OP Malhotra Class 9 Maths Solutions Chapter 16 Area of Plane Figures Chapter Test

Question 1. What is the area of an equilateral triangle having altitude equal to \( 2\sqrt{3} \) cm?
(a) \( \sqrt{3} \)cm²
(b) \( 2\sqrt{3} \)cm²
(c) \( 3\sqrt{3} \)cm²
(d) \( 4\sqrt{3} \)cm²
Answer: (d) \( 4\sqrt{3} \)cm²
Altitude \( h = \frac{\sqrt{3}}{2} \times \text{side} \)
We are given \( h = 2\sqrt{3} \) cm.
So, \( \frac{\sqrt{3}}{2} \times \text{side} = 2\sqrt{3} \)

\( \implies \) Side \( = \frac{2 \times 2\sqrt{3}}{\sqrt{3}} \)

\( \implies \) Side \( = 4 \) cm
Now, the area of an equilateral triangle \( = \frac{\sqrt{3}}{4} \times \text{side}^2 \)

\( \implies \) Area \( = \frac{\sqrt{3}}{4} \times 4^2 \)

\( \implies \) Area \( = \frac{\sqrt{3}}{4} \times 16 \)

\( \implies \) Area \( = 4\sqrt{3} \text{ cm}^2 \) A unique property of equilateral triangles is that their altitudes also serve as medians and angle bisectors.
In simple words: First, use the given altitude to find the length of one side of the triangle. After you have the side length, use the formula for the area of an equilateral triangle to find the final answer.

🎯 Exam Tip: Remember the formulas for the altitude \( \left( \frac{\sqrt{3}}{2} \times \text{side} \right) \) and area \( \left( \frac{\sqrt{3}}{4} \times \text{side}^2 \right) \) of an equilateral triangle. They are key to solving such problems quickly.

Question 2. The altitude of an equilateral triangle is \( \sqrt{3} \) cm. What is its perimeter?
(a) 3 cm
(b) \( 3\sqrt{3} \) cm
(c) 6 cm
(d) \( 6\sqrt{3} \) cm
Answer: (c) 6 cm
The formula for the altitude of an equilateral triangle is \( h = \frac{\sqrt{3}}{2} \times \text{side} \).
Given altitude \( h = \sqrt{3} \) cm.
So, \( \frac{\sqrt{3}}{2} \times \text{side} = \sqrt{3} \)

\( \implies \) Side \( = \frac{2 \times \sqrt{3}}{\sqrt{3}} \)

\( \implies \) Side \( = 2 \) cm
The perimeter of an equilateral triangle is \( 3 \times \text{side} \).

\( \implies \) Perimeter \( = 3 \times 2 \)

\( \implies \) Perimeter \( = 6 \) cm. Equilateral triangles have three equal sides and three equal angles, each 60 degrees.
In simple words: Use the given altitude and its formula to figure out how long one side of the triangle is. Once you have the side length, multiply it by 3 to find the total distance around the triangle.

🎯 Exam Tip: Always recall that all three sides of an equilateral triangle are equal, simplifying the perimeter calculation once a side length is known.

Question 3. The sides of a triangle are in ratio 2: 3: 4. The perimeter of the triangle is 18 cm. The area (in cm²) of the triangle is
(a) 9
(b) 36
(c) \( \sqrt{42} \)
(d) \( 6\sqrt{15} \)
Answer: (d) \( 6\sqrt{15} \)
The sides of the triangle are in the ratio 2 : 3 : 4.
The perimeter of the triangle is 18 cm.
Let the sides be \( 2x, 3x, \) and \( 4x \).
Perimeter \( = 2x + 3x + 4x = 9x \).
We know \( 9x = 18 \) cm.

\( \implies x = \frac{18}{9} = 2 \) cm.
So, the lengths of the sides are:
First side \( (a) = 2x = 2 \times 2 = 4 \) cm
Second side \( (b) = 3x = 3 \times 2 = 6 \) cm
Third side \( (c) = 4x = 4 \times 2 = 8 \) cm
Now, we calculate the semi-perimeter \( s \):
\( s = \frac{\text{Perimeter}}{2} = \frac{18}{2} = 9 \) cm
Using Heron's formula for the area of a triangle:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)

\( \implies \) Area \( = \sqrt{9(9-4)(9-5)(9-6)} \)

\( \implies \) Area \( = \sqrt{9 \times 5 \times 4 \times 3} \)

\( \implies \) Area \( = \sqrt{540} \)
To simplify the square root, find factors of 540:
\( 540 = 36 \times 15 \)

\( \implies \) Area \( = \sqrt{36 \times 15} \)

\( \implies \) Area \( = 6\sqrt{15} \text{ cm}^2 \). Heron's formula is very useful for finding the area of a triangle when only the side lengths are known.
In simple words: First, use the given ratio and perimeter to find the actual lengths of the three sides of the triangle. Then, calculate half of the perimeter (called the semi-perimeter). Finally, use a special formula called Heron's formula with the side lengths and semi-perimeter to find the area.

🎯 Exam Tip: When given a ratio of sides and perimeter, always use a variable (like \( x \)) to find the actual side lengths first. Heron's formula is then applied using the semi-perimeter and side lengths.

Question 4. A parallelogram has sides 15 cm and 7 cm long. The length of one of the diagonals is 20 cm. The area of the parallelogram is
(a) 42 cm²
(b) 60 cm²
(c) 84 cm²
(d) 96 cm²
Answer: (c) 84 cm²
A parallelogram can be divided into two identical triangles by one of its diagonals.
Let the sides of the parallelogram be \( a = 15 \) cm and \( b = 7 \) cm.
Let one diagonal be \( d = 20 \) cm.
We can find the area of one of the triangles formed (e.g., triangle ABC with sides 15 cm, 7 cm, and 20 cm).
First, calculate the semi-perimeter \( s \) of this triangle:
\( s = \frac{\text{sum of sides}}{2} = \frac{15 + 7 + 20}{2} = \frac{42}{2} = 21 \) cm.
Now, use Heron's formula to find the area of the triangle:
Area of \( \Delta ABC = \sqrt{s(s-a)(s-b)(s-c)} \)

\( \implies \) Area \( = \sqrt{21(21-15)(21-7)(21-20)} \)

\( \implies \) Area \( = \sqrt{21 \times 6 \times 14 \times 1} \)

\( \implies \) Area \( = \sqrt{(3 \times 7) \times (2 \times 3) \times (2 \times 7) \times 1} \)

\( \implies \) Area \( = \sqrt{2^2 \times 3^2 \times 7^2} \)

\( \implies \) Area \( = 2 \times 3 \times 7 = 42 \text{ cm}^2 \). The area of a triangle formed by a diagonal is exactly half the area of the parallelogram.
The area of the parallelogram is twice the area of this triangle.
Area of parallelogram \( = 2 \times 42 = 84 \text{ cm}^2 \).
In simple words: A diagonal splits a parallelogram into two identical triangles. Find the area of one of these triangles using its three side lengths (two sides of the parallelogram and the diagonal). Then, double that area to get the total area of the parallelogram.

🎯 Exam Tip: Remember that a diagonal divides a parallelogram into two congruent triangles. This allows you to use Heron's formula on one triangle and then multiply the result by two to find the parallelogram's area.

Question 5. The perimeter of two squares are 68 cm and 60 cm. The perimeter of a third square whose area is equal to the difference of the areas of the two squares.
(a) 64 cm
(b) 60 cm
(c) 32 cm
(d) 8 cm
Answer: (c) 32 cm
For the first square:
Perimeter \( = 68 \) cm.
Side \( = \frac{\text{Perimeter}}{4} = \frac{68}{4} = 17 \) cm.
Area \( = (\text{Side})^2 = 17^2 = 289 \text{ cm}^2 \).
For the second square:
Perimeter \( = 60 \) cm.
Side \( = \frac{\text{Perimeter}}{4} = \frac{60}{4} = 15 \) cm.
Area \( = (\text{Side})^2 = 15^2 = 225 \text{ cm}^2 \).
For the third square:
Its area is equal to the difference of the areas of the first two squares.
Area of third square \( = 289 - 225 = 64 \text{ cm}^2 \).
Side of third square \( = \sqrt{\text{Area}} = \sqrt{64} = 8 \) cm. This calculation uses the fact that area is side squared.
Perimeter of third square \( = 4 \times \text{Side} = 4 \times 8 = 32 \) cm.
In simple words: First, for each of the two given squares, find their side lengths from their perimeters, then calculate their areas. Next, subtract the smaller area from the larger area to find the area of the third square. Finally, find the side length of this third square from its area, and then calculate its perimeter.

🎯 Exam Tip: Always break down multi-step problems into smaller, manageable parts. Calculate side lengths, then areas, then the difference, and finally the new square's perimeter.

Question 6. A square is of area 200 m². A new square is formed in such a way that the length of its diagonal is \( \sqrt{2} \) times of the diagonal of the given square. Then the area of the new square formed is
(a) \( 200\sqrt{2} \) sq m
(b) \( 400\sqrt{2} \) sq m
(c) 400 sq m
(d) 800 sq m
Answer: (c) 400 sq m
Let the given square be Square 1.
Area of Square 1 \( = 200 \text{ m}^2 \).
The area of a square can also be calculated using its diagonal \( d \) as: Area \( = \frac{d^2}{2} \).
So, for Square 1: \( 200 = \frac{d_1^2}{2} \)

\( \implies d_1^2 = 200 \times 2 = 400 \)

\( \implies d_1 = \sqrt{400} = 20 \) m.
For the new square (Square 2):
Its diagonal \( d_2 \) is \( \sqrt{2} \) times the diagonal of Square 1.
\( d_2 = \sqrt{2} \times d_1 = \sqrt{2} \times 20 \) m.
Now, find the area of Square 2 using its diagonal:
Area of Square 2 \( = \frac{d_2^2}{2} \)

\( \implies \) Area \( = \frac{(\sqrt{2} \times 20)^2}{2} \)

\( \implies \) Area \( = \frac{2 \times 20^2}{2} \)

\( \implies \) Area \( = 20^2 = 400 \text{ m}^2 \). This demonstrates the relationship between a square's diagonal and its area.
In simple words: First, find the length of the diagonal of the first square using its area. Then, multiply this diagonal length by \( \sqrt{2} \) to get the diagonal length of the new square. Finally, use the diagonal length of the new square to calculate its area.

🎯 Exam Tip: Remember the relationship between a square's area and its diagonal: Area \( = \frac{d^2}{2} \). This formula often simplifies calculations in problems involving diagonals.

Question 7. The sides of a parallelogram are in the ratio 5 : 4. Its area is 1000 sq. units. Altitude on the greater side is 20 units. Altitude on the smaller side is
(a) 30 units
(b) 25 units
(c) 10 units
(d) 15 units
Answer: (b) 25 units
Let the sides of the parallelogram be \( 5x \) and \( 4x \).
The greater side is \( 5x \).
The altitude on the greater side is 20 units.
The formula for the area of a parallelogram is Base \( \times \) Altitude.
Area \( = \text{Greater Side} \times \text{Altitude on Greater Side} \)
\( 1000 = 5x \times 20 \)

\( \implies 1000 = 100x \)

\( \implies x = \frac{1000}{100} = 10 \)
So, the lengths of the sides are:
Greater side \( = 5x = 5 \times 10 = 50 \) units.
Smaller side \( = 4x = 4 \times 10 = 40 \) units.
Now, we need to find the altitude on the smaller side. Let this be \( h_s \).
Area \( = \text{Smaller Side} \times \text{Altitude on Smaller Side} \)
\( 1000 = 40 \times h_s \)

\( \implies h_s = \frac{1000}{40} \)

\( \implies h_s = 25 \) units. This principle of constant area is fundamental for parallelograms, meaning base times height is constant for any chosen base.
In simple words: Use the given area and the altitude on the greater side to find the actual lengths of the parallelogram's sides. Once you know the smaller side's length, use the area formula again (Area = smaller side multiplied by its altitude) to find the altitude on the smaller side.

🎯 Exam Tip: The area of a parallelogram (Base × Height) remains constant regardless of which side is chosen as the base. This relationship is crucial for finding unknown altitudes or side lengths.

Question 8. One diagonal and perimeter of a rhombus are 24 cm and 52 cm respectively. The other diagonal is
(a) 13 cm
(b) 12 cm
(c) 10 cm
(d) 15 cm
Answer: (c) 10 cm
A rhombus has four equal sides.
Perimeter of rhombus \( = 52 \) cm.
Side of rhombus \( = \frac{\text{Perimeter}}{4} = \frac{52}{4} = 13 \) cm.
Let the diagonals be \( d_1 \) and \( d_2 \). One diagonal \( d_1 = 24 \) cm.
In a rhombus, the diagonals bisect each other at right angles.
So, half of the first diagonal \( = \frac{24}{2} = 12 \) cm.
Consider one of the four right-angled triangles formed by the diagonals. The hypotenuse of this triangle is the side of the rhombus (13 cm), and the two legs are half the lengths of the diagonals (12 cm and \( \frac{d_2}{2} \)).
Using the Pythagorean theorem:
\( (\text{Side})^2 = (\text{half of } d_1)^2 + (\text{half of } d_2)^2 \)
\( 13^2 = 12^2 + \left(\frac{d_2}{2}\right)^2 \)
\( 169 = 144 + \left(\frac{d_2}{2}\right)^2 \)
\( \left(\frac{d_2}{2}\right)^2 = 169 - 144 = 25 \)
\( \frac{d_2}{2} = \sqrt{25} = 5 \) cm.
So, the other diagonal \( d_2 = 2 \times 5 = 10 \) cm. The diagonals of a rhombus are always perpendicular bisectors of each other.
In simple words: First, find the length of one side of the rhombus from its perimeter. Since diagonals of a rhombus cross each other at a right angle and cut each other in half, use half of the given diagonal and the side length in a right-angled triangle. Then, use the Pythagorean theorem to find half of the other diagonal, and finally double that to find its full length.

🎯 Exam Tip: For rhombus problems, remember that its four sides are equal, and its diagonals intersect at a right angle, bisecting each other. This creates four congruent right-angled triangles, allowing the use of the Pythagorean theorem.

Question 9. The area of a field in the shape of a trapezium measures 1440 m². The perpendicular distance between its parallel sides is 24 m. If the ratio of the parallel sides is 5 : 3, the length of the longer parallel sides is:
(a) 45 m
(b) 60 m
(c) 75 cm
(d) 120 m
Answer: (c) 75 m
Area of a trapezium \( = 1440 \text{ m}^2 \).
Perpendicular distance (height) \( = 24 \) m.
The formula for the area of a trapezium is \( \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \).
Let the parallel sides be \( a \) and \( b \).
\( 1440 = \frac{1}{2} \times (a+b) \times 24 \)
\( 1440 = 12 \times (a+b) \)

\( \implies a+b = \frac{1440}{12} = 120 \) m. This is the sum of the two parallel sides.
The ratio of the parallel sides is 5 : 3.
Let the longer side be \( 5x \) and the shorter side be \( 3x \).
Sum of sides \( = 5x + 3x = 8x \).
We know \( 8x = 120 \) m.

\( \implies x = \frac{120}{8} = 15 \) m.
The longer parallel side \( = 5x = 5 \times 15 = 75 \) m. The problem asks for the longer parallel side, which is 5/8 of the total sum.
(Note: Option (c) in the source has "75 cm" but the calculation is in meters. We adopt the numerical value and correct unit.)
In simple words: First, use the trapezium's area and height to find the total length of its two parallel sides. Then, use the given ratio of these parallel sides to calculate the actual length of the longer side.

🎯 Exam Tip: Be careful with units! Although an option might suggest a different unit (like cm here), always derive your answer based on the units given in the problem statement (m in this case).

Question 10. A lawn 30 m long and 16 m wide is surrounded by a path 2 m wide. What is the area of the path?
(a) 200 m²
(b) 280 m²
(c) 300 m²
(d) 320 m²
Answer: (a) 200 m²
Length of the lawn \( (\text{inner length}) = 30 \) m.
Breadth of the lawn \( (\text{inner breadth}) = 16 \) m.
Width of the path around the lawn \( = 2 \) m.
To find the dimensions of the outer rectangle (lawn + path):
Outer length \( = \text{Inner length} + (2 \times \text{path width}) \)
Outer length \( = 30 + (2 \times 2) = 30 + 4 = 34 \) m.
Outer breadth \( = \text{Inner breadth} + (2 \times \text{path width}) \)
Outer breadth \( = 16 + (2 \times 2) = 16 + 4 = 20 \) m.
Area of the path \( = \text{Area of Outer Rectangle} - \text{Area of Inner Rectangle} \).
Area of Outer Rectangle \( = \text{Outer length} \times \text{Outer breadth} = 34 \times 20 = 680 \text{ m}^2 \).
Area of Inner Rectangle \( = \text{Inner length} \times \text{Inner breadth} = 30 \times 16 = 480 \text{ m}^2 \).
Area of path \( = 680 - 480 = 200 \text{ m}^2 \). Calculating areas by multiplying length and breadth is a fundamental skill.
In simple words: First, figure out the total length and width of the lawn including the path. Then, calculate the area of this larger rectangle and subtract the area of the lawn itself. The leftover area is the area of the path.

🎯 Exam Tip: When dealing with paths around a rectangle, always remember to add the path width twice (once for each side) to both the length and breadth of the inner rectangle to get the outer dimensions.