Hard GCSE Area of a triangle Questions

State the formula for the area of a triangle.

For any triangle the area is $\frac{1}{2}ab\sin C$, where $a$ and $b$ are two sides of the triangle and $C$ is the angle between them.

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

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

Work out the area of triangle P.

Triangle $P$ has area $7.5$ cm$^2$, about $7.50$ cm$^2$.

Work out the area of triangle Q.

Triangle $Q$ has area $8\sqrt{3}$ cm$^2$, about $13.86$ cm$^2$.

Work out the area of triangle R.

Triangle $R$ has area $11.25\sqrt{2}$ cm$^2$, about $15.91$ cm$^2$.

Compare the three areas as decimals.

As decimals the three areas are $7.50$, $13.86$ and $15.91$ cm$^2$, so the order is clear.

Put the letters in the order the areas give.

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

Note why the sides alone do not decide the order.

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.

Note the sine can never be more than one.

The sine of any angle of a triangle is between $0$ and $1$, so the area is always positive and never more than $\frac{1}{2}ab$.

Check the factor of one half has been used.

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

Note why no perpendicular height is needed.

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

Note that the third side is not needed.

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

Leave the answer in exact form.

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

Note that exact values make the comparison safe.

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

State the order of the three triangles.

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

State the formula for the area of a triangle.

For any triangle the area is $\frac{1}{2}ab\sin C$, where $a$ and $b$ are two sides of the triangle and $C$ is the angle between them.

Note that every triangle is given in exactly the right form.

Each triangle comes with two sides and the angle BETWEEN them, so the formula applies directly to all five and they can be compared.

Work out the area of triangle A.

Triangle $A$ has area $15$ cm$^2$, which is about $15.00$ cm$^2$.

Work out the area of triangle B.

Triangle $B$ has area $13.5\sqrt{2}$ cm$^2$, which is about $19.09$ cm$^2$.

Work out the area of triangle C.

Triangle $C$ has area $12.25\sqrt{3}$ cm$^2$, which is about $21.22$ cm$^2$.

Work out the area of triangle D.

Triangle $D$ has area $22$ cm$^2$, which is about $22.00$ cm$^2$.

Work out the area of triangle E.

Triangle $E$ has area $16$ cm$^2$, which is about $16.00$ cm$^2$.

Compare the five areas as decimals.

Written as decimals the areas can be put in order, and the biggest is $22.00$ cm$^2$.

Note why the biggest product of sides need not win.

A triangle with longer sides can still have the smaller area if its included angle is small, because $\sin C$ shrinks towards $0$ as the angle closes up.

Note the sine can never be more than one.

The sine of any angle of a triangle is between $0$ and $1$, so the area is always positive and never more than $\frac{1}{2}ab$.

Check the factor of one half has been used.

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

Note why no perpendicular height is needed.

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

Note that the third side is not needed.

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

Note that supplementary angles would not change the ranking.

Replacing any included angle by its supplement leaves that triangle with exactly the same area, so the order of the five areas would be unchanged.

State which triangle has the largest area.

Triangle $D$ has the largest area, $22$ cm$^2$.

State the formula for the area of a triangle.

For any triangle the area is $\frac{1}{2}ab\sin C$, where $a$ and $b$ are two sides of the triangle and $C$ is the angle between them.

Identify the angle between the two given sides.

The sides $YX$ and $YZ$ both start at $Y$, so the angle between them is the angle at $Y$, which is $\angle XYZ$. This is the included angle, and it is the only angle that may be used with these two sides.

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

Doubling the area and dividing by the two sides gives $\sin \angle XYZ = \frac{1}{2}$.

Take the inverse sine to get the acute solution.

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

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

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

Rearrange the area formula to make the sine of the angle the subject.

Multiply both sides by $2$ and divide by the two sides. This gives the SINE of the included angle, not the angle itself.

Check that the obtuse solution really is a solution.

Putting $150^\circ$ back into the area formula gives $15$ cm$^2$, exactly the stated area. So it is a genuine second answer, not an artefact — there really are two different triangles with this area.

Note that nothing in the question rules either angle out.

No diagram is given and the question does not say the angle is acute or obtuse, so both triangles fit every word of the question. Both angles must be given.

Check the answer is not bigger than it could possibly be.

$\sin C$ can never be more than $1$, so the area can never be more than $\frac{1}{2}ab = 30$ cm$^2$. That largest area happens only when the included angle is $90^\circ$.

Note the sine can never be more than one.

The sine of any angle of a triangle is between $0$ and $1$, so the area is always positive and never more than $\frac{1}{2}ab$.

Rule out the other pairs of angles offered.

Each of the other pairs is a genuine pair of supplementary angles, but their sine is not $\frac{1}{2}$, so they do not give the stated area.

Check the factor of one half has been used.

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

Note why no perpendicular height is needed.

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

Note that the third side is not needed.

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

State the two possible angles.

Angle $XYZ$ is either $30^\circ$ or $150^\circ$.

State the formula for the area of a triangle.

For any triangle the area is $\frac{1}{2}ab\sin C$, where $a$ and $b$ are two sides of the triangle and $C$ is the angle between them.

Identify the angle between the two given sides.

The sides $TR$ and $TS$ both start at $T$, so the angle between them is the angle at $T$, which is $\angle RTS$. This is the included angle, and it is the only angle that may be used with these two sides.

Rearrange the area formula to make the missing side the subject.

Multiply both sides by $2$, then divide by the known side and by the sine of the included angle. The missing side is then the subject of the formula.

Write down the exact value of the sine of the included angle.

$\sin 135^\circ = \frac{\sqrt{2}}{2}$ is an exact value, so the area can be left exact instead of being rounded.

Substitute everything that is known into the rearranged formula.

The area is $30\sqrt{2}$ cm$^2$, the known side is $20$ cm and the sine of the included angle is $\frac{\sqrt{2}}{2}$.

Work out the denominator.

Multiplying the known side by the sine of the included angle gives $10\sqrt{2}$.

Divide to find the missing side.

Dividing gives $TS = 6$ cm. Any surds cancel, because the same surd appears in the area and in the sine.

Check the answer by putting it back into the area formula.

Substituting $TS = 6$ cm back gives an area of $30\sqrt{2}$ cm$^2$, which is exactly the area the question states.

Check the units of the answer.

The area is in cm squared and the known side is in cm, so dividing leaves a length in cm, not an area.

Note the standard mistake in this type of question.

Only $\angle RTS$ sits between the two sides used, because only $\angle RTS$ has its vertex at $T$. Putting $\angle TRS$ or $\angle TSR$ into the formula with these two sides gives the area of no triangle at all.

Note that the sine of an angle and of its supplement are equal.

$\sin \theta = \sin (180^\circ - \theta)$, so $135^\circ$ and $45^\circ$ give exactly the same area from the same two sides. That is why a triangle is not fixed by its area and two sides alone.

Note why no perpendicular height is needed.

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

Check the factor of one half has been used.

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

Note the sine can never be more than one.

The sine of any angle of a triangle is between $0$ and $1$, so the area is always positive and never more than $\frac{1}{2}ab$.

State the length of TS.

The length of $TS$ is $6$ cm.

State the formula for the area of a triangle.

For any triangle the area is $\frac{1}{2}ab\sin C$, where $a$ and $b$ are two sides of the triangle and $C$ is the angle between them.

Identify the angle between the two given sides.

The sides $ED$ and $EF$ both start at $E$, so the angle between them is the angle at $E$, which is $\angle DEF$. This is the included angle, and it is the only angle that may be used with these two sides.

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

Multiplying by $2$ and dividing by the two sides gives the SINE of the included angle: $\sin \angle DEF = \frac{\sqrt{2}}{2}$. This is not the angle itself.

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

Between $0^\circ$ and $180^\circ$ there are always TWO angles with the same sine: $\theta$ and $180^\circ - \theta$. Both $45^\circ$ and $135^\circ$ are angles of a genuine triangle with this area, so the sine alone cannot decide between them.

Use the fact that the angle is acute to choose between the two.

$45^\circ$ is acute and $135^\circ$ is not, so the question's extra information is exactly what rules the other solution out. Without it BOTH angles would be correct.

Rearrange the area formula to make the sine of the angle the subject.

Multiply both sides by $2$ and divide by the two sides. This gives the SINE of the included angle, not the angle itself.

Check the chosen angle really does give the stated area.

Putting $45^\circ$ back into the formula gives $36\sqrt{2}$ cm$^2$, the area the question states.

Check the rejected angle also gives the stated area.

$135^\circ$ gives the same area, so it was NOT rejected for being wrong arithmetic. It was rejected only because the question says the angle is acute. This is why the extra word matters.

Note the standard mistake in this type of question.

Only $\angle DEF$ sits between the two sides used, because only $\angle DEF$ has its vertex at $E$. Putting $\angle EDF$ or $\angle EFD$ into the formula with these two sides gives the area of no triangle at all.

Check the answer is not bigger than it could possibly be.

$\sin C$ can never be more than $1$, so the area can never be more than $\frac{1}{2}ab = 72$ cm$^2$. That largest area happens only when the included angle is $90^\circ$.

Note the sine can never be more than one.

The sine of any angle of a triangle is between $0$ and $1$, so the area is always positive and never more than $\frac{1}{2}ab$.

Note that the answer must be a possible angle of a triangle.

An angle of a triangle lies strictly between $0$ and $180$ degrees, so the two solutions of the sine equation are the only candidates.

Check the factor of one half has been used.

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

Note the mistake of dividing by the area instead of by the sides.

The formula rearranges to give the area on top and the two sides underneath. Turning the fraction over usually gives a value bigger than $1$, which no sine can be — a useful warning sign.

State the size of angle DEF.

Angle $DEF$ is $45^\circ$.