CBSE Class 7 Mathematics Perimeter and Area Assignment Set 03

VII - Mathematics Assignment No-08 - Perimeter and Area

M. C. Q.

Question 1. The perimeter of a triangle of side 12.3 cm, 13.3 cm, 10.4 cm is
(i) 36 cm
(ii) 18 cm
(iii) 35 cm
(iv) 17 cm
Answer: (i) 36 cm
Perimeter of a triangle = Sum of all three sides
Perimeter = \(12.3 + 13.3 + 10.4 = 36\) cm
In simple words: Add all the three given side lengths of the triangle to find the perimeter.

Exam Tip: Always check that all given side lengths are in the same unit before adding.

Question 2. The perimeter of a rectangle of side 17.5 cm and 10.3 cm is
(i) 27.8 cm
(ii) 26.8 cm
(iii) 55.6 cm
(iv) 54.6 cm
Answer: (iii) 55.6 cm
Perimeter of a rectangle = \(2 \times (\text{length} + \text{breadth})\)
Perimeter = \(2 \times (17.5 + 10.3)\)
\(= 2 \times 27.8\)
\(= 55.6\) cm
In simple words: Add the length and width together, then multiply the sum by 2.

Exam Tip: Do not forget to multiply by 2 after adding the length and breadth.

Question 3. The perimeter of a square of side 3.2 cm is
(i) 12.7 cm
(ii) 12.6 cm
(iii) 12.8 cm
(iv) 12.9 cm
Answer: (iii) 12.8 cm
Perimeter of a square = \(4 \times \text{side}\)
Perimeter = \(4 \times 3.2 = 12.8\) cm
In simple words: Multiply the side length of the square by 4.

Exam Tip: A square has four equal sides, so multiplying the side length by 4 gives the total perimeter.

Question 4. If the perimeter of an equilateral triangle is 18 cm then its area is
(i) \(\frac{\sqrt{3}}{4} \times 5\) cm\(^2$
(ii) \(9\sqrt{3}\) cm\(^2$
(iii) \(10\sqrt{3}\) cm\(^2$
(iv) \(12\sqrt{3}\) cm\(^2$
Answer: (ii) \(9\sqrt{3}\) cm\(^2$
Side of the equilateral triangle \((a) = \frac{\text{Perimeter}}{3} = \frac{18}{3} = 6\) cm
Area of an equilateral triangle = \(\frac{\sqrt{3}}{4} a^2\)
Area = \(\frac{\sqrt{3}}{4} \times (6)^2 = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3}\) cm\(^2$
In simple words: First find the side length by dividing the perimeter by 3, then use the equilateral triangle area formula.

Exam Tip: Memorize the formula for the area of an equilateral triangle: \(\frac{\sqrt{3}}{4} a^2\).

Question 5. If the length of a rectangle is \((2x+3)\) cm and width is \((2x+1)\) cm, and if the perimeter of rectangle is 168 cm then its area is
(i) 1763 cm\(^2$
(ii) 1764 cm\(^2$
(iii) 1765 cm\(^2$
(iv) 1766 cm\(^2$
Answer: (i) 1763 cm\(^2$
Perimeter = \(2 \times (\text{length} + \text{width})\)
\(168 = 2 \times ((2x+3) + (2x+1))\)
\(168 = 2 \times (4x + 4)\)
\(84 = 4x + 4\)
\(4x = 80\)
\(x = 20\)
Length = \(2(20) + 3 = 43\) cm
Width = \(2(20) + 1 = 41\) cm
Area = Length \(\times\) Width = \(43 \times 41 = 1763\) cm\(^2$
In simple words: Use the perimeter formula to find the value of x, find the dimensions, and then multiply length by width.

Exam Tip: Solve linear equations carefully before substituting values back to find geometric dimensions.

Question 6. If the area of a triangle is 16 cm\(^2\) and its base is 8 cm. Then its height is
(i) 7 cm
(ii) 6 cm
(iii) 5 cm
(iv) 4 cm
Answer: (iv) 4 cm
Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(16 = \frac{1}{2} \times 8 \times \text{height}\)
\(16 = 4 \times \text{height}\)
Height = \(\frac{16}{4} = 4\) cm
In simple words: Rearrange the triangle area formula to solve for height: Height = \(\frac{2 \times \text{Area}}{\text{base}}\).

Exam Tip: Always double the area and divide by the base when finding the height of a triangle.

Question 7. The area of an isosceles triangle of base b unit and each congruent side is a unit is
(i) \(\frac{b}{2} \sqrt{4a^2-b^2}\)
(ii) \(\frac{b}{4} \sqrt{4a^2-b^2}\)
(iii) \(\frac{b}{8} \sqrt{4a^2-b^2}\)
(iv) none of the options
Answer: (ii) \(\frac{b}{4} \sqrt{4a^2-b^2}\)
Height of the isosceles triangle \((h) = \sqrt{a^2 - \left(\frac{b}{2}\right)^2} = \sqrt{a^2 - \frac{b^2}{4}} = \frac{\sqrt{4a^2 - b^2}}{2}\)
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times \frac{\sqrt{4a^2 - b^2}}{2} = \frac{b}{4}\sqrt{4a^2 - b^2}\)
In simple words: Apply Pythagoras theorem to find the altitude from the vertex to the base, then calculate triangle area.

Exam Tip: Memorize standard formulas for special triangles like equilateral and isosceles triangles.

Question 8. The perimeter of a triangle is 36 cm. If its sides are \((x+1)\), \((x+2)\), \((x+3)\) cm. Find the value of x and is
(i) 13
(ii) 12
(iii) 11
(iv) 10
Answer: (iv) 10
Perimeter = Sum of all sides
\(36 = (x+1) + (x+2) + (x+3)\)
\(36 = 3x + 6\)
\(3x = 30\)
\(x = 10\)
In simple words: Add the algebraic expressions for the sides, equate the sum to the perimeter 36, and solve for x.

Exam Tip: Combine like terms carefully when setting up algebraic equations from geometric perimeters.

Question 9. The circumference of a circle of diameter 10 cm is
(i) 31.42 cm
(ii) 31.52 cm
(iii) 30.42 cm
(iv) 30.52 cm
Answer: (i) 31.42 cm
Circumference of circle = \(\pi \times \text{diameter}\)
Circumference = \(3.1416 \times 10 = 31.416 \approx 31.42\) cm
In simple words: Multiply the diameter by \(\pi\) (approx 3.1416) to get the circumference.

Exam Tip: Use \(\pi \approx 3.14\) or \(\frac{22}{7}\) depending on which gives a simpler calculation.

Question 10. If the circumference of a circle is 44 cm then its area is
(i) 164 cm\(^2$
(ii) 144 cm\(^2$
(iii) 154 cm\(^2$
(iv) 134 cm\(^2$
Answer: (iii) 154 cm\(^2$
Circumference = \(2\pi r = 44\)
\(2 \times \frac{22}{7} \times r = 44\)
\(\frac{44}{7} \times r = 44\)
\(r = 7\) cm
Area = \(\pi r^2 = \frac{22}{7} \times (7)^2 = \frac{22}{7} \times 49 = 154\) cm\(^2$
In simple words: Find the radius from the circumference formula first, then use it to find the area of the circle.

Exam Tip: A circle with circumference 44 cm always has a radius of 7 cm and an area of 154 cm\(^2\). This is a very common standard result.

Question 11. A road roller of radius 1.4 m makes 10 revolutions. The total distance covered by it
(i) 78 m
(ii) 88 m
(iii) 98 m
(iv) 108 m
Answer: (ii) 88 m
Distance covered in 1 revolution = Circumference of the roller = \(2\pi r\)
Distance in 1 revolution = \(2 \times \frac{22}{7} \times 1.4 = 2 \times \frac{22}{7} \times \frac{14}{10} = 8.8\) m
Total distance in 10 revolutions = \(8.8 \times 10 = 88\) m
In simple words: Multiply the perimeter of the circular roller by the total number of revolutions (10).

Exam Tip: Distance covered by a rolling wheel in one complete turn is equal to its circumference.

Question 12. If the area of a circle is 1386 cm\(^2\) then its radius is
(i) 21 cm
(ii) 22 cm
(iii) 23 cm
(iv) 24 cm
Answer: (i) 21 cm
Area of circle = \(\pi r^2 = 1386\)
\(\frac{22}{7} \times r^2 = 1386\)
\(r^2 = \frac{1386 \times 7}{22} = 63 \times 7 = 441\)
\(r = \sqrt{441} = 21\) cm
In simple words: Divide the area by \(\pi\), then take the square root to find the radius.

Exam Tip: Memorize squares of numbers up to 30 to quickly find square roots like \(\sqrt{441} = 21\).

Question 13. The area of a parallelogram with base 12 cm and corresponding altitude is 8 cm is
(i) 84 cm\(^2$
(ii) 96 cm\(^2$
(iii) 48 cm\(^2$
(iv) 42 cm\(^2$
Answer: (ii) 96 cm\(^2$
Area of parallelogram = \(\text{base} \times \text{height}\)
Area = \(12 \times 8 = 96\) cm\(^2$
In simple words: Multiply the base length by the height (altitude) of the parallelogram.

Exam Tip: Unlike triangles, the area of a parallelogram does not include the factor of \(\frac{1}{2}\).

Question 14. If the area of a parallelogram is 5760 m\(^2\) and its base is 80 m then its height is
(i) 72 m
(ii) 62 m
(iii) 82 m
(iv) 45 m
Answer: (i) 72 m
Area = \(\text{base} \times \text{height}\)
\(5760 = 80 \times \text{height}\)
Height = \(\frac{5760}{80} = 72\) m
In simple words: Divide the total area by the given base to find the height.

Exam Tip: Cancel out common trailing zeros in numerator and denominator to simplify division quickly.

Question 15. The parallelogram ABCD has base AB as 15 cm and altitude on it is 8 cm. If AD = 10 cm. Find its corresponding altitude
(i) 10 cm
(ii) 11 cm
(iii) 12 cm
(iv) 14 cm
Answer: (iii) 12 cm
Area of parallelogram = \(\text{base} \times \text{corresponding height}\)
Using base AB = 15 cm and height = 8 cm:
Area = \(15 \times 8 = 120\) cm\(^2$
Now using base AD = 10 cm and let corresponding height be \(h\):
\(10 \times h = 120\)
\(h = \frac{120}{10} = 12\) cm
In simple words: Calculate the total area using one base and its altitude, then equate it with the product of the second base and its unknown altitude.

Exam Tip: The area of a parallelogram remains constant regardless of which side is chosen as the base.