OP Malhotra Class 9 Maths Solutions Chapter 16 Area of Plane Figures Exercise 16 (C)

S Chand Class 9 ICSE Maths Solutions Chapter 16 Area of Plane Figures Ex 16(C)

 

Question 1. Find the area of a parallelogram whose base and height are as given below.

Answer:
(i) Given Base \( (b) = 8 \) cm and Height \( (h) = 3 \) cm.
The area of a parallelogram is calculated by multiplying its base and height.
\( \implies \) Area \( = b \times h = 8 \times 3 = 24 \) cm\(^2\).
(ii) Given Base \( (b) = 2.8 \) cm and Height \( (h) = 5 \) cm.
\( \implies \) Area \( = b \times h = 2.8 \times 5 = 14 \) cm\(^2\).
(iii) Given Base \( (b) = 12 \) mm and Height \( (h) = 8.7 \) mm.
\( \implies \) Area \( = b \times h = 12 \times 8.7 = 104.4 \) mm\(^2\).
(iv) Given Base \( (b) = 6.5 \) m and Height \( (h) = 4.8 \) m.
\( \implies \) Area \( = b \times h = 6.5 \times 4.8 = 31.20 \) m\(^2\).
In simple words: To find the area of a parallelogram, you just multiply its base by its perpendicular height. Make sure the units for base and height are the same.

🎯 Exam Tip: Always remember the formula for the area of a parallelogram is base multiplied by height, and ensure all units are consistent before calculating.

 

Question 2. The area of a parallelogram is \( 1\frac { 1 }{ 2 } \) ares. Its base is 20 m. Find its height (1 are = 100 m\(^2\)).
Answer:
First, we need to convert the area from ares to square meters (m\(^2\)).
Given Area of a parallelogram \( = 1\frac { 1 }{ 2 } \) ares.
We know that 1 are \( = 100 \) m\(^2\).
\( \implies \) Area \( = 1.5 \times 100 = 150 \) m\(^2\).
The base of the parallelogram \( (b) = 20 \) m.
The formula for the area of a parallelogram is Area \( = Base \times Height \).
\( \implies \) Height \( (h) = \frac{\text { Area }}{\text { Base }} \)
\( \implies h = \frac { 150 }{ 20 } = 7.5 \) m.
The height of the parallelogram is 7.5 m.
In simple words: First, change the area from 'ares' to 'square meters' using the given conversion. Then, divide the total area by the base length to find the height.

🎯 Exam Tip: Always pay attention to unit conversions (like ares to m\(^2\)) before performing calculations, as incorrect units will lead to wrong answers.

 

Question 3. In a parallelogram ABCD, AB = 8 cm, BC = 5 cm, perp. from A to DC = 3 cm. Find the length of the perp. drawn from B to AD.
Answer:
Let the parallelogram be ABCD.
Given side \(AB = 8\) cm and side \(BC = 5\) cm.
In a parallelogram, opposite sides are equal, so \(AB = DC = 8\) cm and \(BC = AD = 5\) cm.
A perpendicular (height) AL is drawn from A to DC, with length \(AL = 3\) cm.
The area of the parallelogram ABCD can be calculated as Base \( \times \) Height \( = DC \times AL \).
\( \implies \) Area \( = 8 \times 3 = 24 \) cm\(^2\).
Now, we need to find the length of the perpendicular (height) drawn from B to AD. Let this perpendicular be BM and its length be \(x\).
We can also calculate the area of the parallelogram using base AD and height BM.
Area \( = AD \times BM \)
\( \implies 24 = 5 \times x \)
To find \(x\), we divide the area by the base AD.
\( \implies x = \frac { 24 }{ 5 } = 4.8 \) cm.
The length of the perpendicular drawn from B to AD is 4.8 cm.
In simple words: We find the total area of the parallelogram first using the given base and height. Then, we use this same area with the other given base to work out the missing height. This is because the area of a parallelogram is fixed, no matter which side you pick as the base.

🎯 Exam Tip: Remember that the area of a parallelogram remains the same regardless of which side is chosen as the base; you just need to use the corresponding perpendicular height for that base.

 

Question 4. A parallelogram has sides 34 cm and 20 cm. One of its diagonals is 42 cm. Calculate its area.
Answer:
Let the parallelogram be ABCD, with sides \(AB = 34\) cm and \(AD = 20\) cm.
One of its diagonals, say \(BD = 42\) cm.
A diagonal divides a parallelogram into two triangles of equal area. So, Area of parallelogram ABCD \( = 2 \times \) Area of \( \triangle ABD \).
To find the area of \( \triangle ABD \), we can use Heron's formula.
The sides of \( \triangle ABD \) are \(a = AD = 20\) cm, \(b = AB = 34\) cm, and \(c = BD = 42\) cm.
First, calculate the semi-perimeter \( (s) \):
\( s = \frac { a+b+c }{ 2 } = \frac { 20+34+42 }{ 2 } = \frac { 96 }{ 2 } = 48 \) cm.
Now, apply Heron's formula for the area of \( \triangle ABD \):
Area \( (\triangle ABD) = \sqrt{s(s-a)(s-b)(s-c)} \)
\( \implies \) Area \( = \sqrt{48(48-20)(48-34)(48-42)} \)
\( \implies \) Area \( = \sqrt{48 \times 28 \times 14 \times 6} \)
To simplify, we find the prime factors:
\( \implies \) Area \( = \sqrt{(2^4 \times 3) \times (2^2 \times 7) \times (2 \times 7) \times (2 \times 3)} \)
\( \implies \) Area \( = \sqrt{2^8 \times 3^2 \times 7^2} \)
\( \implies \) Area \( = 2^4 \times 3 \times 7 \)
\( \implies \) Area \( = 16 \times 21 = 336 \) cm\(^2\).
The area of the parallelogram ABCD is twice the area of \( \triangle ABD \).
\( \implies \) Area of parallelogram ABCD \( = 2 \times 336 = 672 \) cm\(^2\).
In simple words: A diagonal cuts a parallelogram into two identical triangles. We find the area of one of these triangles using a formula called Heron's formula, which needs the lengths of all three sides. Once we have the area of one triangle, we just double it to get the total area of the parallelogram.

🎯 Exam Tip: When using Heron's formula, make sure to correctly calculate the semi-perimeter and factorize the numbers under the square root to simplify the calculation effectively.

 

Question 5. ABCD is a parallelogram with side AB = 12 cm. Its diagonal AC and BD are of lengths 20 cm and 16 cm respectively. Find the area of ||gm ABCD.
Answer:
Let the parallelogram be ABCD.
Given side \(AB = 12\) cm.
Diagonal \(AC = 20\) cm and Diagonal \(BD = 16\) cm.
In a parallelogram, the diagonals bisect each other. Let the point of intersection be O.
So, in \( \triangle AOB \):
\( AO = \frac { AC }{ 2 } = \frac { 20 }{ 2 } = 10 \) cm.
\( BO = \frac { BD }{ 2 } = \frac { 16 }{ 2 } = 8 \) cm.
Now, we have a triangle \( \triangle AOB \) with sides \(AB = 12\) cm, \(AO = 10\) cm, and \(BO = 8\) cm.
We can find the area of \( \triangle AOB \) using Heron's formula.
First, calculate the semi-perimeter \( (s) \) for \( \triangle AOB \):
\( s = \frac { AB+AO+BO }{ 2 } = \frac { 12+10+8 }{ 2 } = \frac { 30 }{ 2 } = 15 \) cm.
Now, apply Heron's formula:
Area \( (\triangle AOB) = \sqrt{s(s-AB)(s-AO)(s-BO)} \)
\( \implies \) Area \( = \sqrt{15(15-12)(15-10)(15-8)} \)
\( \implies \) Area \( = \sqrt{15 \times 3 \times 5 \times 7} \)
\( \implies \) Area \( = \sqrt{(3 \times 5) \times 3 \times 5 \times 7} \)
\( \implies \) Area \( = \sqrt{3^2 \times 5^2 \times 7} \)
\( \implies \) Area \( = 3 \times 5 \times \sqrt{7} = 15\sqrt{7} \) cm\(^2\).
The diagonals of a parallelogram divide it into four triangles of equal area.
Therefore, Area of parallelogram ABCD \( = 4 \times \) Area \( (\triangle AOB) \).
\( \implies \) Area \( = 4 \times 15\sqrt{7} = 60\sqrt{7} \) cm\(^2\).
In simple words: First, use the fact that the diagonals of a parallelogram cut each other in half to find the sides of one of the four small triangles formed inside. Then, calculate the area of this small triangle using Heron's formula. Finally, multiply this small triangle's area by four to get the total area of the parallelogram.

🎯 Exam Tip: Remember that parallelogram diagonals bisect each other, forming four triangles of equal area. This property is key to solving problems when side lengths and diagonals are given.

 

Question 6. What is the area of a rhombus which has diagonals of 8 cm and 10 cm.
Answer:
Let the rhombus be ABCD.
Given the lengths of the diagonals are \(d_1 = 8\) cm and \(d_2 = 10\) cm.
The formula for the area of a rhombus when its diagonals are known is:
Area \( = \frac { 1 }{ 2 } \times d_1 \times d_2 \)
\( \implies \) Area \( = \frac { 1 }{ 2 } \times 8 \times 10 \)
\( \implies \) Area \( = 4 \times 10 = 40 \) cm\(^2\).
The area of the rhombus is 40 cm\(^2\).
In simple words: To find the area of a rhombus, you simply multiply the lengths of its two diagonals together, and then divide the result by two.

🎯 Exam Tip: Make sure to remember the specific formula for the area of a rhombus using its diagonals, as it's a common shortcut.

 

Question 7. The area of a rhombus is 98 cm\(^2\). If one of its diagonals is 14 cm, what is the length of the other diagonal ?
Answer:
Given the area of the rhombus \( = 98 \) cm\(^2\).
The length of one diagonal \( (d_1) = 14 \) cm.
Let the length of the other diagonal be \(d_2\).
The formula for the area of a rhombus is:
Area \( = \frac { 1 }{ 2 } \times d_1 \times d_2 \)
We can rearrange this formula to find the second diagonal:
\( d_2 = \frac { 2 \times \text{Area} }{ d_1 } \)
\( \implies d_2 = \frac { 2 \times 98 }{ 14 } \)
\( \implies d_2 = \frac { 196 }{ 14 } \)
\( \implies d_2 = 14 \) cm.
The length of the other diagonal is 14 cm.
In simple words: To find the missing diagonal of a rhombus, you first double its area, then divide that by the length of the diagonal you already know.

🎯 Exam Tip: When working with area formulas, remember how to rearrange them to find any missing dimension. Double-check your arithmetic, especially with division.

 

Question 8. PQRS is a rhombus.
(i) If it is given that PQ = 3 cm, calculate the perimeter of PQRS.
(ii) If the height of the rhombus is 2.5 cm, calculate its area.
(iii) The diagonals of a rhombus are 8 cm and 6 cm respectively. Find its perimeter.
Answer:
(i) In a rhombus PQRS, all sides are equal.
Given side \(PQ = 3\) cm.
The perimeter of a rhombus \( = 4 \times \text{side} \).
\( \implies \) Perimeter \( = 4 \times 3 = 12 \) cm.
(ii) Given height of the rhombus \( = 2.5 \) cm.
The base of the rhombus is its side, which is \(3\) cm from part (i).
The area of a rhombus \( = \text{base} \times \text{height} \).
\( \implies \) Area \( = 3 \times 2.5 = 7.5 \) cm\(^2\).
(iii) Given the diagonals of the rhombus are 8 cm and 6 cm.
In a rhombus, the diagonals bisect each other at right angles.
Let the diagonals be \(AC = 8\) cm and \(BD = 6\) cm. Let them intersect at O.
Consider the right-angled triangle formed by half of each diagonal and one side (e.g., \( \triangle POQ \)).
\( PO = \frac { 8 }{ 2 } = 4 \) cm and \( QO = \frac { 6 }{ 2 } = 3 \) cm.
Using the Pythagorean Theorem in \( \triangle POQ \):
\( PQ^2 = PO^2 + QO^2 \)
\( \implies PQ^2 = 4^2 + 3^2 \)
\( \implies PQ^2 = 16 + 9 \)
\( \implies PQ^2 = 25 \)
\( \implies PQ = \sqrt{25} = 5 \) cm.
So, the side of the rhombus is 5 cm.
The perimeter of the rhombus \( = 4 \times \text{side} \).
\( \implies \) Perimeter \( = 4 \times 5 = 20 \) cm.
In simple words: (i) A rhombus has four equal sides, so its perimeter is four times the length of one side. (ii) Its area can also be found by multiplying its base (side length) by its perpendicular height. (iii) If you know the diagonals, they cut each other in half at 90 degrees, forming four right-angled triangles. Use Pythagoras to find the side length of the rhombus from these smaller triangles, then calculate the perimeter.

🎯 Exam Tip: Remember the properties of a rhombus: all sides are equal, diagonals bisect each other at right angles, and use the appropriate formulas for perimeter and area based on the given information.

 

Question 9. The sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate,
(i) The length of the other diagonal and
(ii) The area of the rhombus.
Answer:
Let the rhombus be ABCD, with side length \(AB = 5\) cm.
Let one diagonal \(AC = 8\) cm.
(i) In a rhombus, the diagonals bisect each other at right angles. Let them intersect at O.
Consider \( \triangle AOB \). It is a right-angled triangle.
\( AO = \frac { AC }{ 2 } = \frac { 8 }{ 2 } = 4 \) cm.
Using the Pythagorean Theorem in \( \triangle AOB \):
\( AB^2 = AO^2 + BO^2 \)
\( \implies 5^2 = 4^2 + BO^2 \)
\( \implies 25 = 16 + BO^2 \)
\( \implies BO^2 = 25 - 16 \)
\( \implies BO^2 = 9 \)
\( \implies BO = \sqrt{9} = 3 \) cm.
The other diagonal \(BD = 2 \times BO = 2 \times 3 = 6\) cm.
(ii) Now we have both diagonals: \(d_1 = AC = 8\) cm and \(d_2 = BD = 6\) cm.
The area of a rhombus \( = \frac { 1 }{ 2 } \times d_1 \times d_2 \).
\( \implies \) Area \( = \frac { 1 }{ 2 } \times 8 \times 6 \)
\( \implies \) Area \( = 4 \times 6 = 24 \) cm\(^2\).
In simple words: (i) Use the side length and half of the known diagonal in a right-angled triangle formed by the bisecting diagonals. Apply Pythagoras theorem to find half of the other diagonal, then double it. (ii) Once both diagonals are known, multiply them and divide by two to get the area of the rhombus.

🎯 Exam Tip: Remember to halve the diagonal length before applying Pythagoras theorem in the right-angled triangle formed by the bisecting diagonals. This is a common mistake point.

 

Question 10. In the figure, ABCX is a rhombus of side 5 cm. Angles BAD and ADC are right angles. If DC = 8 cm, calculate the area of ABCX.
Answer:
From the given information and figure, we understand the setup as follows:
We have a figure `ABCD` where `AB` is parallel to `DC`. `AB = 5` cm, `DC = 8` cm.
We are told `ABCX` is a rhombus of side 5 cm. This means `AB = BC = CX = XA = 5` cm.
Let's consider the region `ABX` within the larger figure such that `X` lies on `DC`.
Since `ABCX` is a rhombus with side 5 cm, `AB = CX = 5` cm.
Given `DC = 8` cm and `XC = 5` cm, the length `DX` can be calculated as `DC - XC`.
\( \implies DX = 8 - 5 = 3 \) cm.
Now, consider the right-angled triangle \( \triangle ADX \), where `AX` is the perpendicular height from `A` to `DC` (or `DX`).
The side `AD` (which is a side of the rhombus `ABCX`) is 5 cm. `DX` is 3 cm.
Using the Pythagorean Theorem in \( \triangle ADX \):
\( AD^2 = AX^2 + DX^2 \)
\( \implies 5^2 = AX^2 + 3^2 \)
\( \implies 25 = AX^2 + 9 \)
\( \implies AX^2 = 25 - 9 \)
\( \implies AX^2 = 16 \)
\( \implies AX = \sqrt{16} = 4 \) cm.
The height of the rhombus `ABCX` (corresponding to base `XC` or `AB`) is `AX = 4` cm.
The area of a rhombus is given by `base × height`. For rhombus `ABCX`, we can use `XC` as the base and `AX` as the height.
\( \implies \) Area of rhombus `ABCX` \( = XC \times AX = 5 \times 4 = 20 \) cm\(^2\).
In simple words: We first find the length of the segment `DX` by subtracting `XC` from `DC`. Then, using `AD` as the hypotenuse and `DX` as one side in the right-angled triangle `ADX`, we find the height `AX` using Pythagoras. Finally, we calculate the area of the rhombus `ABCX` by multiplying its base (`XC` or `AB`) by this height `AX`.

🎯 Exam Tip: In complex figures, carefully identify the relevant triangles and apply the Pythagorean theorem to find missing perpendicular heights. The area of a rhombus can be found using `base × height` if the perpendicular height is known.