Class VII Mathematics worksheet on Perimeter and Area with 12 problems and a diagram of triangle ABC.

Let’s solve each problem one by one, step by step. I’ll make sure to check my calculations carefully before giving the final answer for each.

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Problem 1:

The length and breadth are in ratio 5:2.
Let length = 5x, breadth = 2x.

Perimeter of rectangle = 2 × (length + breadth) = 2 × (5x + 2x) = 2 × 7x = 14x

Cost of fencing = Rs. 12.5 per meter → Total cost = 12.5 × perimeter = 12.5 × 14x = 175x

Given total cost = Rs. 25000
So, 175x = 25000
→ x = 25000 ÷ 175 = 142.857... Let me calculate exactly:

25000 ÷ 175
Divide numerator and denominator by 25:
1000 ÷ 7 = 142.857... Hmm, but let’s do exact division:

175 × 142 = 175×140 + 175×2 = 24500 + 350 = 24850
25000 - 24850 = 150 → so 142 + 150/175 = 142 + 6/7 ≈ 142.857

But maybe we made a mistake? Wait — let’s recheck:

Actually, 175x = 25000
x = 25000 / 175
Simplify fraction: divide numerator and denominator by 25 → 1000 / 7 → x = 1000/7

Then:

Length = 5x = 5 × 1000/7 = 5000/7 ≈ 714.29 m
Breadth = 2x = 2000/7 ≈ 285.71 m

Wait — that seems too big for a field with fencing cost 25000 at 12.5/m? Let me verify:

Perimeter = 14x = 14 × 1000/7 = 2000 meters
Cost = 2000 × 12.5 = 25000 → YES, correct.

So answers are:

Length = 5000/7 m or approximately 714.29 m
Breadth = 2000/7 m or approximately 285.71 m

But since the question doesn’t specify decimal or fraction, we can leave as fractions.

✔ Final Answer for Q1: Length = 5000/7 m, Breadth = 2000/7 m

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Problem 2:

Rectangular park: 100m × 60m
Path 2m wide laid INSIDE along boundary.

So, inner rectangle (without path) will have:

Length = 100 - 2×2 = 96m (since path on both sides)
Breadth = 60 - 2×2 = 56m

Area of path = Area of outer rectangle - Area of inner rectangle
= (100×60) - (96×56)

Calculate:

100×60 = 6000
96×56: Let’s compute → 96×50 = 4800, 96×6=576 → total 4800+576=5376

So area of path = 6000 - 5376 = 624 sq.m

Cost of gravelling = Rs. 30 per sq.m → Total cost = 624 × 30

624 × 30 = 624 × 3 × 10 = 1872 × 10 = 18720

✔ Final Answer for Q2: Area of path = 624 m², Cost = Rs. 18,720

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Problem 3:

Lawn: 70m × 50m
Walking path 1m wide laid OUTSIDE on all sides.

So, new dimensions including path:

Length = 70 + 2×1 = 72m
Breadth = 50 + 2×1 = 52m

Area of walking path = Area of outer rectangle - Area of lawn
= (72×52) - (70×50)

Compute:

72×52: 70×52 = 3640, 2×52=104 → total 3640+104=3744
70×50 = 3500

Area = 3744 - 3500 = 244 sq.m

✔ Final Answer for Q3: Area of walking path = 244 m²

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Problem 4:

Square garden area = 2500 sq.m → side = √2500 = 50m

Uniform width 2.5m runs around it ON THE INSIDE.

So, inner square side = 50 - 2×2.5 = 50 - 5 = 45m

Area of path = Outer area - Inner area = 2500 - (45×45) = 2500 - 2025 = 475 sq.m

Cost = Rs. 25 per sq.m → Total cost = 475 × 25

475 × 25 = 475 × 100 / 4 = 47500 / 4 = 11875

Or: 400×25=10000, 75×25=1875 → total 11875

✔ Final Answer for Q4: Cost = Rs. 11,875

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Problem 5:

Right triangle, perimeter = 30cm, hypotenuse = 13cm, base = 5cm

We need to find area.

In right triangle, area = (1/2) × base × height

We know base = 5cm, hypotenuse = 13cm → use Pythagoras to find other leg (height):

Let height = h
Then: 5² + h² = 13² → 25 + h² = 169 → h² = 144 → h = 12cm

Check perimeter: 5 + 12 + 13 = 30 → matches!

Area = (1/2) × 5 × 12 = 30 cm²

✔ Final Answer for Q5: Area = 30 cm²

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Problem 6:

Area of right triangle = 100 cm²
One perpendicular leg = 25 cm → let other leg be b

Area = (1/2) × 25 × b = 100
→ 25b / 2 = 100
→ 25b = 200
→ b = 8 cm

✔ Final Answer for Q6: Other leg = 8 cm

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Problem 7:

△ABC, AC=25cm, BC=7cm, AE=10cm (AE is perpendicular from A to EC, which includes B and C)

From diagram: E-B-C are colinear, AE ⊥ EC, D is foot from B to AC? Wait, diagram shows:

A
| \
| \
E----B----C
\
D (on AC), BD ⊥ AC

Given: AC=25, BC=7, AE=10

(i) Area of △ABC

Note: Base BC = 7cm, but height from A to BC? Not directly given.

But AE is perpendicular to EC, and E-B-C are straight line → so AE is height from A to line EC, which contains base BC.

So if we take BC as base, then height is AE = 10cm? But only if AE is perpendicular to BC — which it is, since AE ⊥ EC and BC is part of EC.

Yes! So area of △ABC = (1/2) × base BC × height AE = (1/2) × 7 × 10 = 35 cm²

(ii) Find DB

DB is perpendicular from B to AC → so DB is height from B to side AC.

We can use area again: Area of △ABC = 35 cm² = (1/2) × AC × DB

AC = 25cm → 35 = (1/2) × 25 × DB
→ 35 = 12.5 × DB
→ DB = 35 / 12.5 = 350 / 125 = 14/5 = 2.8 cm

✔ Final Answer for Q7: (i) Area = 35 cm², (ii) DB = 2.8 cm

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Problem 8:

Isosceles triangle, perimeter = 32cm, base = 12cm

Equal sides: let each be x → 2x + 12 = 32 → 2x = 20 → x = 10cm

Now, to find area: drop perpendicular from apex to base → splits base into two 6cm parts.

Height h: by Pythagoras → h² + 6² = 10² → h² + 36 = 100 → h² = 64 → h = 8cm

Area = (1/2) × base × height = (1/2) × 12 × 8 = 48 cm²

✔ Final Answer for Q8: Area = 48 cm²

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Problem 9:

Parallelogram ABCD, sides 12cm and 8cm. Height corresponding to base CD = 6cm.

Assume CD is one side — say CD = 12cm (since height corresponds to it).

(i) Area = base × height = 12 × 6 = 72 cm²

(ii) Now, same area, but base AD = 8cm → find height corresponding to AD.

Area = 72 = 8 × h → h = 72 / 8 = 9 cm

✔ Final Answer for Q9: (i) Area = 72 cm², (ii) Height = 9 cm

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Problem 10:

Area of parallelogram = 450 cm², altitude = 20cm → find corresponding side (base)

Area = base × height → 450 = base × 20 → base = 450 / 20 = 22.5 cm

✔ Final Answer for Q10: Side = 22.5 cm

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Problem 11:

Area of parallelogram = area of square.

Perimeter of square = 160m → side = 160/4 = 40m → area = 40×40 = 1600 m²

So area of parallelogram = 1600 m²

Height of parallelogram = 20m → find base

Area = base × height → 1600 = base × 20 → base = 1600 / 20 = 80 m

✔ Final Answer for Q11: Base = 80 m

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Problem 12:

Field is parallelogram. One diagonal = 42m long. Perpendicular distance from either outlying vertex to this diagonal = 80 cm.

Wait — units mismatch: diagonal in meters, distance in cm → convert!

80 cm = 0.8 m

In a parallelogram, a diagonal divides it into two congruent triangles.

So area of parallelogram = 2 × area of one triangle formed by diagonal.

For one triangle: base = diagonal = 42m, height = perpendicular distance from opposite vertex = 0.8m

Area of one triangle = (1/2) × 42 × 0.8 = 21 × 0.8 = 16.8 m²

So area of parallelogram = 2 × 16.8 = 33.6 m²

✔ Final Answer for Q12: Area = 33.6 m²

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## Final Answers:

1. Length = 5000/7 m, Breadth = 2000/7 m
2. Area of path = 624 m², Cost = Rs. 18,720
3. Area of walking path = 244 m²
4. Cost = Rs. 11,875
5. Area = 30 cm²
6. Other leg = 8 cm
7. (i) Area = 35 cm², (ii) DB = 2.8 cm
8. Area = 48 cm²
9. (i) Area = 72 cm², (ii) Height = 9 cm
10. Side = 22.5 cm
11. Base = 80 m
12. Area = 33.6 m²