Math worksheet for calculating the area and perimeter of various triangles with given dimensions.

Let’s solve each triangle problem step by step.

We need to find:
- Area = (base × height) ÷ 2
- Perimeter = sum of all three sides

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Problem 1:
Sides: 4ft, 8ft, 12ft
Height = 6ft (given, perpendicular to base 12ft)

→ Area = (12 × 6) ÷ 2 = 72 ÷ 2 = 36 sq ft
→ Perimeter = 4 + 8 + 12 = 24 ft

✔ Check: Height is given as 6ft — that’s fine even if not matching Pythagoras? Wait — let’s verify if this triangle makes sense. If base is 12ft and height is 6ft, but sides are 4ft and 8ft — those can’t both reach from top to ends of 12ft base with only 6ft height. Actually, in the diagram, it looks like the height splits the base into two parts. But since the worksheet gives us the height directly, we use it for area. For perimeter, just add outer sides. So we’ll go with what’s labeled.

Actually — looking again: The triangle has sides 4ft, 8ft, and base 12ft. The height is drawn inside, 6ft. But geometrically, if you have a triangle with sides 4, 8, 12 — that’s degenerate! Because 4+8=12 — it would be flat. That doesn’t make sense. Probably, the 4ft and 8ft are not the full sides — wait, no, in the diagram, they are the two slanted sides. Hmm. Maybe it's mislabeled? Or perhaps the 4ft and 8ft are segments? No — standard interpretation: the triangle has three sides: left side 4ft, right side 8ft, bottom 12ft. Then height is 6ft — but that contradicts geometry because max height possible for such triangle would be less than 4ft or 8ft.

Wait — maybe the 4ft and 8ft are NOT the sides? Let me re-express based on typical worksheets: Often, when a height is drawn, the “sides” shown are the actual side lengths. But here, 4+8=12 — impossible triangle. So likely, the 4ft and 8ft are NOT the side lengths — perhaps they are something else? No — label says "4ft" along left edge, "8ft" along right edge, "12ft" at bottom. This must be an error in the problem, OR — perhaps the 4ft and 8ft are not the full sides? Unlikely.

Alternative idea: Maybe the 4ft and 8ft are the legs of right triangles formed by the height? But then total base would be sqrt(4² - 6²)... which is imaginary. Not possible.

This suggests there might be a typo in the original worksheet. But since this is a student task, and we’re told to solve as given, I’ll proceed using the numbers provided, assuming the height is correct for area, and sides are as labeled for perimeter — even if geometrically inconsistent. Many elementary worksheets do this to focus on formula application.

So for Problem 1:
Area = ½ × 12 × 6 = 36 sq ft
Perimeter = 4 + 8 + 12 = 24 ft

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Problem 2:
Sides: 15ft, 20ft, 25ft
Height = 12ft (to base 25ft)

→ Area = (25 × 12) ÷ 2 = 300 ÷ 2 = 150 sq ft
→ Perimeter = 15 + 20 + 25 = 60 ft

Check: 15-20-25 is a multiple of 3-4-5 → valid right triangle? 15² + 20² = 225 + 400 = 625 = 25² → yes! So it’s a right triangle with legs 15 and 20, hypotenuse 25. But height to hypotenuse should be (15×20)/25 = 300/25 = 12ft — matches! Perfect.

✔ Valid.

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Problem 3:
Isosceles triangle: two sides 10m, base 6m, height 8m

→ Area = (6 × 8) ÷ 2 = 48 ÷ 2 = 24 sq m
→ Perimeter = 10 + 10 + 6 = 26 m

Check: With base 6m, half-base = 3m. Height 8m → side should be sqrt(3² + 8²) = sqrt(9+64)=sqrt(73)≈8.54m — but labeled as 10m. Inconsistency again. But worksheet gives 10m, so we use it. Again, probably for simplicity, ignoring geometric accuracy. Proceed as labeled.

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Problem 4:
Triangle with sides 8cm, 6cm, 5cm? Wait — labels: one side 8cm, another 6cm, base 12cm, and a vertical side 5cm with right angle. Looking at diagram: It appears to be a large triangle split by a line. The right angle is at bottom right, between 12cm base and 5cm vertical side. Then the other side is 8cm? And there’s a 6cm segment inside? Confusing.

Actually, re-examining: The triangle has vertices connected: bottom left to top (8cm), top to bottom right (which is 6cm?), and bottom right to bottom left (12cm). Also, from top down to base, there’s a perpendicular of length... not labeled? Wait — no, the 5cm is the vertical leg on the right, and 12cm is horizontal base, forming a right triangle on the right part. But the whole shape is one triangle? Or composite?

Looking carefully: The figure shows a triangle with base 12cm. From the top vertex, a line goes down to the base, splitting it. On the right part, there’s a right angle marked, and the vertical side is 5cm, and the slanted side from top to bottom-right corner is labeled 6cm? But also, the left slanted side is 8cm. And the segment from top to foot of perpendicular is labeled 6cm? This is messy.

Perhaps it’s a single triangle with sides: left side 8cm, right side 6cm, base 12cm. And the height is not given — but there’s a 5cm labeled vertically on the right. Maybe the 5cm is the height? But it’s not from the apex.

Another interpretation: The triangle is composed of two right triangles sharing the height. Left part: base x, height h, hypotenuse 8cm. Right part: base y, height h, hypotenuse 6cm. Total base x+y=12cm. Also, the vertical side on far right is 5cm — but that might be separate.

I think there’s confusion in labeling. To resolve: In many such worksheets, when a height is drawn and a right angle is shown, and some sides are labeled, we take the base and corresponding height for area. Here, the base is 12cm, and if we assume the height is the perpendicular distance — but it’s not labeled. However, on the right, there’s a 5cm vertical side with right angle — perhaps that’s meant to be the height? But it’s not from the apex.

Wait — look at the diagram description: There’s a triangle with base 12cm. From the top vertex, a line drops to the base, making a right angle on the right side, and that vertical segment is 5cm. Also, the side from top to bottom-left is 8cm, and from top to bottom-right is 6cm? But then the 5cm is part of it.

Actually, I recall that in some diagrams, the 5cm might be the height. Let me calculate area using base 12cm and height 5cm? But is 5cm the height? Only if it’s perpendicular to base — which it is, since right angle is shown. But is it from the apex? The diagram shows the right angle at the bottom right corner, so the 5cm is adjacent to the base, not necessarily the altitude from apex.

This is ambiguous. Perhaps the intended height is 5cm for base 12cm? But then area = 0.5*12*5=30. Perimeter = 8+6+12=26. But earlier calculation showed inconsistency.

Alternatively, maybe the 6cm and 5cm form a right triangle on the right, so the height is 5cm, and the base of that small triangle is sqrt(6^2 - 5^2)=sqrt(11)≈3.32cm. Then the left part has base 12 - 3.32=8.68cm, and hypotenuse 8cm, so height would be sqrt(8^2 - 8.68^2) — negative! Impossible.

This suggests the labeling is problematic. Given that this is a worksheet for students, and to keep it simple, perhaps the intended solution is:

Use base = 12cm, height = 5cm (since it's perpendicular and labeled), so area = 0.5 * 12 * 5 = 30 cm²
Perimeter = 8 + 6 + 12 = 26 cm

Even though geometrically it may not close perfectly, we'll go with that for consistency with worksheet style.

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Problem 5:
Equilateral? Sides all 3cm? But there's a height of 7cm? That can't be — in equilateral triangle with side 3cm, height is (√3/2)*3 ≈ 2.598cm, not 7cm. Also, base is 6cm? Labels: two sides 3cm, base 6cm, and height 7cm? Again, inconsistent.

Diagram: Triangle with sides 3cm, 3cm, 6cm — but 3+3=6, degenerate. Height 7cm — impossible. Probably, the 3cm are not the sides — or perhaps it's a different configuration.

Looking back: "3cm" on left side, "3cm" on top side, "6cm" on bottom, and "7cm" as height from top to base? But if base is 6cm, and height 7cm, area = 0.5*6*7=21. Perimeter = 3+3+6=12. But again, geometrically impossible since max height for sides 3,3,6 is 0.

This is frustrating. Likely, the "3cm" labels are not the side lengths but something else? Or perhaps it's a typo, and the sides are different.

Another possibility: The triangle has base 6cm, and the two other sides are not 3cm each — but the diagram shows "3cm" on each of the other sides. Perhaps it's an isosceles triangle with equal sides 3cm, base 6cm — but that's flat.

I think for the sake of completing the worksheet, we'll use the given numbers:

Area = 0.5 * base * height = 0.5 * 6 * 7 = 21 cm²
Perimeter = 3 + 3 + 6 = 12 cm

Even though it's not geometrically valid, many worksheets prioritize formula practice over geometric accuracy at this level.

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Problem 6:
Right triangle: legs 16m and 12m, hypotenuse 18m? But 16² + 12² = 256 + 144 = 400, sqrt(400)=20, not 18. So hypotenuse should be 20m, but labeled 18m. Inconsistency.

But worksheet says 18m. So for area, since it's right-angled, area = 0.5 * leg1 * leg2 = 0.5 * 16 * 12 = 96 sq m
Perimeter = 16 + 12 + 18 = 46 m

Again, geometrically incorrect, but we follow labels.

Note: If it were correct, hypotenuse would be 20m, perimeter 48m, but it's labeled 18m, so we use 18m.

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Problem 7:
Base 2.2yd, height 2.5yd (perpendicular), and sides 1.5yd and 3.3yd.

→ Area = 0.5 * 2.2 * 2.5 = 0.5 * 5.5 = 2.75 sq yd
→ Perimeter = 1.5 + 3.3 + 2.2 = 7.0 yd

Check: Is the height consistent? Half-base = 1.1yd, height 2.5yd, so left side should be sqrt(1.1² + 2.5²) = sqrt(1.21 + 6.25) = sqrt(7.46) ≈ 2.73yd, but labeled 1.5yd — not match. Similarly, right side: half-base same? No, the height may not bisect. Assuming the 2.5yd is the height to base 2.2yd, we use it for area. Sides are given, so perimeter is sum.

Proceed.

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Problem 8:
Right triangle: legs 8yd and 6yd, hypotenuse 10yd.

Check: 6² + 8² = 36 + 64 = 100 = 10² — perfect right triangle.

→ Area = 0.5 * 6 * 8 = 24 sq yd
→ Perimeter = 6 + 8 + 10 = 24 yd

Perfect.

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Now, compiling all answers:

1. Area = 36 sq ft, Perimeter = 24 ft
2. Area = 150 sq ft, Perimeter = 60 ft
3. Area = 24 sq m, Perimeter = 26 m
4. Area = 30 cm², Perimeter = 26 cm [assuming height=5cm for base=12cm]
5. Area = 21 cm², Perimeter = 12 cm [despite geometric issues]
6. Area = 96 sq m, Perimeter = 46 m [using given hypotenuse 18m]
7. Area = 2.75 sq yd, Perimeter = 7.0 yd
8. Area = 24 sq yd, Perimeter = 24 yd

For problems 4,5,6,7, there are geometric inconsistencies, but since the worksheet provides specific numbers, we use them as given for calculation purposes.

Final Answer:

1. Area = 36 sq ft, Perimeter = 24 ft
2. Area = 150 sq ft, Perimeter = 60 ft
3. Area = 24 sq m, Perimeter = 26 m
4. Area = 30 cm², Perimeter = 26 cm
5. Area = 21 cm², Perimeter = 12 cm
6. Area = 96 sq m, Perimeter = 46 m
7. Area = 2.75 sq yd, Perimeter = 7.0 yd
8. Area = 24 sq yd, Perimeter = 24 yd