50, 13.8613.86 and 15.9115.91 cm2^2, so the order is clear.

  7. Put the letters in the order the areas give.

    P, Q, R

    Smallest first, the triangles come in the order P,Q,RP, Q, R.

  8. Note why the sides alone do not decide the order.

    12absin⁡C depends on the angle too\frac{1}{2}ab\sin C \text{ depends on the angle too}

    The product of the two sides is only part of the story: the sine of the included angle can be as small as it likes, so a triangle with long sides can still be the smallest.

  9. Note the sine can never be more than one.

    0<sin⁡C≤1 for 0∘<C<180∘0 < \sin C \le 1 \text{ for } 0^\circ < C < 180^\circ

    The sine of any angle of a triangle is between 00 and 11, so the area is always positive and never more than 12ab\frac{1}{2}ab.

  10. Check the factor of one half has been used.

    12absin⁡C≠absin⁡C\frac{1}{2}ab\sin C \ne ab\sin C

    Forgetting the 12\frac{1}{2} doubles the area — it gives the area of the parallelogram built on the two sides, not of the triangle.

  11. Note why no perpendicular height is needed.

    12absin⁡C=12×a×(bsin⁡C)\frac{1}{2}ab\sin C = \frac{1}{2} \times a \times (b\sin C)

    The factor bsin⁡Cb\sin C IS the perpendicular height onto the side aa. The formula does the work of dropping the perpendicular, which is why two sides and the angle between them are enough.

  12. Note that the third side is not needed.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    Two sides and the angle between them fix the triangle completely, so its area is settled without ever working out the third side.

  13. Leave the answer in exact form.

    32≠0.87\frac{\sqrt{3}}{2} \ne 0.87

    Replacing a surd by a rounded decimal part-way through throws accuracy away. An exact answer is a surd, not a decimal.

  14. Note that exact values make the comparison safe.

    7.5≠837.5 \ne 8\sqrt{3}

    Comparing the exact areas avoids any chance that two rounded decimals come out equal when the true areas are not.

  15. State the order of the three triangles.

    P, Q, R

    Smallest area first, the order is P,Q,RP, Q, R.

Answer
P, Q, R

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