Perimeter and Area Worksheet for Students - Calculate the Perimeter and Area of Triangles and Rectangles.

Let’s solve each problem one by one. We’ll find the perimeter (add all sides) and area (use the right formula for each shape).

---

Problem 1: Right Triangle

Sides: 42 cm, 49 cm, 65 cm
→ Perimeter = 42 + 49 + 65 = 156 cm
→ Area of triangle = (base × height) ÷ 2
We can use 42 and 49 as base and height (since it’s a right triangle — they’re perpendicular).
Area = (42 × 49) ÷ 2 = 2058 ÷ 2 = 1029 cm²

✔ Check: 42×49 = let’s compute: 40×49=1960, 2×49=98 → 1960+98=2058 → ÷2 = 1029 ✔️

---

Problem 2: Right Triangle

Sides: 86 cm, 79 cm, 117 cm
→ Perimeter = 86 + 79 + 117 = 282 cm
→ Area = (86 × 79) ÷ 2
First, 86 × 79:
Break it down: 86 × 80 = 6880, minus 86 = 6880 - 86 = 6794
Then divide by 2: 6794 ÷ 2 = 3397 cm²

✔ Check: 86×79 = (80+6)(80-1) = better to just do 86×70=6020, 86×9=774 → 6020+774=6794 → ÷2=3397 ✔️

---

Problem 3: Two Rectangles? Wait — looks like an L-shape made of two rectangles.

Actually, looking at the diagram: It’s a composite shape — probably meant to be treated as one polygon with given side lengths.

But wait — the figure shows:

Top rectangle: width 18 cm, height 35 cm
Bottom rectangle: width 42 cm, height 25 cm
And they are joined so that total shape has outer sides: left = 35+25=60 cm, bottom=42 cm, right=25 cm, top part=18 cm, then inner step? Hmm.

Actually, this is tricky. But since it's labeled with specific outer dimensions, maybe we should treat it as a single polygon with these sides:

Looking at the drawing: The full shape has these outer edges:

Left vertical: 35 cm
Top horizontal: 18 cm
Right vertical (top part): ? Actually, from the diagram, it seems like:

It’s a rectilinear shape. Let me trace the perimeter:

Start at bottom-left corner:

→ Go up 35 cm
→ Go right 18 cm
→ Go down (35 - 25) = 10 cm? Wait no — actually, the bottom rectangle is 25 cm tall, so if top is 35 cm, then the drop on the right is 35 - 25 = 10 cm? But the diagram doesn’t show that.

Wait — perhaps it’s simpler: Maybe the shape is composed of two separate rectangles? But the problem says “find perimeter and area” for #3 — likely one shape.

Alternatively, maybe it’s a big rectangle missing a piece? Or perhaps it’s drawn as:

Imagine a large rectangle 42 cm wide and 35 cm high, but with a smaller rectangle cut out from the top-right? No — the labels suggest:

Actually, re-examining: The figure has:

- Left side: 35 cm
- Bottom: 42 cm
- Right side: 25 cm (but only partway up?)
- Then a horizontal segment of 25 cm going left?
- Then up 18 cm?
- Then left 18 cm?

This is confusing without seeing exact connections.

Wait — perhaps it’s two separate shapes? But the numbering suggests one item per number.

Another idea: Maybe it’s a single polygon with vertices connected such that the sides are:

From start: up 35, right 18, down 10 (to match 25 cm height), right 24 (since 42-18=24), down 25, left 42, up 60? That doesn’t make sense.

I think there might be a misinterpretation. Let me look again.

Actually, in many worksheets, when they show an L-shape like this with those numbers, they mean:

The entire shape has:

Outer dimensions: height = 35 + 25 = 60 cm? No — the 35 and 25 are heights of different parts.

Perhaps it’s better to calculate area first.

Area = area of top rectangle + area of bottom rectangle
Top: 18 cm × 35 cm = 630 cm²
Bottom: 42 cm × 25 cm = 1050 cm²
Total area = 630 + 1050 = 1680 cm²

Now perimeter: To find perimeter of composite shape, we add all outer sides.

Assume the shape is like this:

```
_______
| |
| | 35 cm
|_____|______
| |
| | 25 cm
|______|
42 cm
```

But the top rectangle is only 18 cm wide, sitting on the left part of the bottom rectangle which is 42 cm wide.

So the full outline:

Start at bottom-left:

→ Up 25 cm (left side of bottom rect)
→ Right 42 cm (bottom)
→ Up 25 cm? No — already went up 25.

Better path:

Start at bottom-left corner:

1. Move right along bottom: 42 cm
2. Move up right side: 25 cm
3. Move left along top of bottom rect: but only until under the top rect? Since top rect is 18 cm wide, and assuming aligned left, then from right end of bottom, move left 42 - 18 = 24 cm? But that would be internal.

Actually, standard way for L-shape:

Perimeter = sum of all external sides.

If top rectangle is 18x35 attached to top-left of bottom rectangle 42x25, then:

External sides:

- Bottom: 42 cm
- Right side of bottom: 25 cm
- Top of bottom (right part): 42 - 18 = 24 cm (horizontal, going left)
- Right side of top rectangle: 35 cm (going up)
- Top of top rectangle: 18 cm (going left)
- Left side: 35 + 25 = 60 cm (going down)

Wait, that doubles the left side.

Let me list unique outer edges:

Imagine walking around the shape clockwise:

Start at bottom-left:

→ Right 42 cm (bottom edge)
→ Up 25 cm (right edge of bottom part)
→ Left 24 cm (top edge of bottom part, since top rect covers only 18 cm of the 42 cm width)
→ Up 35 cm (right edge of top rectangle)
→ Left 18 cm (top edge of top rectangle)
→ Down 60 cm (left edge, which is 35 + 25) back to start.

Is that correct? From top-left of top rectangle, going down the entire left side: yes, 35 + 25 = 60 cm.

So perimeter = 42 + 25 + 24 + 35 + 18 + 60

Calculate:
42 + 25 = 67
67 + 24 = 91
91 + 35 = 126
126 + 18 = 144
144 + 60 = 204 cm

Area we had: 18*35 + 42*25 = 630 + 1050 = 1680 cm²

But is the left side really 60 cm? In the diagram, it might be shown as two segments, but for perimeter, we add the actual path.

In some interpretations, if the left side is continuous, it's 60 cm, but in the drawing, it might be split, but still same length.

I think 204 cm is correct for perimeter.

But let me verify with another method.

Total perimeter if both separate:
Top rect: 2*(18+35)=2*53=106
Bottom rect: 2*(42+25)=2*67=134
Sum: 240 cm
But where they join, we subtract twice the overlapping part. They share a segment of 18 cm (the top of bottom rect under the top rect). So we subtract 2*18 = 36 cm.
240 - 36 = 204 cm ✔️

Yes! So perimeter = 204 cm, area = 1680 cm²

---

Problem 4: Triangle

Sides: 33 cm, 22 cm, 40 cm
→ Perimeter = 33 + 22 + 40 = 95 cm
→ Area: Not a right triangle obviously. But we can use Heron's formula.

Heron's formula:
s = semi-perimeter = 95 / 2 = 47.5 cm
Area = √[s(s-a)(s-b)(s-c)] = √[47.5(47.5-33)(47.5-22)(47.5-40)]
= √[47.5 * 14.5 * 25.5 * 7.5]

This is messy. Maybe it's intended to be calculated differently? Or perhaps it's a right triangle? Check: 22² + 33² = 484 + 1089 = 1573, 40²=1600 — not equal. Close but not right.

Perhaps the height is given? In the diagram, it might be implied that 22 cm is base and 33 cm is height? But it's not marked as right angle.

Looking back at the image description — for problem 4, it's a triangle with sides 33, 22, 40, and no indication of right angle. But in many school problems, if not specified, sometimes they expect you to assume or use given info.

Wait — in the original image, for problem 4, is there a height marked? The user didn't provide visual, but from text, it's just "4. 40 cm, 33 cm, 22 cm" — probably just sides.

But for area, without height or angle, we must use Heron's.

Let me calculate numerically.

s = 47.5
s-a = 47.5 - 33 = 14.5
s-b = 47.5 - 22 = 25.5
s-c = 47.5 - 40 = 7.5

Product: 47.5 * 14.5 * 25.5 * 7.5

First, 47.5 * 7.5 = (50-2.5)(7.5) = 50*7.5 - 2.5*7.5 = 375 - 18.75 = 356.25

14.5 * 25.5 = (14+0.5)(25+0.5) = 14*25 + 14*0.5 + 0.5*25 + 0.5*0.5 = 350 + 7 + 12.5 + 0.25 = 369.75

Now multiply 356.25 * 369.75 — this is getting too big.

Perhaps approximate or see if it's nice number.

Maybe the triangle is meant to have base 22 and height 33? But that would be if right-angled, which it's not.

Another thought: in some diagrams, the height is drawn inside. For problem 4, if it's scalene, but perhaps for this level, they expect to use the formula with given base and height if implied.

Looking at other problems, for example problem 5 has a height marked (66 cm for base 76 cm), so likely for problem 4, no height is given, so we must use Heron's or it's a mistake.

Perhaps I can leave it, but let's try to calculate.

Note that 47.5 = 95/2, etc.

Area = √[ (95/2) * (95/2 - 33) * (95/2 - 22) * (95/2 - 40) ] = √[ (95/2) * (29/2) * (51/2) * (15/2) ] because 95/2 - 33 = 95/2 - 66/2 = 29/2, similarly 95/2 - 22 = 95/2 - 44/2 = 51/2, 95/2 - 40 = 95/2 - 80/2 = 15/2

So area = √[ (95 * 29 * 51 * 15) / 16 ]

Compute numerator: 95 * 29 * 51 * 15

First, 95 * 15 = 1425
29 * 51 = 29*50 + 29*1 = 1450 + 29 = 1479
Then 1425 * 1479

This is large. Perhaps factor.

Notice that 95 = 5*19, 29 prime, 51=3*17, 15=3*5, so overall: 5*19 * 29 * 3*17 * 3*5 = 3^2 * 5^2 * 17 * 19 * 29

So inside sqrt: [3^2 * 5^2 * 17 * 19 * 29] / 16

So area = (3*5 / 4) * √(17*19*29) = (15/4) * √(9367) since 17*19=323, 323*29=323*30 - 323=9690-323=9367

√9367 ≈ ? 96^2=9216, 97^2=9409, 9409 - 9367=42, so approx 96.8, not nice.

This suggests that perhaps for this problem, it's intended to be a right triangle or something else. Maybe I misread the sides.

Another possibility: in the diagram, the 33 cm and 22 cm are legs, and 40 cm is hypotenuse, but 33^2 + 22^2 = 1089 + 484 = 1573, 40^2=1600, not equal.

Difference is small, perhaps rounding, but unlikely.

Perhaps for this worksheet, they expect students to use the formula with base and height if available, but here no height is given for problem 4.

Let's skip and come back, or assume that the height is not given, so perhaps it's not required, but that can't be.

Looking at problem 6: it has sides 45, 74, 87, and likely not right triangle.

Perhaps for all triangles, if not right-angled, they provide height, but in problem 4, no height is mentioned in the text.

In the user's initial description, for problem 4: "4. 40 cm, 33 cm, 22 cm" — no additional info.

But in the image, perhaps there is a height drawn. Since I don't have the image, I need to infer.

For the sake of progress, and since this is for a student, perhaps in context, for problem 4, it's assumed that 22 cm is base and 33 cm is height, even though not right-angled, but that would be incorrect.

Another idea: perhaps the 33 cm is the height corresponding to base 22 cm. In many diagrams, they draw the height inside.

Let me check problem 5: it has "76 cm" base and "66 cm" height marked, so likely for problem 4, if no height is marked, we can't assume.

Perhaps for problem 4, it's a typo, or perhaps it's 33 and 22 as legs.

Let's calculate the area using Heron's with approximation.

s = 47.5
s-a=14.5, s-b=25.5, s-c=7.5
Product = 47.5 * 14.5 = let's calculate: 47*14.5 = 47*14 + 47*0.5 = 658 + 23.5 = 681.5, plus 0.5*14.5=7.25, so 688.75? Better: 47.5 * 14.5 = (50-2.5)(14.5) = 50*14.5 - 2.5*14.5 = 725 - 36.25 = 688.75

Then 25.5 * 7.5 = 25*7.5 + 0.5*7.5 = 187.5 + 3.75 = 191.25

Then 688.75 * 191.25

This is tedious. 688.75 * 190 = 688.75*100=68875, *90=61987.5, total 68875+61987.5=130862.5, then 688.75*1.25=860.9375, so total approximately 131723.4375

Then sqrt of that / 4? No, the product is for s(s-a)(s-b)(s-c) = 47.5*14.5*25.5*7.5 = as above, but earlier I have 688.75 * 191.25 for the product of the four terms? No.

s(s-a)(s-b)(s-c) = 47.5 * 14.5 * 25.5 * 7.5

I did 47.5*14.5 = 688.75
25.5*7.5 = 191.25
Then 688.75 * 191.25

Let me compute 688.75 * 191.25

Note that 688.75 = 68875/100, 191.25 = 19125/100, so product = (68875 * 19125) / 10000

But perhaps use calculator in mind.

688.75 * 190 = 688.75 * 100 = 68875, *90 = 61987.5, sum 130862.5
688.75 * 1.25 = 688.75 * 5/4 = (688.75 * 5)/4 = 3443.75 / 4 = 860.9375
So total 130862.5 + 860.9375 = 131723.4375

So s(s-a)(s-b)(s-c) = 131723.4375

Then area = sqrt(131723.4375) ≈ ? 363^2 = 131769, 362^2 = 131044, 363^2=131769, 131769 - 131723.4375 = 45.5625, so sqrt ≈ 363 - a bit, say 362.9, not nice.

This is not suitable for a school problem. Likely, for problem 4, the 33 cm is the height to the base 22 cm, or vice versa.

Perhaps the triangle is right-angled at the vertex between 33 and 22, but as calculated, 33^2 + 22^2 = 1089 + 484 = 1573, 40^2=1600, close but not equal. Difference of 27, so not.

Another possibility: the sides are 33, 22, and the third side is not 40, but in the text it's "40 cm", so probably is.

Perhaps for this worksheet, they intend for students to use the formula with the given numbers as base and height if possible.

Let's look at problem 6: sides 45, 74, 87 — similarly, 45^2 + 74^2 = 2025 + 5476 = 7501, 87^2=7569, not equal.

Problem 7: isosceles triangle with base 53 cm, equal sides 73 cm, and height 50 cm marked? In the text, "7. 73 cm, 53 cm, 50 cm" — likely 50 cm is the height to the base 53 cm.

Similarly, for problem 4, perhaps 33 cm is the height to base 22 cm, or something.

To resolve, I'll assume that for problem 4, since no height is specified, but in the context, perhaps it's not required, but that can't be.

Perhaps the 40 cm is the base, and 33 cm is the height, but then area = (40*33)/2 = 660 cm², and perimeter 95 cm.

Or if 22 cm is base, 33 cm height, area = (22*33)/2 = 363 cm².

Let's see which makes sense.

Perhaps from the diagram, the height is drawn, but since not specified, I'll proceed with the calculation for others and come back.

For now, let's do problem 5.

Problem 5: Isosceles Triangle

Base: 76 cm, height: 66 cm, equal sides: 101 cm each? Sides are 101, 101, 76? But in text: "5. 101 cm, 76 cm, 66 cm" — likely 66 cm is the height, not a side.

In the description, for problem 5: "5. 101 cm, 76 cm, 66 cm" — and from context, 76 cm is base, 66 cm is height, and 101 cm is the equal side.

So perimeter = 101 + 101 + 76 = 278 cm
Area = (base × height) / 2 = (76 × 66) / 2
76/2 = 38, so 38 × 66 = 38*60 + 38*6 = 2280 + 228 = 2508 cm²

✔ Check: 76*66 = 76*60=4560, 76*6=456, total 5016, /2=2508 ✔️

---

Problem 6: Triangle

Sides: 45 cm, 74 cm, 87 cm
→ Perimeter = 45 + 74 + 87 = 206 cm
→ Area: Again, not right triangle. 45^2 + 74^2 = 2025 + 5476 = 7501, 87^2=7569, not equal.

Use Heron's formula.

s = 206 / 2 = 103 cm
s-a = 103-45=58
s-b = 103-74=29
s-c = 103-87=16
Area = √[103 * 58 * 29 * 16]

Calculate step by step.

First, 103 * 58 = 103*50=5150, 103*8=824, total 5974
29 * 16 = 464
Then 5974 * 464

This is large. Note that 16 is square, so √16 = 4, so area = 4 * √(103 * 58 * 29)

103 * 29 = 103*30 - 103 = 3090 - 103 = 2987
Then 2987 * 58 = 2987*50=149350, 2987*8=23896, total 173246
So area = 4 * √173246

√173246 ≈ ? 416^2 = 173056, 417^2=173889, 173056 + 190 = 173246, so 416 + 190/(2*416) ≈ 416 + 190/832 ≈ 416 + 0.228 = 416.228, so area ≈ 4*416.228 = 1664.912, not nice.

Again, likely in the diagram, there is a height given. For example, perhaps 45 cm is base, and height is given, but not specified.

Perhaps for this problem, the 66 cm in problem 5 was height, so for problem 6, maybe 45 cm is base, and height is something else.

To save time, and since this is for a student, perhaps the intention is to use the formula with base and height when available, and for others, assume.

Let's do problem 7.

Problem 7: Isosceles Triangle

Sides: 73 cm, 73 cm, 53 cm, and height 50 cm (likely to the base 53 cm)
So perimeter = 73 + 73 + 53 = 199 cm
Area = (base × height) / 2 = (53 × 50) / 2 = 2650 / 2 = 1325 cm²

✔ Check: 53*50=2650, /2=1325 ✔️

---

Problem 8: Right Triangle?

Sides: 31 cm, 26 cm, 17 cm
Check if right-angled: 17^2 + 26^2 = 289 + 676 = 965, 31^2=961, close but not equal. 965 - 961 = 4, so not.

Perimeter = 31 + 26 + 17 = 74 cm
Area: If we assume it's right-angled at the vertex between 17 and 26, then area = (17*26)/2 = 442/2 = 221 cm², but since 17^2 + 26^2 = 965 ≠ 961 = 31^2, it's not exact.

Perhaps 31 is hypotenuse, so area = (leg1 * leg2)/2, but which are legs? If 17 and 26 are legs, area = (17*26)/2 = 221 cm², and perimeter 74 cm.

Or if 26 and 31 are legs, but 26^2 + 31^2 = 676 + 961 = 1637, 17^2=289, not.

So likely, the right angle is between 17 and 26, and 31 is approximately hypotenuse, but for school, they might accept area = (17*26)/2 = 221 cm².

Perhaps the height is given, but not specified.

Another idea: in the diagram, the 26 cm might be the height to base 17 cm or something.

To be consistent, for problem 8, let's assume that the two shorter sides are legs, so area = (17 * 26) / 2 = 221 cm².

Perimeter 74 cm.

---

Problem 9: Rectangle

Sides: 23 cm and 18 cm
→ Perimeter = 2*(23 + 18) = 2*41 = 82 cm
→ Area = 23 * 18 = 414 cm² (since 20*18=360, 3*18=54, total 414)

✔ Check: 23*18 = 23*20 - 23*2 = 460 - 46 = 414 ✔️

---

Now back to problem 4 and 6.

For problem 4, perhaps the 33 cm is the height to the base 22 cm. In many worksheets, they draw the height inside the triangle.

Similarly for problem 6.

Let me assume that for problem 4, base is 22 cm, height is 33 cm, so area = (22*33)/2 = 363 cm², perimeter 95 cm.

For problem 6, perhaps base is 45 cm, height is given as something, but not specified. In the text, only three numbers: 45, 74, 87.

Perhaps for problem 6, it's not a right triangle, but in the diagram, the height is marked. Since not, I'll use Heron's or skip.

Another thought: in problem 6, the sides are 45, 74, 87, and perhaps 74 is the base, and height is 45 or something, but that doesn't make sense.

Perhaps the 87 cm is the base, and height is 45 cm, but then area = (87*45)/2 = 1957.5 cm², but is that accurate? Only if height corresponds.

To make it work, and since this is for a student, I'll provide the answers as per common interpretation.

For problem 4: perimeter 95 cm, area 363 cm² (assuming base 22, height 33)

For problem 6: let's calculate with Heron's roughly, or assume.

Notice that in problem 6, if we take base 74 cm, and if height is h, but not given.

Perhaps the triangle is divided, but I think for consistency, I'll use the following:

After re-thinking, in the original image, for problem 4, it might be that the 33 cm is the height to the base 22 cm, as it's common in such diagrams.

Similarly for problem 6, perhaps the 45 cm is the height to the base 74 cm or something.

Let's look at the numbers.

For problem 6: sides 45, 74, 87. If we take base 74 cm, and suppose the height is h, then area = (74*h)/2 = 37h.

From Heron's, s=103, area = sqrt(103*58*29*16) = as before.

103*58*29*16 = let's calculate numerical value.

103*58 = 5974
29*16 = 464
5974*464.

5974*400 = 2,389,600
5974*64 = 5974*60=358,440, 5974*4=23,896, total 382,336
Sum 2,389,600 + 382,336 = 2,771,936

So area = sqrt(2,771,936)

What is square root? 1664^2 = ? 1600^2=2,560,000, 64^2=4096, 2*1600*64=204,800, so (1600+64)^2 = 1600^2 + 2*1600*64 + 64^2 = 2,560,000 + 204,800 + 4,096 = 2,768,896

2,771,936 - 2,768,896 = 3,040, so not integer.

1665^2 = (1664+1)^2 = 1664^2 + 2*1664 +1 = 2,768,896 + 3,328 +1 = 2,772,225

2,772,225 - 2,771,936 = 289, so sqrt(2,771,936) = sqrt(2,772,225 - 289) ≈ 1665 - 289/(2*1665) ≈ 1665 - 289/3330 ≈ 1665 - 0.0868 = 1664.9132, so area ≈ 1664.91 cm², not nice.

This suggests that for problem 6, perhaps the 45 cm is the height to the base 74 cm, so area = (74*45)/2 = 1665 cm², and perimeter 206 cm.

Oh! 1665 is very close to our calculation 1664.91, so likely it's intended to be 1665 cm², with base 74 cm and height 45 cm.

Similarly for problem 4, if base 22 cm, height 33 cm, area = (22*33)/2 = 363 cm², and 33^2 + 22^2 = 1089 + 484 = 1573, while 40^2=1600, difference 27, so not exact, but perhaps accepted.

For problem 8, if base 17 cm, height 26 cm, area = (17*26)/2 = 221 cm², and 17^2 + 26^2 = 289 + 676 = 965, 31^2=961, difference 4, so close.

So for school level, they might ignore the slight discrepancy and use the given numbers as base and height.

So I'll go with that.

Summary:

1. Perimeter: 156 cm, Area: 1029 cm²
2. Perimeter: 282 cm, Area: 3397 cm²
3. Perimeter: 204 cm, Area: 1680 cm²
4. Perimeter: 95 cm, Area: 363 cm² (assuming base 22, height 33)
5. Perimeter: 278 cm, Area: 2508 cm²
6. Perimeter: 206 cm, Area: 1665 cm² (assuming base 74, height 45)
7. Perimeter: 199 cm, Area: 1325 cm²
8. Perimeter: 74 cm, Area: 221 cm² (assuming base 17, height 26)
9. Perimeter: 82 cm, Area: 414 cm²

Now for problem 3, we have perimeter 204 cm, area 1680 cm².

Let me write the final answers.

Final Answer:
1. Perimeter: 156 cm, Area: 1029 cm²
2. Perimeter: 282 cm, Area: 3397 cm²
3. Perimeter: 204 cm, Area: 1680 cm²
4. Perimeter: 95 cm, Area: 363 cm²
5. Perimeter: 278 cm, Area: 2508 cm²
6. Perimeter: 206 cm, Area: 1665 cm²
7. Perimeter: 199 cm, Area: 1325 cm²
8. Perimeter: 74 cm, Area: 221 cm²
9. Perimeter: 82 cm, Area: 414 cm²