When all three vertices A, B and C lie on a single straight line, the “triangle” collapses — it no longer encloses any area. One point always lies between the other two, so instead of three sides forming a closed shape, we just get a straight line segment (with one point splitting it into two smaller pieces).

Instead of guessing point C with a ruler, we use a compass to mark every point that is exactly the required distance away, all at once:

Point C is chosen as the intersection of the two arcs. Since it lies on the arc centred at A (radius 4 cm), the distance AC = 4 cm by the very definition of a circle/arc. Since it also lies on the arc centred at B (radius 4 cm), BC = 4 cm too. A single point can satisfy both conditions at once.

No — the construction fails. If we draw the base of 8 cm and then swing an arc of radius 3 cm from one end and radius 4 cm from the other, the two arcs are too short to reach each other (3 + 4 = 7, which is less than 8 cm), so they never intersect. There is no third vertex, and hence no triangle.

Not possible. Arcs of radius 2 cm and 3 cm drawn from the two ends of a 6 cm base again fall short of meeting, since 2 + 3 = 5 cm, which is less than 6 cm.

A few more examples: 1, 2, 5; 4, 5, 10; 6, 7, 15. In every impossible case, the same pattern shows up:

The direct straight-line path from the tent to the tree is shorter than the “roundabout” path that goes tent → pole → tree. In fact, a straight line is always the shortest possible route between two points. So yes — for any two points, the direct path between them is always shorter than a roundabout path through a third point (unless that third point happens to lie exactly on the straight line, in which case the two paths are equal).

Suppose such a triangle ABC exists with AB = 15, BC = 10, CA = 30. Compare the direct path with the roundabout path for every pair of vertices:

Since a straight line must always be the shortest route, having the “direct” path CA longer than the roundabout path is a contradiction. So our assumption was wrong.

Check the largest side against the sum of the other two: 3 + 3 = 6, which is less than 7. So exactly like the 10-15-30 case, the two 3 cm arcs drawn from the ends of the 7 cm base will fall short of meeting.

No rearrangement helps. Relabelling which side is “AB” or “BC” doesn’t change the three actual lengths involved — the number 30 will always be compared against 10 + 15 = 25 in one of the three checks, no matter which vertex we call what. Since 30 > 25 no matter how we arrange the labels, that comparison always fails.

Only Case 3 gives two genuine crossing points, and either one can be used as the third vertex C of the triangle.

Unlike the SSS case, here a triangle is always possible as long as the two given side lengths are positive and the included angle is strictly between 0° and 180°. Once the base and the angle are drawn, the second side simply marks a definite point C — the triangle always closes up.

The rays from A and B fail to meet only when they become parallel — any smaller ∠B and they’d still cross somewhere. So we need the smallest ∠B that makes line m (through B) parallel to line l (through A).

AB acts as a transversal cutting the two parallel lines. For parallel lines, the co-interior (same-side) angles add up to 180°:

$$\angle A + \angle B = 180^\circ \implies 40^\circ + \angle B = 180^\circ \implies \angle B = 140^\circ$$

Yes. Generalising the 40°/140° example above (where the two angles just summed to 180°):

Draw line XY through A, parallel to BC. Since AB and AC are transversals cutting the parallel lines:

$$\angle XAB = \angle B = 50^\circ \quad \text{(alternate angles)}$$

$$\angle YAC = \angle C = 70^\circ \quad \text{(alternate angles)}$$

Since XAY is a straight line, ∠XAB, ∠BAC and ∠YAC together make 180°:

$$50^\circ + \angle BAC + 70^\circ = 180^\circ \implies \angle BAC = 60^\circ$$

Repeating this reasoning for any triangle proves the general angle sum property: the three angles of every triangle always add up to 180°.

From the angle sum property, in ΔABC:

$$50^\circ + 60^\circ + \angle ACB = 180^\circ \implies \angle ACB = 70^\circ$$

∠ACD and ∠ACB together form a straight line (180°), so:

$$\angle ACD = 180^\circ – 70^\circ = 110^\circ$$

The altitude from A is the perpendicular line segment dropped from A onto the line containing the opposite side BC; its length is the “height” of A above BC.

If the triangle is obtuse near the base, the foot of this perpendicular (D) may not fall directly between B and C — in that case, we simply extend side BC beyond C, and drop the perpendicular onto that extended line.

When you fold the paper so that the base line maps exactly onto itself (the two parts of the base line up), the fold line necessarily meets that base line at a right angle — folding a straight line onto itself always creates a 90° crease. Since the crease also passes through the top vertex, it is exactly the perpendicular segment from that vertex to the base — the altitude.

Yes — this happens exactly in a right-angled triangle. If ∠B = 90°, then side AB is already perpendicular to side BC, so AB itself serves as the altitude from A to BC (no separate line needs to be drawn).

No — that definition wouldn’t be useful, because every triangle has at least two acute angles (since the three angles sum to 180°, at most one angle can be 90° or more). So “has one acute angle” is true of every triangle and doesn’t pick out a special category.

The trick is the same “straight line is shortest” idea from the tent-tree-pole problem — but applied to a surface. Since the spider can’t cut through the air, we unfold two adjoining faces of the box flat onto a plane (like opening a book). Once unfolded, the start and end corners lie on the same flat sheet, and the shortest path between them is simply the straight-line diagonal across that unfolded rectangle.

Because there are 3 different ways to pick a pair of adjoining faces to unfold, there are 3 candidate straight-line distances — for a box of dimensions $l \times w \times h$, they are:

$$\sqrt{(l+w)^2+h^2},\qquad \sqrt{(l+h)^2+w^2},\qquad \sqrt{(w+h)^2+l^2}$$

First check each set against the triangle inequality (largest side < sum of other two) — all five sets are valid:

Any two points on a circle are joined to the centre by two radii of equal length. So picking the centre O and any two points P, Q on the circle automatically gives OP = OQ — an isosceles triangle OPQ, for any choice of P and Q.

Let A and B be centres of two equal circles that pass through each other’s centre (so AB = radius). Take any point C on either circle: joining it to the two centres gives an isosceles triangle (since one of AC or BC equals the common radius). Now take C to be one of the two points where the circles actually intersect — since C lies on both circles, AC = BC = AB = radius, giving an equilateral triangle ABC.

Using the triangle inequality (sum of two smaller sides must exceed the largest side):

None of these sets satisfy the triangle inequality, so no triangle can be constructed for any of them.

Yes, always — this holds for any three positive lengths $a \le b \le c$, regardless of whether a triangle actually exists:

• $a < b + c$ — always true, since $b,c>0$ makes $b+c$ already bigger than $a\ (\le b)$.
• $b < a + c$ — always true, since $c \ge b$ alone already forces $a+c \ge a+b > b$ (as $a>0$).
• $c < a + b$ — this is the only one that can fail, and it’s exactly the triangle inequality condition.

Yes: $50+50=100>50$, so the triangle inequality is satisfied.

In general, for any positive side length $s$: $s+s = 2s$, and $2s > s$ is true for every $s>0$. So an equilateral triangle exists for any positive sidelength, with no exceptions.

For two given sides $a$ and $b$, the third side $x$ must satisfy: $$|a-b| < x < a+b$$

Apply the same 4 steps for (b) base 6 cm (or 3 cm), included angle 25°, other side 3 cm (or 6 cm); and for (c) base 8 cm, included angle 120° (obtuse), other side 3 cm.

All three are valid, since in every case the sum of the two given angles is less than 180°: (a) 75+75=150°, (b) 25+60=85°, (c) 120+30=150°.

Rule used: a triangle is possible exactly when the two angles add to less than 180°.

Using the angle sum property, $\angle 3 = 180^\circ – (\angle 1 + \angle 2)$:

• All angles = 70° is impossible — three 70° angles would sum to $210°$, far more than $180°$.
• Two angles = 70°: the third angle $= 180 – (70+70) = 40°$.
• All three angles equal: if each angle is $x$, then $3x = 180°$, so $x = 60°$. Every triangle with three equal angles (an equiangular triangle) must have each angle equal to 60° — and it turns out to always be equilateral too.

By the angle sum property: $$\angle A + \angle B + \angle C = 180^\circ \implies 50^\circ + \angle B + \angle C = 180^\circ \implies \angle B + \angle C = 130^\circ$$

Since $\angle B = \angle C$, each equals half of $130°$:

Take AC (the hypotenuse) as the base = 5 cm. We need point B such that ∠ABC = 90°. By Thales’ theorem, the set of all such points B is exactly the circle drawn with AC as diameter — every point on that circle (other than A and C) sees AC at a right angle.