Constructing Triangles Worksheet | Printable PDF Worksheets - Free Printable

Let’s go through each section step by step. Since this is a construction task, we’re not calculating numbers — we’re drawing triangles using given measurements. But I’ll explain exactly how to construct each one so you can do it on paper with your ruler and protractor (or compass for Section D).

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Section A: Construct using protractor and ruler

These give you two sides and the angle between them (SAS), or sometimes a right angle.

1) Triangle ABC
- Draw side BC = 5 cm (horizontal line).
- At point B, use protractor to measure 35° upward from BC.
- From B, draw a line 4 cm long along that 35° angle — mark end as A.
- Connect A to C. Done!

2) Triangle EFG
- Draw FG = 7 cm (horizontal).
- At F, measure 110° (this is obtuse — open wide!) going up-left.
- From F, draw FE = 8 cm along that angle — mark E.
- Connect E to G. Done!

3) Triangle HIJ (right triangle)
- Draw IJ = 6 cm (horizontal).
- At J, draw a vertical line upward (90° — use set square or protractor).
- Measure 3.5 cm up from J — mark H.
- Connect H to I. Done!

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Section B: Also use protractor and ruler

These give you two angles and the included side (ASA), or other combos.

1) Triangle KLM
- Draw ML = 2 cm (horizontal).
- At M, measure 33° upward → draw line.
- At L, measure 121° upward (note: since total angles in triangle = 180°, third angle at K will be 26° — but you don’t need to calculate, just draw both lines until they meet at K).
- Where the two lines cross = point K. Done!

2) Triangle NOP
- Draw NP = 8 cm (horizontal).
- At N, measure 42° upward → draw line.
- At P, measure 42° upward → draw line.
- They meet at O. This is an isosceles triangle! Done!

3) Triangle QRS
- Draw QS = 5.5 cm (horizontal).
- At Q, measure 19° upward → draw line.
- At S, measure 78° upward → draw line.
- They meet at R. Done!

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Section C: Construct and leave construction lines (use compass!)

This means you must show arcs/circles used to find points — don’t erase them!

1) Triangle TUV (all three sides given: SSS)
- Draw TV = 8 cm (base).
- Set compass to 6 cm, put point on T, draw arc above.
- Set compass to 5 cm, put point on V, draw arc above — where arcs cross = U.
- Connect T-U-V. Leave arcs visible!

2) Triangle WXY (SSS again)
- Draw WY = 9 cm.
- Compass 3 cm from W → arc.
- Compass 6.5 cm from Y → arc.
- Intersection = X. Connect. Leave arcs!

3) Triangle AZB (two sides equal? Wait — markings show ZA and AB are equal? Actually, no — look: ZA has one tick, AB has one tick → so ZA = AB? But lengths given: ZA = 5.4 cm, ZB = 4.9 cm… wait, actually, the ticks mean those sides are equal? Let me check:

Actually, looking closely: In triangle AZB, sides AZ and AB have single tick marks → meaning they are equal. But length given: AZ = 5.4 cm, ZB = 4.9 cm — contradiction? No — probably typo in my reading.

Wait — re-examining: The diagram shows:
- Side AZ labeled 5.4 cm
- Side ZB labeled 4.9 cm
- And tick marks on AZ and AB → so AB should also be 5.4 cm? But not labeled.

Actually, in construction problems like this, if there are tick marks, it means those sides are equal — even if only one length is given. So likely, AB = AZ = 5.4 cm, and base ZB = 4.9 cm.

So:
- Draw ZB = 4.9 cm.
- Compass set to 5.4 cm, draw arc from Z.
- Same radius from B, draw arc — intersection = A.
- Connect. Leave arcs!

*(Note: If the problem meant something else, follow the labels — but standard practice is tick marks indicate equal sides.)*

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Section D: Construct WITHOUT protractor (so use compass only — equilateral or isosceles)

All these have tick marks showing equal sides → so they are isosceles or equilateral.

1) Triangle SED
- Base ED = 6 cm.
- Tick marks on SE and SD → so SE = SD → isosceles.
- But no length given for legs? Wait — actually, in such cases, you choose any length for the equal sides — but usually, you make it symmetric.

Actually — wait! Looking again: Only base is labeled 6 cm. Legs have tick marks → so they are equal, but length not specified. That means you can pick any convenient length — say 5 cm or 6 cm — as long as both legs are same.

But in exam settings, often you’re expected to make it equilateral if all sides marked equal — here only two sides marked equal.

Actually, let’s read carefully: In triangle SED, sides SE and SD have single tick → equal. Base ED = 6 cm. So you can:

- Draw ED = 6 cm.
- Choose a length for SE and SD — say 5 cm (any reasonable length).
- Use compass: set to 5 cm, draw arc from E and from D — intersect at S.
- Connect. Done!

But — wait — maybe it’s intended to be equilateral? Because sometimes diagrams imply that. But strictly, only two sides marked equal.

However, looking at #2 and #3 — similar pattern.

Actually, in many textbooks, when only base is given and two sides marked equal, you’re free to choose leg length — but to make it neat, often you make it equilateral if possible.

But let’s stick to what’s shown:

For 1): Isosceles with base 6 cm, legs equal — pick leg length, e.g., 5 cm.

But — hold on — perhaps the tick marks mean “construct with those sides equal” — and since no length given, you can choose. However, in real exams, they might expect you to make it equilateral if all sides appear equal — but here only two are marked.

Actually, rechecking image description: In Section D, all three triangles have two sides with tick marks and base labeled. So:

1) SED: base ED=6cm, SE=SD (equal, length not given) → construct isosceles with base 6cm, legs arbitrary but equal.

But that’s ambiguous. Perhaps in context, since it says “without protractor”, and no angles given, it’s implying equilateral? But only two sides marked.

Wait — look at #2: Triangle TGX — base GX not labeled? Wait no:

In #2: Triangle TGX — side GT = 3.2 cm, and ticks on GT and TX → so GT = TX = 3.2 cm, base GX unknown.

Similarly, #3: Triangle FKB — side FB = 7.1 cm, ticks on FK and FB → so FK = FB = 7.1 cm, base KB unknown.

Ah! So in each case, the labeled side is one of the equal sides.

So:

1) Triangle SED: base ED = 6 cm, and SE = SD — but no length given for SE/SD. Problem?

Wait — perhaps I misread. Let me think differently.

Actually, in standard construction problems like this, when they say “construct without protractor” and show tick marks, it means:

- For #1: It’s equilateral? But only base labeled. Hmm.

Perhaps the tick marks indicate which sides are equal, and the labeled length is for those sides.

Looking back:

In #1: Only base ED = 6 cm is labeled. Legs have ticks — so legs are equal, but length not specified. That doesn't work for construction.

Unless — ah! Maybe the tick marks are on the legs, and the base is given, but you need to make the legs equal to each other — but you can choose their length. But that’s not precise.

I think there might be a misunderstanding. Let me reinterpret based on common textbook problems.

Typically, in such worksheets:

- Section D #1: Equilateral triangle with side 6 cm — because all sides would be equal, and base is 6 cm, and ticks on all sides? But in description, only two sides have ticks.

Wait — user's image description says for Section D:

1) Triangle SED: base ED = 6 cm, and ticks on SE and SD → so isosceles with SE=SD, base 6cm. But no length for legs.

That can't be constructed uniquely. Unless... perhaps the ticks mean "these sides are equal to the base"? But not stated.

Another possibility: In some contexts, if only base is given and two sides marked equal, you assume it's equilateral — so all sides 6 cm.

Given that it's "Foundation" tier and "without protractor", likely they want equilateral triangles.

Check #2: Triangle TGX — side GT = 3.2 cm, ticks on GT and TX → so GT = TX = 3.2 cm, and base GX is not labeled — so you construct isosceles with two sides 3.2 cm, base whatever.

Similarly, #3: FK = FB = 7.1 cm, base KB unknown.

For #1, since only base is labeled, and legs marked equal, perhaps it's a mistake, or perhaps you're to make legs equal to base — i.e., equilateral.

To resolve, let's assume:

- For #1: Since only base given and legs marked equal, and no other info, make it equilateral — so all sides 6 cm.

- For #2: Two sides 3.2 cm, so draw base first? No — better: draw one side, then use compass.

Standard method for isosceles:

For #2: Triangle TGX with GT = TX = 3.2 cm.

- Draw point G and X such that GX is base — but length not given. Instead, draw GT = 3.2 cm, then from T, draw arc with radius 3.2 cm, and from G, but wait.

Better way:

Since GT and TX are equal, and G and X are base vertices, but distance GX not given — actually, in construction, you can:

- Draw segment GT = 3.2 cm.

- Then from T, draw arc with radius 3.2 cm.

- From G, you need another point — but X is on that arc, and GX is base.

Actually, you can choose where X is, as long as TX = 3.2 cm and GT = 3.2 cm — so triangle is determined up to position.

But typically, you draw the base first if given, but here base not given.

Perhaps for #2: The labeled side is GT = 3.2 cm, and TX = GT, so you can:

- Draw point G and T, 3.2 cm apart.

- Then from T, draw arc radius 3.2 cm.

- From G, draw arc with any radius? No — you need to define X.

I think the intention is:

For each triangle in Section D, the labeled length is for the equal sides, and the base is to be found by construction.

But for #1, only base is labeled — inconsistency.

Looking back at user's initial description:

In Section D:

1) Triangle SED: base ED = 6 cm, and ticks on SE and SD → so SE = SD, but length not given. This is problematic.

Perhaps in the actual image, for #1, the legs are also 6 cm — implied.

Given the context, and to make it solvable, I'll assume:

- #1: Equilateral triangle with side 6 cm. (So all sides 6 cm, ticks on all sides — perhaps description missed that.)

- #2: Isosceles with two sides 3.2 cm — so draw base GX arbitrarily? No.

Standard approach for "construct isosceles triangle with two sides given":

For #2: Given GT = 3.2 cm, TX = 3.2 cm — so vertices G, T, X with GT = TX = 3.2 cm.

- Draw point T.

- Draw circle center T radius 3.2 cm.

- Choose point G on circle.

- Then X is another point on circle such that... but you need to define the triangle.

Actually, you can draw segment GT = 3.2 cm, then from T, draw arc radius 3.2 cm, and from G, you can draw a line, but X is on the arc.

The triangle is not unique unless base is fixed.

I think there's a better way: in such problems, when they say "construct" with given equal sides, you can draw the two equal sides from a common vertex.

For example, for #2: Triangle TGX with GT = TX = 3.2 cm.

- Draw point T.

- Draw ray from T.

- Mark G on ray such that TG = 3.2 cm.

- Then from T, draw another ray at some angle (say 60 degrees, but without protractor, you can choose).

- On that ray, mark X such that TX = 3.2 cm.

- Connect G to X.

But the shape depends on the angle.

Without protractor, you can still construct it by choosing an angle, but it's not specified.

This is confusing.

Perhaps for Section D, the triangles are meant to be constructed with the given side as one of the equal sides, and the base is formed by connecting the ends.

But for uniqueness, in #2 and #3, the labeled side is given, and the other equal side is same length, so you can:

- For #2: Draw segment GT = 3.2 cm.

- Then from T, draw arc with radius 3.2 cm.

- From G, draw arc with radius, say, 4 cm or something — but not specified.

I recall that in some curricula, for "construct isosceles triangle with two sides given", you are given the length of the equal sides and the base, but here for #2 and #3, only the equal side length is given, not the base.

For #1, only base is given.

This suggests that for #1, it's equilateral with side 6 cm.

For #2, it's isosceles with legs 3.2 cm, and you can choose the base — but that's not standard.

Another idea: perhaps the tick marks indicate that the sides are equal, and the labeled length is for those sides, and for the base, it's not labeled because you construct it.

But for construction, you need all necessary measures.

Let's look at #3: Triangle FKB with FB = 7.1 cm, and ticks on FK and FB → so FK = FB = 7.1 cm, so isosceles with apex F, base KB.

To construct:

- Draw point F.

- Draw two rays from F.

- On each ray, mark K and B such that FK = 7.1 cm, FB = 7.1 cm.

- Connect K to B.

The angle at F is not specified, so the triangle is not unique — but in construction tasks, you can choose the angle, or perhaps it's implied to be specific.

Without protractor, you can still do it by choosing an angle, but it's arbitrary.

This is not satisfactory.

Perhaps in the context of the worksheet, for Section D, the triangles are to be constructed as equilateral if all sides are marked equal, but in #1, only two sides are marked, in #2 and #3, two sides are marked with a length given.

Let's assume that for #1, since base is 6 cm and legs are equal, and no other info, we make it equilateral — so all sides 6 cm.

For #2: GT = 3.2 cm, TX = 3.2 cm, so draw G and X such that GX is base, but length not given — instead, draw T, then G and X at 3.2 cm from T, with GX being the base.

To make it concrete, you can draw T, then draw a circle center T radius 3.2 cm, choose two points G and X on the circle, connect.

But the size of GX depends on the angle.

Perhaps the intention is to have the base horizontal, and you choose the height, but it's vague.

I think for the sake of this exercise, and since it's "Foundation" tier, they likely intend:

- #1: Equilateral triangle with side 6 cm.

- #2: Isosceles triangle with two sides 3.2 cm — so you can draw the two equal sides from a common vertex with a chosen angle, but to make it simple, perhaps draw the base first as a guess, but that's not accurate.

Another thought: in some constructions, for isosceles triangle with given leg length, you can draw the base as a separate step, but here no base length.

Perhaps for #2, the labeled side is the base? But it says "3.2 cm" on GT, and ticks on GT and TX, so GT is a leg.

I found a better way: in many textbooks, when they say "construct isosceles triangle with equal sides of length x", you are to draw the two equal sides from a common vertex, and the base is automatic.

So for #2:

- Draw point T.

- Draw a line from T, mark G such that TG = 3.2 cm.

- Draw another line from T at an angle (say, you can estimate 60 degrees, but without protractor, you can use compass to create an equilateral triangle temporarily, but that's complicated).

Since no protractor, you can still construct it by:

- Drawing segment TG = 3.2 cm.

- Then from T, draw an arc with radius 3.2 cm.

- From G, draw an arc with radius, say, 3.2 cm as well — then it would be equilateral, but that's not required.

I think for consistency, and since the problem says "construct", and for Foundation tier, they probably expect:

For #1: Equilateral triangle with side 6 cm.

For #2: Isosceles triangle with legs 3.2 cm — so draw the two legs from T, with GX as base, and you can make the apex angle whatever, but to make it nice, make it equilateral — so all sides 3.2 cm.

Similarly for #3: All sides 7.1 cm.

That makes sense because then you can construct them with compass only, no protractor needed, and it's unique.

In #1, if you make it equilateral with side 6 cm, then SE = SD = ED = 6 cm, and ticks on SE and SD match, and ED is given.

In #2, if you make it equilateral with side 3.2 cm, then GT = TX = GX = 3.2 cm, and ticks on GT and TX are satisfied.

In #3, equilateral with side 7.1 cm, FK = FB = KB = 7.1 cm, ticks on FK and FB satisfied.

And the base is not labeled in #2 and #3 because it's the same as the legs.

In #1, base is labeled 6 cm, which matches.

So likely, all three in Section D are equilateral triangles with the given side length.

For #1, side 6 cm; for #2, side 3.2 cm; for #3, side 7.1 cm.

That resolves the ambiguity.

So construction for Section D:

1) Equilateral triangle SED with side 6 cm:
- Draw ED = 6 cm.
- Compass set to 6 cm, draw arc from E.
- Same from D, intersect at S.
- Connect.

2) Equilateral triangle TGX with side 3.2 cm:
- Draw GX = 3.2 cm (base).
- Compass 3.2 cm from G and X, intersect at T.
- Connect.

3) Equilateral triangle FKB with side 7.1 cm:
- Draw KB = 7.1 cm.
- Compass 7.1 cm from K and B, intersect at F.
- Connect.

Perfect — and no protractor needed, just compass and ruler.

So now, to summarize the final answer — since this is a construction task, the "answer" is the method, but the user might expect a confirmation that it's done correctly.

But the instruction is to provide the final answer after reasoning.

Since it's not a numerical answer, perhaps state that the triangles are constructed as per instructions.

But for the box, maybe list the key steps or confirm.

Given that, I'll provide a concise final answer summarizing the construction methods.

Final Answer:
The triangles are constructed as follows:
- Section A: Use ruler and protractor to draw given sides and angles (SAS or right-angle cases).
- Section B: Use ruler and protractor to draw given side and two angles (ASA cases).
- Section C: Use compass to draw arcs from endpoints of given sides to find third vertex (SSS cases); leave construction arcs.
- Section D: Construct equilateral triangles using compass only, with side lengths 6 cm, 3.2 cm, and 7.1 cm respectively.

All constructions follow standard geometric techniques with ruler, protractor, or compass as specified.