Solved Exercises on Area of Triangles in Grade 7

Master area of triangles: base-height method, right triangles, and composite figures through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Triangle Area
Exercise 1
Find the area of a triangle with base 12 cm and height 8 cm.
Definition:

Height of a Triangle: The perpendicular distance from the base to the opposite vertex.

Method for triangle area:
  1. Identify the base and corresponding height
  2. Apply the area formula: A = ½ × base × height
  3. Substitute the values
  4. Calculate the result
  5. Include the units (cm²)
Given
Base = 12 cm, Height = 8 cm
Formula
A = ½bh
Result
48 cm²
Step 1: Identify the base and height

Base = 12 cm, Height = 8 cm

Step 2: Apply the area formula

Area = ½ × base × height

Step 3: Substitute values

Area = ½ × 12 × 8

Step 4: Calculate

Area = ½ × 96 = 48 cm²

Step 5: Verify the calculation

Check: 12 × 8 = 96, then 96 ÷ 2 = 48 ✓

Area = 48 cm²
Final answer:

The area of the triangle is 48 cm².

Applied rules:

• Triangle Area Formula: A = ½bh

• Perpendicular Height: Height must be perpendicular to base

• Units: Area units are squared (cm²)

2 Finding Missing Base or Height
Exercise 2
The area of a triangle is 60 cm² and its height is 10 cm. Find the length of the base.
Definition:

Rearranging the Area Formula: When area and one dimension are known, rearrange A = ½bh to find the missing dimension.

Given
Area = 60 cm², Height = 10 cm
Formula
A = ½bh → b = 2A/h
Result
12 cm
Step 1: Start with the area formula

A = ½bh

Step 2: Rearrange to solve for base

2A = bh

b = 2A/h

Step 3: Substitute known values

b = (2 × 60) / 10

b = 120 / 10

Step 4: Calculate

b = 12 cm

Step 5: Verify the solution

Area = ½ × 12 × 10 = ½ × 120 = 60 cm² ✓

Base = 12 cm
Final answer:

The length of the base is 12 cm.

Applied rules:

• Formula Rearrangement: Isolate the unknown variable

• Algebraic Manipulation: Multiply both sides by 2, then divide by height

• Verification: Substitute back to check the answer

3 Right Triangle Area
Exercise 3
Find the area of a right triangle with legs measuring 6 cm and 8 cm.
Definition:

Right Triangle: A triangle with one 90° angle. The legs of a right triangle can serve as base and height.

Given
Legs = 6 cm, 8 cm
Formula
A = ½bh
Result
24 cm²
Step 1: Identify the legs as base and height

In a right triangle, the two legs are perpendicular to each other

Base = 6 cm, Height = 8 cm (or vice versa)

Step 2: Apply the area formula

Area = ½ × base × height

Step 3: Substitute values

Area = ½ × 6 × 8

Step 4: Calculate

Area = ½ × 48 = 24 cm²

Step 5: Verify with Pythagorean theorem

For a right triangle, hypotenuse = √(6² + 8²) = √(36 + 64) = √100 = 10 cm

This confirms we have a valid right triangle

Area = 24 cm²
Final answer:

The area of the right triangle is 24 cm².

Applied rules:

• Right Triangle Property: Legs are perpendicular (serve as base and height)

• Area Formula: A = ½bh

• Pythagorean Theorem: a² + b² = c² for verification

Triangle Area Properties and Methods
\(A = \frac{1}{2} \times \text{base} \times \text{height}\)
Triangle Area Formula
Right Triangle
A = ½ab
Legs as base and height
Equilateral
A = (s²√3)/4
Side length s
Height
h = 2A/b
From area formula
Key definitions:

Base: Any side of the triangle that serves as the reference for measuring height.

Height: The perpendicular distance from the base to the opposite vertex.

Legs: The two sides of a right triangle that form the right angle.

Hypotenuse: The longest side of a right triangle, opposite the right angle.

Area: The amount of space inside a two-dimensional shape, measured in square units.

Complete methodology:
  1. Identify the Base and Height: Ensure height is perpendicular to base
  2. Apply the Formula: A = ½ × base × height
  3. Substitute Values: Plug in the known measurements
  4. Calculate: Perform the multiplication and division
  5. Check Units: Ensure answer is in square units
  6. Verify: Check calculation and reasonableness
Tip 1: The height must be perpendicular to the base (at 90°).
Tip 2: In right triangles, the legs can serve as base and height.
Tip 3: Any side can be the base if you use the corresponding perpendicular height.
Tip 4: Always include units in your final answer (cm², m², etc.).
Common errors: Using slanted height instead of perpendicular height, forgetting to divide by 2, misidentifying base and height.
Exam preparation: Practice with different triangle orientations, work with fractional measurements, master formula rearrangement.
Formulas to know by heart:

• Triangle Area: A = ½bh

• Base: b = 2A/h

• Height: h = 2A/b

• Right Triangle Area: A = ½ab (where a, b are legs)

• Equilateral Triangle: A = (s²√3)/4

• Pythagorean Theorem: a² + b² = c²

Solution: Exercises 4 to 5
4 Composite Figure
Exercise 4
A rectangular garden has dimensions 20 m by 12 m. A triangular flower bed is cut out from one corner, with legs of 6 m and 8 m. Find the remaining area of the garden.
Definition:

Composite Figure: A shape made up of two or more simple geometric figures. To find the area, calculate the area of each part and combine as needed.

Rectangle
20m × 12m = 240m²
Triangle
½(6)(8) = 24m²
Remaining
240 - 24 = 216m²
Step 1: Calculate the area of the rectangle

Rectangle Area = length × width

Rectangle Area = 20 × 12 = 240 m²

Step 2: Calculate the area of the triangular flower bed

Since the triangle is cut from a corner of the rectangle, it's a right triangle

Triangle Area = ½ × base × height

Triangle Area = ½ × 6 × 8 = 24 m²

Step 3: Subtract the triangle area from the rectangle area

Remaining Area = Rectangle Area - Triangle Area

Remaining Area = 240 - 24 = 216 m²

Step 4: Verify the calculation

Check: 240 - 24 = 216 ✓

The remaining area should be less than the original area ✓

Remaining area = 216 m²
Final answer:

The remaining area of the garden is 216 m².

Applied rules:

• Composite Area: Total Area = Sum of parts or Difference of parts

• Rectangle Area: A = length × width

• Right Triangle Area: A = ½ab

5 Multi-Step Problem
Exercise 5
A triangle has an area of 45 cm². If the base is increased by 2 cm and the height is decreased by 1 cm, the area remains the same. Find the original base and height.
Definition:

Simultaneous Equations: When two conditions must be satisfied at the same time, we create a system of equations to solve for multiple unknowns.

Original
A = 45, b = ?, h = ?
Modified
(b+2)(h-1) = 45
Solution
b=10cm, h=9cm
Step 1: Set up the first equation

Original area: ½bh = 45

Therefore: bh = 90

Step 2: Set up the second equation

New dimensions: (b + 2) and (h - 1)

New area: ½(b + 2)(h - 1) = 45

Therefore: (b + 2)(h - 1) = 90

Step 3: Expand the second equation

(b + 2)(h - 1) = bh - b + 2h - 2 = 90

Since bh = 90: 90 - b + 2h - 2 = 90

Simplify: -b + 2h - 2 = 0

Therefore: b = 2h - 2

Step 4: Substitute into first equation

Since b = 2h - 2 and bh = 90:

(2h - 2)h = 90

2h² - 2h = 90

2h² - 2h - 90 = 0

h² - h - 45 = 0

Step 5: Solve the quadratic equation

Using the quadratic formula: h = [1 ± √(1 + 180)]/2 = [1 ± √181]/2

Since √181 ≈ 13.45: h = (1 + 13.45)/2 ≈ 7.23 or h = (1 - 13.45)/2 ≈ -6.23

Since height must be positive: h ≈ 7.23 cm

Then b = 90/h ≈ 90/7.23 ≈ 12.45 cm

Step 6: Recalculate more precisely

Actually, let's solve h² - h - 45 = 0 more carefully

h² - h - 45 = 0

Using quadratic formula: h = (1 ± √(1 + 180))/2 = (1 ± √181)/2

Since √181 is not a perfect square, let's try integer solutions

If h = 10, then b = 9: Area = ½(9)(10) = 45 ✓

Modified: (9+2)(10-1) = 11×9 = 99, so ½(99) = 49.5 ≠ 45

If h = 9, then b = 10: Area = ½(10)(9) = 45 ✓

Modified: (10+2)(9-1) = 12×8 = 96, so ½(96) = 48 ≠ 45

Let me solve: bh = 90 and (b+2)(h-1) = 90

bh = (b+2)(h-1)

bh = bh - b + 2h - 2

0 = -b + 2h - 2

b = 2h - 2

(2h-2)h = 90

2h² - 2h - 90 = 0

h² - h - 45 = 0

h = (1 + √181)/2 ≈ 7.23 cm

b = 90/h ≈ 12.45 cm

Original base ≈ 12.45 cm, height ≈ 7.23 cm
Final answer:

The original base was approximately 12.45 cm and the height was approximately 7.23 cm.

Applied rules:

• Simultaneous Equations: Set up equations based on given conditions

• Quadratic Formula: Solve equations of the form ax² + bx + c = 0

• Algebraic Manipulation: Expand and simplify expressions

Complete Guide: Area of Triangles
\(\text{Height} = \frac{2 \times \text{Area}}{\text{Base}}\)
Height from Area Formula
Key definitions:

Triangle: A polygon with three sides and three angles. The sum of interior angles is always 180°.

Area: The measure of the surface enclosed by a shape, expressed in square units (cm², m², etc.).

Base: The side of a triangle that is used as a reference for measuring the height.

Height: The perpendicular distance from the base to the opposite vertex.

Right Triangle: A triangle with one 90° angle. The sides forming the right angle are called legs.

Altitude: Another term for height in a triangle.

Complete methodology:
  1. Identify the Triangle Type: Determine if it's right, equilateral, isosceles, or scalene
  2. Find Base and Height: Ensure the height is perpendicular to the base
  3. Select Appropriate Formula: Use standard A = ½bh or specialized formulas
  4. Substitute Values: Replace variables with given measurements
  5. Calculate Carefully: Perform multiplication and division in correct order
  6. Verify Solution: Check calculations and ensure answer is reasonable
Tip 1: The height of a triangle is always perpendicular to the base, not necessarily a side of the triangle.
Tip 2: In obtuse triangles, the height may lie outside the triangle.
Tip 3: You can use any side as the base as long as you use the corresponding perpendicular height.
Tip 4: Always check that your answer makes sense by comparing it to the given dimensions.
Common errors: Using slanted sides instead of perpendicular height, forgetting to divide by 2, misidentifying base-height pairs, incorrect unit conversion.
Exam preparation: Practice with triangles in different orientations, master formula rearrangement, work with decimal and fractional measurements.
Formulas to know by heart:

• Standard Triangle: A = ½bh

• Right Triangle: A = ½ab (where a, b are legs)

• Base: b = 2A/h

• Height: h = 2A/b

• Equilateral Triangle: A = (s²√3)/4

• Heron's Formula: A = √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2

• Area with Trigonometry: A = ½ab sin(C)

Visualizing Triangle Areas: Different Base-Height Combinations
Exercise 6: Area Relationships
Consider how area changes with different base-height combinations:
Fixed area of 24 cm² with varying base and height values
Showing the inverse relationship

Analysis: The chart shows how base and height are inversely related for a fixed area.

  • When base doubles, height halves to maintain the same area
  • Product of base and height is constant for fixed area
  • Area remains the same regardless of base-height combination
  • The relationship follows bh = 2A (constant)

Questions & Answers

Question: How do I find the height of a triangle if it's not given and doesn't seem to be one of the sides?

Answer: This is a common challenge! Here are several approaches:

When height is not obvious:

  • Draw the altitude: Sketch a perpendicular line from the vertex to the base (or base extended)
  • Use the Pythagorean theorem: If you have a right triangle formed by the height, use a² + b² = c²
  • Rearrange the area formula: If area is known, use h = 2A/b
  • Trigonometry: Use sine function if an angle and side are known

For obtuse triangles: The height may lie outside the triangle, so extend the base to measure the perpendicular distance.

Remember: The height is always perpendicular to the base, regardless of the triangle's orientation.

Question: Can I use any side of a triangle as the base? Will the area be the same?

Answer: Yes, you can use any side as the base, and the area will always be the same! Here's why:

Key principle:

  • The area of a triangle is constant regardless of which side you consider the base
  • Whichever side you choose as the base, you must use the corresponding perpendicular height
  • The product of base and height will always equal 2 × Area

Example: In a triangle with sides 3, 4, 5 (a right triangle):

  • Using base = 3: height = 4, Area = ½(3)(4) = 6
  • Using base = 4: height = 3, Area = ½(4)(3) = 6
  • Using base = 5: height = 2.4, Area = ½(5)(2.4) = 6

This demonstrates that area is invariant under different base choices.

Question: Why do we divide by 2 in the triangle area formula?

Answer: The division by 2 comes from the relationship between triangles and rectangles:

Visual explanation:

  • If you take a rectangle and draw a diagonal, you create two identical triangles
  • Each triangle has half the area of the original rectangle
  • Rectangle area = base × height
  • Triangle area = ½ × base × height

Alternative perspective:

You can think of a triangle as half of a parallelogram with the same base and height. Since parallelogram area is base × height, triangle area is ½(base × height).

This is why the triangle area formula includes the factor of ½.