Which side lengths can even be a triangle — and once they are, what kind.
The problem
Can 5, 7, 13 be the sides of a triangle?
The two shorter sides must together beat the longest: 5 + 7 = 12, which is less than 13, so they can never close up into a triangle. Only that one comparison matters — checking the two smallest against the largest is always enough. “Yes” comes from checking an easy pair instead, like 7 + 13 = 20 > 5, which always passes; and the sides need not differ — 5, 5, 5 is a perfectly good triangle.
Otherwise they cannot reach across to meet.
5 + 7 = 12
the two shorter sides
12 < 13
too short to close
not a triangle
no such shape exists
Strictly the rule applies to all three pairs, but only the two shortest can ever fail. Check that pair and you are done.
a real triangle
a flat line, not a triangle
With the sum exactly equal, the two short sides lie flat along the long one. No height, no area, no triangle.
Given two sides of 7 and 10, the third is boxed in.
x < 7 + 10 = 17
less than the sum
x > 10 − 7 = 3
more than the difference
3 < x < 17
the whole range
Difference on the low side, sum on the high side. Both ends are strict.
If the question wants whole numbers, that is 4 through 16 — thirteen possible values, not seventeen. If it asks for their sum, pair them off: count × middle value = 13 × 10 = 130.
Compare the longest side squared with the other two squared.
| If | The triangle is |
|---|---|
| c² = a² + b² | right |
| c² < a² + b² | acute |
| c² > a² + b² | obtuse |
6, 8, 9 → c = 9
the longest
81 vs 36 + 64 = 100
compare
81 < 100 → acute
all angles under 90°
Putting the wrong side on the left flips the verdict. Sort the three numbers first, then test.
equilateral — all three equal
isosceles — two equal
scalene — none equal
acute — all under 90°
right — one exactly 90°
obtuse — one over 90°
A triangle gets one name from each column — “isosceles right”, for instance, which is the 45-45-90.
Only one angle can be 90° or more. Two would already use up 180° before the third angle even exists.
Sides x + 1, 2x − 3 and 10, with 10 the longest. Find the smallest whole-number x.
(x + 1) + (2x − 3) > 10
two shorter beat the longest
3x − 2 > 10, x > 4
+ 2, ÷ 3
x = 5
strict, so 4 itself fails
Test it: x = 5 gives 6, 7, 10 and 6 + 7 = 13 > 10 ✓. x = 4 gives 5, 5, 10 — flat ✗.
“Two sides of an isosceles triangle are 3 and 7.” The equal pair could be 3, 3 or 7, 7. But 3 + 3 = 6 is less than 7, so 3-3-7 does not exist. Only 7-7-3 is left, and the perimeter is 17.
Two sides are 5 and 12. For which whole numbers x is the triangle acute? We do not know which side is longest, so we try both.
x longest: x² < 5² + 12² = 169
x < 13, and x ≥ 12 → x = 12
12 longest: 12² < 5² + x²
144 < 25 + x²
x² > 119 → x ≥ 11
x = 11 or 12
x = 11 or 12
join the two cases
The edges: x = 13 is right (5-12-13), and x = 10 is obtuse (25 + 100 = 125 < 144). Both cases also sit inside the plain range 7 < x < 17.
The plain test is easy marks; UPCAT-type items twist it — count the whole-number third sides, put x inside the sides, or ask for acute rather than possible. Count whole numbers as (top − bottom − 1) because both ends are excluded, and for acute or obtuse always decide first which side is longest. Ten-second check: test the answer’s neighbours, which should fail.
The problem
Can 8, 15, 17 form a triangle? If so, what kind?
8 + 15 = 23 > 17, so it is a triangle. Now compare: 8² + 15² = 64 + 225 = 289, and 17² = 289 — exactly equal, so it is right-angled. Bigger on the left would mean acute; smaller would mean obtuse.
