This MCQ module is based on: Triangle Inequality and SSS Construction
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Triangle Inequality and SSS Construction
🎓 Class 7
Mathematics
CBSE
Theory
Ch 7 — A Tale of Three Intersecting Lines
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Triangle Inequality and SSS Construction
Targeting Class 7 level in Geometry, with Basic difficulty.
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7.2 Constructing a Triangle When its Sides are Given (SSS)
Suppose we want to build a triangle with three different side lengths — say 4 cm, 5 cm and 6 cm. Let AB = 5 cm be the base, so A and B are two vertices. The third vertex C has to be 4 cm from A and 6 cm from B.
The set of points at distance 4 cm from A is a circle of radius 4 cm with centre A. Similarly, the set of points at distance 6 cm from B is a circle of radius 6 cm with centre B. So C lies where these two circles meet.
Fig 7.3 — Two arcs of radius 4 cm (from A) and 6 cm (from B) meet at C.
Four-step SSS procedure.
Step 1. Draw base AB = 5 cm.
Step 2. With centre A, draw an arc of radius 4 cm.
Step 3. With centre B, draw an arc of radius 6 cm so that it meets the first arc. Call the meeting point C.
Step 4. Join AC and BC.
Are Triangles Possible for Any Lengths?
Try to construct a triangle with sides 3 cm, 4 cm, 8 cm. Draw AB = 8 cm, then an arc of 3 cm at A and an arc of 4 cm at B. The two arcs do not meet — together their reach is only \(3+4=7\) cm, less than the base of 8 cm. No triangle exists.
Rough reasoning (tent & tree). Consider a straight path of 15 m, 10 m and 30 m forming a loop. The direct edge from tent to tree (15 m + 10 m = 25 m going round) can never equal 30 m. The direct line between two points is always the shortest.
The Triangle Inequality
Triangle Inequality
A triangle with sides \(a, b, c\) exists if and only if the sum of any two sides is greater than the third: \(a+b>c,\ b+c>a,\ c+a>b\). Equivalently, the sum of the two smaller lengths must be greater than the longest length.With the equivalent shortcut, we only need to check one inequality (smallest two vs. largest). For sides 3, 4, 8: the two smaller sum to \(3+4=7\), which is less than 8, so no triangle.
Fig 7.5 — Three cases for two arcs: touch, apart, cross.
Figure it Out
Q1. Which of the following can be the sides of a triangle? (a) 2, 2, 5 (b) 3, 4, 6 (c) 5, 4, 8 (d) 3, 4, 5 (e) 10, 20, 25 (f) 10, 20, 35
Check smallest two > largest: (a) \(2+2=4<5\) ❌; (b) \(3+4=7>6\) ✅; (c) \(5+4=9>8\) ✅; (d) \(3+4=7>5\) ✅; (e) \(10+20=30>25\) ✅; (f) \(10+20=30<35\) ❌. Triangles possible: (b), (c), (d), (e).
Q2. Check if a triangle exists with sides: (a) 1, 100, 100 (b) 3, 6, 9 (c) 1, 1, 5 (d) 5, 10, 12
(a) \(1+100=101>100\) ✅ very thin isosceles. (b) \(3+6=9\) — only equal, degenerate, so no. (c) \(1+1=2<5\) ❌. (d) \(5+10=15>12\) ✅.
Q3. Does an equilateral triangle exist for every side length between 99 and 101?
Yes. For an equilateral triangle with side \(s\), the inequality \(s+s>s\) reduces to \(s>0\), which is true for all positive \(s\). So every length — integer or decimal — between 99 and 101 gives a valid equilateral triangle.
Q4. For the third side of a triangle with two sides 5 and 5, list two valid lengths (decimals allowed).
The third side \(x\) must satisfy \(|5-5|
Activity: Stick Triangle Test
Materials: Seven sticks of lengths 2, 3, 5, 6, 7, 10, 12 cm.
- Pick any three sticks and try to form a closed triangle on your desk.
- Note which combinations close up and which leave a gap.
- For each combination, check \( \text{smallest} + \text{middle} > \text{longest}\).
- Record 6 combinations that succeed and 3 that fail.
Succeed: (3,5,6), (5,6,7), (6,7,10), (5,7,10), (7,10,12), (6,10,12). Fail: (2,3,6) [2+3=5<6], (2,5,10), (3,5,10).
Competency-Based Questions
Scenario: A park has three benches A, B, C. Paths are straight. Path AB = 40 m, path BC = 25 m. A new path AC is to be built.
Q1. What is the smallest whole-metre length the path AC can have?
L3 Apply\(AC>|40-25|=15\) m. Smallest whole number > 15 is 16 m.
Q2. Analyse: can AC be 65 m?
L4 Analyse\(AC<40+25=65\). Since 65 is not strictly less than 65, the triangle collapses into a line — no real triangle. So AC cannot be 65 m.
Q3. Evaluate the claim: "Two sides of a triangle are 12 and 18, so the third side can be anywhere between 6 and 30."
L5 EvaluateCorrect range concept — but the endpoints must be excluded. Actual range: \(6 < x < 30\) (strict inequalities). At 6 or 30, the figure is degenerate.
Q4. Create a real-world word problem (e.g. a delivery route) where the triangle inequality decides whether a path is possible.
L6 CreateSample: "A drone flies from warehouse W to relay R (8 km), then to customer C (5 km). Can the direct return W→C be longer than 13 km?" Answer: No — WC < 8 + 5 = 13 km by triangle inequality.
Assertion–Reason Questions
A: A triangle with sides 3 cm, 4 cm, 7 cm cannot be drawn.
R: In any triangle the sum of the two smaller sides must be strictly greater than the longest side.
R: In any triangle the sum of the two smaller sides must be strictly greater than the longest side.
(a) — \(3+4=7\) is not greater than 7, so no triangle. R explains A.
A: If each side of a triangle is less than 1 cm, the triangle still exists.
R: The triangle inequality depends only on the relative sizes of the sides, not on how small they are.
R: The triangle inequality depends only on the relative sizes of the sides, not on how small they are.
(a) — e.g. 0.4, 0.5, 0.6 cm all satisfy the inequality. R explains A.
A: The sides 5, 5, 9 form an isosceles triangle.
R: \(5 + 5 = 10 > 9\), so the triangle inequality holds.
R: \(5 + 5 = 10 > 9\), so the triangle inequality holds.
(a) — two sides equal ⇒ isosceles; and inequality is satisfied.
Frequently Asked Questions
How do you know if three lengths can form a triangle?
Check the triangle inequality: the sum of every pair of lengths must be strictly greater than the third length. If 3 cm, 4 cm and 8 cm are given, 3 plus 4 is 7, which is less than 8, so no triangle exists.
What is SSS construction of a triangle?
SSS (Side-Side-Side) construction builds a triangle when all three side lengths are known. Draw one side, then use the compass with the other two side lengths as radii from the endpoints to locate the third vertex.
Why is SSS a valid construction method?
Because three fixed side lengths that satisfy the triangle inequality determine a unique triangle (up to reflection). No ambiguity is possible, which is why SSS is also a congruence condition.
Does the order of sides matter in SSS construction?
The final triangle's shape and size are the same regardless of which side is chosen as the base. Only the orientation on the paper changes.
What mistake can cause SSS construction to fail?
Setting the compass to the wrong radius, letting the radius slip between arcs, or picking three lengths that violate the triangle inequality—each produces no valid triangle.
AI Tutor
Mathematics Class 7 — Ganita Prakash
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