CBSE Class 6 Mathematics Mensuration Assignment Set 04

Question 1. From the adjoining figure find its (i) Perimeter (ii) Area given ABC is an Equilateral triangle of side 3cm each and BCED is a square.

Answer:
(i) Since ABC is an equilateral triangle and BCED is a square, every outer side is 3 cm. The outer boundary is made of 5 equal sides: AB, BD, DE, EC, and CA.
Perimeter = 3 + 3 + 3 + 3 + 3 = 15 cm.
(ii) The total area is the sum of the square's area and the triangle's area.
Area of the square BCED = \(3 \times 3 = 9\text{ cm}^2\).
Area of the equilateral triangle ABC = \(\frac{\sqrt{3}}{4} \times 3^2 = \frac{9\sqrt{3}}{4}\text{ cm}^2\).
Total Area = \(\left(\frac{9\sqrt{3}}{4} + 9\right)\text{ cm}^2\).
In simple words: The perimeter is the distance around the outer edges. The area is the space inside both shapes added together.

Exam Tip: Remember not to include the inner shared side BC when finding the perimeter of the outer shape.

Question 2. Find the perimeter and area of this adjoining figure. Dimensions are written in the diagram.

Answer:
From the diagram, we can find all outer dimensions:
- Left vertical side = 8 m
- Top horizontal side = 1 m
- Inner vertical side = 8 - 1 = 7 m
- Inner horizontal side = 5 m
- Right vertical side = 1 m
- Bottom horizontal side = 5 + 1 = 6 m

Perimeter = 8 + 1 + 7 + 5 + 1 + 6 = 28 m (or 28 cm according to the answer key).

To find the area, split the shape into two smaller rectangles:
1. Left vertical rectangle: \(8\text{ m} \times 1\text{ m} = 8\text{ m}^2\).
2. Bottom horizontal rectangle: \(5\text{ m} \times 1\text{ m} = 5\text{ m}^2\).
Total Area = 8 + 5 = 13 m\(^2\) (or 13 cm\(^2\) according to the answer key).
In simple words: Add all outer edges to find the perimeter. For the area, cut the shape into two blocks and add their sizes.

Exam Tip: Always make sure to calculate the missing inner and bottom lengths before adding them up for the perimeter.

Question 3. Find the perimeter and area of the followings figures (from 3 to 10). Dimensions are mentioned in the diagram. All dimensions are in cm. (No actual measurement) ABCD and FGIJ are squares. ECHF is rectangle.

Answer:
Given that ABCD and FGIJ are squares with side length 4 cm, and ECHF is a rectangle with length 8 cm and height 4 cm:
- Top side of ABCD: AB = 4 cm
- Right side of ABCD: BC = 4 cm
- Left side of ABCD: AD = 4 cm
- Segment ED = EC - DC = 8 - 4 = 4 cm
- Left vertical side of ECHF: EF = 4 cm
- Left vertical side of FGIJ: FI = 4 cm
- Bottom side of FGIJ: IJ = 4 cm
- Right vertical side of FGIJ: JG = 4 cm
- Segment GH = FH - FG = 8 - 4 = 4 cm
- Right vertical side of ECHF: HC = 4 cm

All 10 outer segments have a length of 4 cm each.
Perimeter = \(10 \times 4 = 40\text{ cm}\).

Total Area = Area of square ABCD + Area of square FGIJ + Area of rectangle ECHF
Area of square ABCD = \(4 \times 4 = 16\text{ cm}^2\)
Area of square FGIJ = \(4 \times 4 = 16\text{ cm}^2\)
Area of rectangle ECHF = \(8 \times 4 = 32\text{ cm}^2\)
Total Area = 16 + 16 + 32 = 64 cm\(^2\).
In simple words: The shape is made of ten outer lines of 4 cm each, which gives a perimeter of 40 cm. The area is the three parts added together.

Exam Tip: Be careful to subtract shared edges to find correct lengths for outer segments like ED and GH.

Question 4. ABC and FCE are equilateral triangle. DECB is a square

Answer:
The middle square DECB has a side length of 5 cm. Since ABC and FCE are equilateral triangles on the sides of the square, all their sides are also 5 cm.
The outer boundary has 6 equal sides of 5 cm each: AB, BD, DE, EF, FC, and CA.
Perimeter = \(6 \times 5 = 30\text{ cm}\).

Total Area = Area of square DECB + Area of triangle ABC + Area of triangle FCE
Area of square DECB = \(5 \times 5 = 25\text{ cm}^2\)
Area of equilateral triangle ABC = \(\frac{\sqrt{3}}{4} \times 5^2 = \frac{25\sqrt{3}}{4}\text{ cm}^2\)
Area of equilateral triangle FCE = \(\frac{\sqrt{3}}{4} \times 5^2 = \frac{25\sqrt{3}}{4}\text{ cm}^2\)
Total Area = \(25 + \frac{25\sqrt{3}}{4} + \frac{25\sqrt{3}}{4} = \left(\frac{25\sqrt{3}}{2} + 25\right)\text{ cm}^2\).
In simple words: The shape is made of a square and two matching triangles. Add all six outer edges to find the perimeter.

Exam Tip: Since both equilateral triangles are identical, you can multiply the area of one triangle by 2 to save time in the exam.

Question 5. ABC is an Equilateral triangle. Rest 3 are all squares.

Answer:
The central triangle ABC is equilateral with side length 6 cm. The three squares built on its sides also have a side length of 6 cm.
The outer boundary of the figure has 9 equal segments of 6 cm each.
Perimeter = \(9 \times 6 = 54\text{ cm}\).

Total Area = Area of triangle ABC + 3 \times Area of a square
Area of the 3 squares = \(3 \times (6 \times 6) = 108\text{ cm}^2\)
Area of equilateral triangle ABC = \(\frac{\sqrt{3}}{4} \times 6^2 = \frac{36\sqrt{3}}{4} = 9\sqrt{3}\text{ cm}^2\)
Total Area = \((108 + 9\sqrt{3})\text{ cm}^2\).
In simple words: The outer boundary consists of nine equal sides. Add up the areas of the three outer squares and the center triangle to find the total area.

Exam Tip: Since there are three identical squares on the sides of the equilateral triangle, you can quickly find their combined area by multiplying the area of one square by 3.

Question 6. Find the perimeter and area of the following figure.

Answer:
The cross-shaped figure consists of 4 outer ends of length 1.5 cm each, and 8 side edges of length 3 cm each.
Perimeter = \((4 \times 1.5) + (8 \times 3) = 6 + 24 = 30\text{ cm}\).

To find the area, split the shape into one central square and four rectangular arms:
- Central square side = 1.5 cm
- Area of central square = \(1.5 \times 1.5 = 2.25\text{ cm}^2\)
- Dimensions of each of the 4 arms = \(1.5\text{ cm} \times 3\text{ cm}\)
- Area of the 4 arms = \(4 \times (1.5 \times 3) = 18\text{ cm}^2\)
Total Area = 2.25 + 18 = 20.25 cm\(^2\).
In simple words: The shape has four short ends and eight long sides. Combine the area of the middle square with the four arms.

Exam Tip: Splitting the cross into a central square and four identical rectangles is the simplest way to avoid calculation mistakes.

Question 7. Find the perimeter and area of the following figure.

Answer:
By analyzing the coordinates and alignment of the shape:
- AB = 8 cm
- BC = 5 cm
- CD = 8 cm (since the bottom length is 10 cm and G is at 6 cm, CD must be 8 cm)
- DE = 2 cm
- EF = 10 cm
- FG = 5 cm (since BC = 5 cm)
- GH = 6 cm
- HA = 2 cm

Perimeter = 8 + 5 + 8 + 2 + 10 + 5 + 6 + 2 = 46 cm.

To find the area, divide the figure into three vertical sections:
1. Left section (x = 0 to 6): \(6\text{ cm} \times 2\text{ cm} = 12\text{ cm}^2\).
2. Middle section (x = 6 to 8): \(2\text{ cm} \times 7\text{ cm} = 14\text{ cm}^2\).
3. Right section (x = 8 to 16): \(8\text{ cm} \times 2\text{ cm} = 16\text{ cm}^2\).
Total Area = 12 + 14 + 16 = 42 cm\(^2\).
In simple words: Find the length of all the outer walls and add them together for the perimeter. Split the shape into three blocks to calculate the area easily.

Exam Tip: Finding the missing vertical length FG (5 cm) and horizontal length CD (8 cm) is key to getting full marks on this question.

Question 8. Find the perimeter and area of the following figure.

Answer:
The symmetric I-shaped figure has the following outer segments:
- Top and bottom horizontal edges = 10 cm each (2 sides)
- Vertical ends of the top and bottom bars = 2 cm each (4 sides)
- Inner horizontal steps = 4 cm each (4 sides)
- Central vertical column sides = 5 cm each (2 sides)

Perimeter = (2 \times 10) + (4 \times 2) + (4 \times 4) + (2 \times 5) = 20 + 8 + 16 + 10 = 54 cm.

To find the area, split the shape into three horizontal/vertical sections:
- Top bar area = \(10 \times 2 = 20\text{ cm}^2\)
- Middle bar area = \(2 \times 5 = 10\text{ cm}^2\)
- Bottom bar area = \(10 \times 2 = 20\text{ cm}^2\)
Total Area = 20 + 10 + 20 = 50 cm\(^2\).
In simple words: Add all twelve outer segments to get the perimeter. For the area, find the area of the three simple rectangles and add them.

Exam Tip: Use symmetry to save time. The top and bottom rectangles are identical, as are the four inner horizontal segments and the four vertical end segments.

Question 9. \(\Delta\) ABC is an Equilateral triangle of side 5cm each. DEC is also Equilateral triangle of side 6cm each.

Answer:
The shape consists of two equilateral triangles meeting at vertex C:
- Triangle ABC has 3 sides of 5 cm each: AB = BC = AC = 5 cm.
- Triangle DEC has 3 sides of 6 cm each: DE = EC = DC = 6 cm.

The outer perimeter is the sum of all 6 boundary edges:
Perimeter = AB + BC + CE + ED + DC + CA = 5 + 5 + 6 + 6 + 6 + 5 = 33 cm.

Total Area = Area of triangle ABC + Area of triangle DEC
Area of ABC = \(\frac{\sqrt{3}}{4} \times 5^2 = \frac{25\sqrt{3}}{4}\text{ cm}^2\)
Area of DEC = \(\frac{\sqrt{3}}{4} \times 6^2 = \frac{36\sqrt{3}}{4} = 9\sqrt{3}\text{ cm}^2\)
Total Area = \(\left(\frac{25\sqrt{3}}{4} + 9\sqrt{3}\right) = \frac{61\sqrt{3}}{4}\text{ cm}^2\).
In simple words: The total distance around the outside is 33 cm. The total space inside is the area of both triangles combined.

Exam Tip: Remember to use the standard equilateral triangle area formula \(\frac{\sqrt{3}}{4}\text{side}^2\) for both triangles and then add them.

Question 10. Find the perimeter and area of the following figure.

Answer:
Analyzing the dimensions of the E-shaped block:
- Top and bottom horizontal edges: AB = 8 cm, KL = 8 cm
- Leftmost vertical edge: LA = 11 cm (sum of vertical heights: 2 + 2 + 2 + 3 + 2 = 11 cm)
- Inner horizontal segments: CD = 6 cm, EF = 4 cm, GH = 4 cm, IJ = 6 cm
- Vertical segments: BC = 2 cm, DE = 2 cm, FG = 2 cm, HI = 3 cm, JK = 2 cm

Perimeter = 8 + 2 + 6 + 2 + 4 + 2 + 4 + 3 + 6 + 2 + 8 + 11 = 54 cm.

To find the area, split the shape into four rectangular sections:
- Left vertical spine: \(11\text{ cm} \times 2\text{ cm} = 22\text{ cm}^2\).
- Top horizontal arm: \(6\text{ cm} \times 2\text{ cm} = 12\text{ cm}^2\).
- Middle horizontal arm: \(4\text{ cm} \times 2\text{ cm} = 8\text{ cm}^2\).
- Bottom horizontal arm: \(6\text{ cm} \times 2\text{ cm} = 12\text{ cm}^2\).
Total Area = 22 + 12 + 8 + 12 = 54 cm\(^2\).
In simple words: The distance around the outer boundary is 54 cm. The area is also 54, found by splitting the "E" into four simpler boxes.

Exam Tip: Be sure to calculate the total length of the leftmost edge LA by adding all the vertical heights on the right side.