Theorems on triangles, parallelograms and tangents
A camogie club is marking out a triangular training zone with three poles. Aoife has ropes 3 m, 4 m and 8 m long. She says, "These will make a triangle." Will they? By the end of this page you will be able to decide, and also to find areas and lengths using the theorems on triangles, parallelograms and tangents.
Sides and angles of a triangle
Draw a triangle and look at the largest angle. The side opposite it, the side it "faces", is the longest. The smallest angle faces the shortest side.
Remember also that the three angles of a triangle add up to $180^\circ$. This lets you find the third angle first, then compare.
In triangle $ABC$, $|\angle A| = 70^\circ$, $|\angle B| = 50^\circ$. a) Find $|\angle C|$. b) Name the longest and the shortest side, giving the reason.
In triangle $ABC$, $|\angle A| = 70^\circ$, $|\angle B| = 50^\circ$.
- a) Find $|\angle C|$.
- b) Name the longest and the shortest side, giving the reason.
The triangle inequality
Why? The straight line between two points is the shortest route. Going from $A$ to $B$ via $C$ cannot be shorter than going directly.
Back to Aoife's ropes: $3 + 4 = 7$, which is less than $8$. The two short ropes cannot reach across, so no triangle is possible.
Can a triangle have sides of lengths 3 cm, 4 cm and 8 cm? Use the triangle inequality to explain.
Can a triangle have sides of lengths 3 cm, 4 cm and 8 cm? Use the triangle inequality to explain.
Two sides of a triangle are 7 cm and 12 cm long. a) Between which two values must the third side $x$ lie? b) If the triangle is isosceles, find the possible values of…
Two sides of a triangle are 7 cm and 12 cm long.
- a) Between which two values must the third side $x$ lie?
- b) If the triangle is isosceles, find the possible values of $x$.
Areas of triangles and parallelograms
The height is measured at right angles to the base, not along a slanted side.
A diagonal cuts a parallelogram into two triangles with the same base and height, so each has half the area. This is why the triangle area has the $\tfrac{1}{2}$.
The diagonals of a parallelogram bisect each other: they cross at their common midpoint. So if $ABEC$ is a parallelogram, then $[AE]$ and $[BC]$ bisect each other, and the diagonal $[BC]$ splits the parallelogram into triangles $ABC$ and $EBC$ (or $[AE]$ into $ABE$ and $ACE$) of equal area, each half of the parallelogram. So the parallelogram has twice the area of triangle $ABC$.
Triangles with the same base and the same height have the same area, whatever their shape. If a point $D$ lies on $[BC]$, triangles $ABD$ and $ADC$ share the height from $A$, so their areas are in the same ratio as their bases $|BD| : |DC|$.
A triangle has base 10 cm and perpendicular height 6 cm. Calculate its area.
A triangle has base 10 cm and perpendicular height 6 cm. Calculate its area.
A parallelogram $ABCD$ has $|AB| = 9$ cm, and the perpendicular distance between $AB$ and $DC$ is 5 cm. a) Calculate the area of the parallelogram. b) The diagonal $AC$…
A parallelogram $ABCD$ has $|AB| = 9$ cm, and the perpendicular distance between $AB$ and $DC$ is 5 cm.
- a) Calculate the area of the parallelogram.
- b) The diagonal $AC$ divides the parallelogram into two triangles. State the area of each and give a reason.
In triangle $PQR$, $M$ is the midpoint of $[QR]$. Prove that triangles $PQM$ and $PMR$ have equal areas. Hence, if the area of triangle $PQR$ is $48\text{ cm}^2$, state…
In triangle $PQR$, $M$ is the midpoint of $[QR]$. Prove that triangles $PQM$ and $PMR$ have equal areas. Hence, if the area of triangle $PQR$ is $48\text{ cm}^2$, state the area of $PQM$.
In triangle $ABC$, the point $D$ lies on $[BC]$. Triangles $ABC$ and $ABD$ share the vertex $A$. a) If $|BD| : |DC| = 2 : 3$, show that the area of $ABD$ is…
In triangle $ABC$, the point $D$ lies on $[BC]$. Triangles $ABC$ and $ABD$ share the vertex $A$.
- a) If $|BD| : |DC| = 2 : 3$, show that the area of $ABD$ is $\frac{2}{5}$ of the area of $ABC$.
- b) The parallelogram $ABEC$ is completed with $E$ so that $[AE]$ and $[BC]$ bisect each other. Show that the area of the parallelogram is twice the area of triangle $ABC$.
Tangents to a circle
A tangent is a line that touches a circle at exactly one point.
Because the radius meets the tangent at $90^\circ$, triangle $OAP$ is right-angled at $A$. That lets you use the theorem of Pythagoras, with $[OP]$ as the hypotenuse. The two triangles $OAP$ and $OBP$ are congruent, which is why $|PA| = |PB|$.
A circle has centre $O$ and a tangent touching it at point $T$. The radius is $5$ cm. State the angle between $OT$ and the tangent, and find the length of the tangent…
A circle has centre $O$ and a tangent touching it at point $T$. The radius is $5$ cm. State the angle between $OT$ and the tangent, and find the length of the tangent from a point $P$ with $|OP| = 13$ cm.
Two tangents from an external point $P$ touch a circle with centre $O$ at $A$ and $B$. The radius is 6 cm and $|OP| = 10$ cm. a) Find $|PA|$ and explain why…
Two tangents from an external point $P$ touch a circle with centre $O$ at $A$ and $B$. The radius is 6 cm and $|OP| = 10$ cm.
- a) Find $|PA|$ and explain why $|PA| = |PB|$.
- b) Find the area of the quadrilateral $OAPB$.
- c) Find $|\angle APB|$ to the nearest degree.
Back to Aoife's ropes: $3 + 4 = 7 < 8$, so the triangle inequality tells us they cannot form a triangle.