Practice worksheet for finding the area of triangles with given dimensions.

To find the area of a triangle, we use the formula:

Area = (base × height) ÷ 2

But — not all triangles give you the base and height directly. Some are equilateral (all sides equal), some are isosceles (two sides equal), and some are just regular triangles with three different side lengths.

For problems like this worksheet, if only side lengths are given (no height), we can’t always use the simple base-height formula unless we know which side is the base and what the corresponding height is.

Wait — looking at the problems again: many of them show only side lengths, no heights marked. That means for most of these, we’re expected to recognize that they might be equilateral or isosceles, and perhaps use Heron’s Formula? But that’s usually too advanced for elementary worksheets.

Actually — let me check each problem carefully.

Looking at Problem 1:
Sides: 7.1 ft, 7.1 ft, 4.8 ft → Isosceles triangle. Base = 4.8 ft, two equal sides = 7.1 ft. We need height to compute area.

We can find the height by splitting the base in half (since it’s isosceles), forming two right triangles.

Half-base = 4.8 ÷ 2 = 2.4 ft

Then use Pythagoras:
height² + 2.4² = 7.1²
height² + 5.76 = 50.41
height² = 50.41 - 5.76 = 44.65
height ≈ √44.65 ≈ 6.682 ft

Then Area = (base × height) ÷ 2 = (4.8 × 6.682) ÷ 2 ≈ 32.0736 ÷ 2 ≈ 16.04 sq ft

But wait — maybe the worksheet expects us to assume that one side is the base and another is the height? No — because in triangle diagrams, unless a right angle is shown, we can’t assume that.

Hold on — let’s look at Problem 9: sides 7.6 mm, 7.6 mm, 9.0 mm — again isosceles.

Problem 10: 6.7 mm, 6.7 mm, 6.7 mm — equilateral!

Ah — here’s the key: For equilateral triangles, there IS a formula using only side length:

Area = (√3 / 4) × side²

And for isosceles triangles, we can drop a perpendicular from the apex to the base (which bisects the base) and use Pythagoras to find height, then use (base × height)/2.

But this seems complicated for a worksheet titled “Find the Area of Triangles” without mentioning Heron’s formula or square roots.

Wait — maybe I’m overcomplicating. Let me re-examine the image description.

The user said: “Do not describe the image.” So I must rely on the text provided in the OCR-like input.

Looking back at the original problem list:

1. 7.1 ft, 7.1 ft, 4.8 ft → isosceles
2. 14.2 m, 14.2 m, 13.3 m → isosceles
3. 10.3 mm, 12.5 mm, 14.2 mm → scalene — all sides different! Can't use simple method.
4. 7.5 mm, 7.5 mm, 5.2 mm → isosceles
5. 12.7 ft, 12.7 ft, 12.7 ft → equilateral
6. 12.3 cm, 7.8 cm, 12.5 cm → scalene
7. 14.8 dm, 14.8 dm, 14.8 dm → equilateral
8. 5.7 in, 3.0 in, 6.5 in → scalene
9. 7.6 mm, 7.6 mm, 9.0 mm → isosceles
10. 6.7 mm, 6.7 mm, 6.7 mm → equilateral

This is inconsistent. Some are equilateral, some isosceles, some scalene. Without heights, we cannot compute area for scalene triangles without Heron’s formula.

But Heron’s formula is:

s = (a+b+c)/2
Area = √[s(s-a)(s-b)(s-c)]

That’s probably what’s expected for scalene triangles.

Given that this is likely a middle school worksheet, and since some triangles are equilateral or isosceles, but others are scalene, I think we are meant to use Heron’s formula for all, or perhaps the worksheet has a typo? Or maybe in the actual image, some have heights marked?

But the user instruction says: “Do not describe the image.” And the text provided doesn’t mention any heights — only side lengths.

So I must proceed with Heron’s formula for all triangles, as it works for any triangle when all three sides are known.

Let me verify with Problem 5: equilateral triangle with side 12.7 ft.

Using Heron’s formula:

s = (12.7 + 12.7 + 12.7)/2 = 38.1/2 = 19.05

Area = √[19.05(19.05-12.7)(19.05-12.7)(19.05-12.7)]
= √[19.05 × 6.35 × 6.35 × 6.35]

Calculate step by step:

6.35 × 6.35 = 40.3225
40.3225 × 6.35 ≈ 256.047875
Then 19.05 × 256.047875 ≈ let's compute:

19 × 256.047875 = 4864.909625
0.05 × 256.047875 = 12.80239375
Total ≈ 4877.71201875

Now sqrt(4877.712) ≈ ?

Well, 69.8^2 = 4872.04
69.9^2 = 4886.01
So approx 69.84

But for equilateral triangle, area should be (√3/4)*side² = (1.732/4)*(12.7)^2 = (0.433)*161.29 ≈ 69.84 — yes, matches.

So Heron’s formula works.

Therefore, for consistency, I will use Heron’s formula for all 10 problems.

Let’s solve each one carefully.

---

Problem 1: Sides 7.1 ft, 7.1 ft, 4.8 ft

s = (7.1 + 7.1 + 4.8)/2 = 19.0/2 = 9.5

Area = √[9.5(9.5-7.1)(9.5-7.1)(9.5-4.8)]
= √[9.5 × 2.4 × 2.4 × 4.7]

Compute inside:

2.4 × 2.4 = 5.76
5.76 × 4.7 = 27.072
9.5 × 27.072 = 257.184

√257.184 ≈ 16.037 → 16.0 sq ft (rounded to nearest tenth)

---

Problem 2: 14.2 m, 14.2 m, 13.3 m

s = (14.2+14.2+13.3)/2 = 41.7/2 = 20.85

Area = √[20.85(20.85-14.2)(20.85-14.2)(20.85-13.3)]
= √[20.85 × 6.65 × 6.65 × 7.55]

First, 6.65 × 6.65 = 44.2225
44.2225 × 7.55 ≈ 333.879875
20.85 × 333.879875 ≈ let's compute:

20 × 333.879875 = 6677.5975
0.85 × 333.879875 ≈ 283.79789375
Total ≈ 6961.39539375

√6961.395 ≈ 83.435 → 83.4 sq m

Check with isosceles method:

Base = 13.3, half-base = 6.65
Height = √(14.2² - 6.65²) = √(201.64 - 44.2225) = √157.4175 ≈ 12.547
Area = (13.3 × 12.547)/2 ≈ 166.8751/2 ≈ 83.437 — same. Good.

---

Problem 3: 10.3 mm, 12.5 mm, 14.2 mm — scalene

s = (10.3+12.5+14.2)/2 = 37.0/2 = 18.5

Area = √[18.5(18.5-10.3)(18.5-12.5)(18.5-14.2)]
= √[18.5 × 8.2 × 6.0 × 4.3]

Compute step by step:

8.2 × 6.0 = 49.2
49.2 × 4.3 = 211.56
18.5 × 211.56 = 3913.86

√3913.86 ≈ 62.56 → 62.6 sq mm

---

Problem 4: 7.5 mm, 7.5 mm, 5.2 mm — isosceles

s = (7.5+7.5+5.2)/2 = 20.2/2 = 10.1

Area = √[10.1(10.1-7.5)(10.1-7.5)(10.1-5.2)]
= √[10.1 × 2.6 × 2.6 × 4.9]

2.6 × 2.6 = 6.76
6.76 × 4.9 = 33.124
10.1 × 33.124 = 334.5524

√334.5524 ≈ 18.29 → 18.3 sq mm

Isosceles check: base=5.2, half=2.6, height=√(7.5² - 2.6²)=√(56.25-6.76)=√49.49≈7.035, area=(5.2×7.035)/2≈18.291 — good.

---

Problem 5: 12.7 ft, 12.7 ft, 12.7 ft — equilateral

s = (12.7*3)/2 = 38.1/2 = 19.05

As before, Area ≈ 69.84 → 69.8 sq ft

Or using formula: (√3/4)*12.7² = (1.73205/4)*161.29 ≈ 0.4330125 * 161.29 ≈ 69.84 — same.

---

Problem 6: 12.3 cm, 7.8 cm, 12.5 cm — scalene

s = (12.3+7.8+12.5)/2 = 32.6/2 = 16.3

Area = √[16.3(16.3-12.3)(16.3-7.8)(16.3-12.5)]
= √[16.3 × 4.0 × 8.5 × 3.8]

Compute:

4.0 × 8.5 = 34
34 × 3.8 = 129.2
16.3 × 129.2 = 2105.96

√2105.96 ≈ 45.89 → 45.9 sq cm

---

Problem 7: 14.8 dm, 14.8 dm, 14.8 dm — equilateral

s = (14.8*3)/2 = 44.4/2 = 22.2

Area = √[22.2(22.2-14.8)^3] = √[22.2 × 7.4^3]

7.4^2 = 54.76
7.4^3 = 54.76 × 7.4 = 405.224
22.2 × 405.224 = 8995.9728

√8995.9728 ≈ 94.85 → 94.9 sq dm

Formula: (√3/4)*14.8² = 0.4330125 * 219.04 ≈ 94.85 — good.

---

Problem 8: 5.7 in, 3.0 in, 6.5 in — scalene

s = (5.7+3.0+6.5)/2 = 15.2/2 = 7.6

Area = √[7.6(7.6-5.7)(7.6-3.0)(7.6-6.5)]
= √[7.6 × 1.9 × 4.6 × 1.1]

Compute:

1.9 × 4.6 = 8.74
8.74 × 1.1 = 9.614
7.6 × 9.614 = 73.0664

√73.0664 ≈ 8.548 → 8.5 sq in

---

Problem 9: 7.6 mm, 7.6 mm, 9.0 mm — isosceles

s = (7.6+7.6+9.0)/2 = 24.2/2 = 12.1

Area = √[12.1(12.1-7.6)(12.1-7.6)(12.1-9.0)]
= √[12.1 × 4.5 × 4.5 × 3.1]

4.5 × 4.5 = 20.25
20.25 × 3.1 = 62.775
12.1 × 62.775 = 759.5775

√759.5775 ≈ 27.56 → 27.6 sq mm

Isosceles check: base=9.0, half=4.5, height=√(7.6² - 4.5²)=√(57.76 - 20.25)=√37.51≈6.125, area=(9.0×6.125)/2≈27.5625 — good.

---

Problem 10: 6.7 mm, 6.7 mm, 6.7 mm — equilateral

s = (6.7*3)/2 = 20.1/2 = 10.05

Area = √[10.05(10.05-6.7)^3] = √[10.05 × 3.35^3]

3.35^2 = 11.2225
3.35^3 = 11.2225 × 3.35 ≈ 37.595375
10.05 × 37.595375 ≈ 377.83351875

√377.8335 ≈ 19.44 → 19.4 sq mm

Formula: (√3/4)*6.7² = 0.4330125 * 44.89 ≈ 19.44 — good.

---

Now, compiling all answers with units:

1. 16.0 sq ft
2. 83.4 sq m
3. 62.6 sq mm
4. 18.3 sq mm
5. 69.8 sq ft
6. 45.9 sq cm
7. 94.9 sq dm
8. 8.5 sq in
9. 27.6 sq mm
10. 19.4 sq mm

I rounded all to one decimal place as appropriate for the given measurements (most inputs had one decimal).

Final Answer:
1. 16.0 ft²
2. 83.4 m²
3. 62.6 mm²
4. 18.3 mm²
5. 69.8 ft²
6. 45.9 cm²
7. 94.9 dm²
8. 8.5 in²
9. 27.6 mm²
10. 19.4 mm²