CBSE Class 9 Mathematics Areas Of Parallelogram And Triangle Worksheet Set 01

Question 1. Calculate the area of quadrilateral ABCD.
Answer: From the given figure of quadrilateral ABCD, the diagonal AC divides the quadrilateral into two triangles: triangle ABC and triangle ADC.
Let base \(b_1 = BC = 17\text{ cm}\) and height \(h_1 = 8\text{ cm}\) for triangle ABC.
Let base \(b_2 = DC\) (or using the perpendicular height \(h_2 = 9\text{ cm}\) drawn to diagonal AC).
Area of triangle ABC = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 17 \times 8 = 68\text{ cm}^2\)
Assuming the perpendicular height from vertex D to diagonal AC is 9 cm and the length of diagonal AC (acting as base for both triangles) is 17 cm:
Area of triangle ADC = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 17 \times 9 = 76.5\text{ cm}^2\)
Total Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC = \(68 + 76.5 = 144.5\text{ cm}^2\)
In simple words: Split the quadrilateral into two triangles using the diagonal, find the area of each triangle separately, and add them together.
Exam Tip: Always identify the base and the corresponding perpendicular height clearly before applying the triangle area formula \(\frac{1}{2} \times \text{base} \times \text{height}\).

Question 2. In the given figure ABCD is a trapezium in which AB || DC and its diagonal AC and BD intersect at O. Prove that \(\text{ar}(\Delta AOD) = \text{ar}(\Delta BOC)\).
Answer: We know that triangles on the same base and between the same parallels have equal areas.
Triangle ABC and triangle ABD lie on the same base AB and between the same parallel lines AB and DC.
Therefore, \(\text{ar}(\Delta ABC) = \text{ar}(\Delta ABD)\)
Subtracting \(\text{ar}(\Delta AOB)\) from both sides:
\(\text{ar}(\Delta ABC) - \text{ar}(\Delta AOB) = \text{ar}(\Delta ABD) - \text{ar}(\Delta AOB)\)
\(\text{ar}(\Delta BOC) = \text{ar}(\Delta AOD)\)
Hence proved: \(\text{ar}(\Delta AOD) = \text{ar}(\Delta BOC)\).
In simple words: Two triangles sharing a base between two parallel lines have equal areas. Subtracting the common triangle area from both gives the required result.

Exam Tip: Remember the standard theorem: Triangles on the same base and between the same parallels are equal in area.

Question 3. In the given figure ABCD is a parallelogram. If \(\text{ar}(\Delta BAP) = 10\text{ cm}^2\) and \(\text{ar}(\Delta CPD) = 30\text{ cm}^2\), then find \(\text{ar}(\text{parallelogram } ABCD)\).
Answer: Let P be any point inside or outside the parallelogram. Draw a line through P parallel to the sides AB and DC.
Using the property of triangles and parallelograms on the same base and between the same parallels, if a point P is inside parallelogram ABCD, the sum of areas of triangles formed with opposite sides is equal to half the area of the parallelogram.
Here, \(\text{ar}(\Delta BAP) + \text{ar}(\Delta CPD) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD)\)
Given \(\text{ar}(\Delta BAP) = 10\text{ cm}^2\) and \(\text{ar}(\Delta CPD) = 30\text{ cm}^2\):
\(10 + 30 = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD)\)
\(40 = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD)\)
\(\text{ar}(\text{parallelogram } ABCD) = 40 \times 2 = 80\text{ cm}^2\)
In simple words: The sum of the areas of the two triangles formed by an interior point and opposite sides is always half the area of the parallelogram.

Exam Tip: This is a standard direct result for interior points in parallelograms: \(\text{ar}(\Delta PAB) + \text{ar}(\Delta PCD) = \frac{1}{2} \text{ar}(\text{parallelogram})\).

Question 4. ABCD is a parallelogram. If \(PC : PA = 1 : 3\) and \(\text{ar}(\Delta BPC) = 16\text{ cm}^2\), then find \(\text{ar}(\Delta ADP)\).
Answer: Given \(PC : PA = 1 : 3\). Let \(PC = x\) and \(PA = 3x\).
Consider triangles BPC and BPA having vertices at B and bases along line AC.
The ratio of their areas is equal to the ratio of their bases:
\(\frac{\text{ar}(\Delta BPC)}{\text{ar}(\Delta BPA)} = \frac{PC}{PA} = \frac{1}{3}\)
Given \(\text{ar}(\Delta BPC) = 16\text{ cm}^2\):
\(\frac{16}{\text{ar}(\Delta BPA)} = \frac{1}{3}\)
\(\text{ar}(\Delta BPA) = 16 \times 3 = 48\text{ cm}^2\)
Now, diagonal AC divides the parallelogram ABCD into two triangles of equal area:
\(\text{ar}(\Delta ABC) = \text{ar}(\Delta ADC) = \text{ar}(\Delta BPC) + \text{ar}(\Delta BPA) = 16 + 48 = 64\text{ cm}^2\)
Since \(\text{ar}(\Delta ADC) = 64\text{ cm}^2\) and point P lies on line AC, triangle ADC is composed of triangle ADP and triangle CDP.
Using the ratio of bases along AC where \(PC : PA = 1 : 3\):
\(\frac{\text{ar}(\Delta CDP)}{\text{ar}(\Delta ADP)} = \frac{PC}{PD}\) (or proportional to segments), or using heights from D to line AC:
Since triangles ADC and ABC share vertex D and B with base AC, and triangle APC is split by line DP into triangles with base ratio \(1:3\):
\(\text{ar}(\Delta ADP) = \frac{3}{4} \text{ar}(\Delta ADC) = \frac{3}{4} \times 64 = 48\text{ cm}^2\)
In simple words: Use the property that triangles with the same height have areas proportional to their bases.

Exam Tip: When a vertex is shared, the ratio of triangle areas equals the ratio of their respective bases on a straight line.

Question 5. In the given figure, O is any point on the diagonal BD of parallelogram ABCD. Prove that \(\text{ar}(\Delta AOD) = \text{ar}(\Delta COD)\).
Answer: Consider triangle ABD and triangle CBD. Diagonal BD divides the parallelogram ABCD into two triangles of equal area:
\(\text{ar}(\Delta ABD) = \text{ar}(\Delta CBD)\)
O is a point on diagonal BD. Join AO and CO.
In triangle ABD, AO is a line segment from vertex A to side BD. Similarly, in triangle CBD, CO is a line segment from vertex C to side BD.
Alternatively, consider triangle ABC and triangle ADC, or use triangles with base OD:
Since O lies on diagonal BD, triangle AOD and triangle COD have the same base OD.
Also, AB || DC, and the height from A and C to diagonal BD are equal because ABCD is a parallelogram.
Therefore, \(\text{ar}(\Delta AOD) = \text{ar}(\Delta COD)\).
In simple words: Triangles standing on the same base OD and between the same parallels have equal areas.

Exam Tip: Highlight that the perpendicular distances from opposite vertices A and C to the diagonal BD are equal in a parallelogram.

Question 6. AD is the median of \(\Delta ABC\). If \(\text{ar}(\Delta ABD) = x\text{ cm}^2\) and \(\text{ar}(\Delta ABC) = y\text{ cm}^2\), find the relation between x and y.
Answer: A median of a triangle divides it into two triangles of equal area.
Since AD is the median of \(\Delta ABC\), it divides \(\Delta ABC\) into two triangles: \(\Delta ABD\) and \(\Delta ADC\).
Therefore, \(\text{ar}(\Delta ABD) = \text{ar}(\Delta ADC) = \frac{1}{2} \text{ar}(\Delta ABC)\)
Given \(\text{ar}(\Delta ABD) = x\) and \(\text{ar}(\Delta ABC) = y\):
\(x = \frac{1}{2}y\)
or \(y = 2x\)
In simple words: A median cuts a triangle in half, so the area of triangle ABD is half the total area of triangle ABC.
Exam Tip: Remember the key theorem: A median of a triangle divides the triangle into two triangles of equal areas.

Question 7. In \(\Delta ABC\), D and E are two points that trisect base BC. Show that \(\text{ar}(\Delta ADE) = \frac{1}{3} \text{ar}(\Delta ABC)\).
Answer: Since D and E trisect the base BC, we have:
\(BD = DE = EC = \frac{1}{3} BC\)
Consider triangles ABD, ADE, and AEC. All three triangles share the same vertex A and their bases BD, DE, and EC lie along the same straight line BC.
Since their bases are equal in length (\(BD = DE = EC\)) and they share the same height from vertex A to line BC, their areas are equal:
\(\text{ar}(\Delta ABD) = \text{ar}(\Delta ADE) = \text{ar}(\Delta AEC)\)
The total area of triangle ABC is the sum of these three triangular areas:
\(\text{ar}(\Delta ABC) = \text{ar}(\Delta ABD) + \text{ar}(\Delta ADE) + \text{ar}(\Delta AEC)\)
Since all three areas are equal to \(\text{ar}(\Delta ADE)\):
\(\text{ar}(\Delta ABC) = 3 \times \text{ar}(\Delta ADE)\)
Therefore, \(\text{ar}(\Delta ADE) = \frac{1}{3} \text{ar}(\Delta ABC)\).
In simple words: Trisecting the base into three equal parts divides the triangle into three smaller triangles of equal area.

Exam Tip: State clearly that triangles with equal bases on the same line and a common vertex have equal areas.

Question 8. ABCD is a rhombus whose one angle is \(60^\circ\). Prove that the ratio of the lengths of its diagonals is \(\sqrt{3} : 1\).
Answer: Let ABCD be a rhombus with side length \(a\). Let diagonals AC and BD intersect at O at a right angle (\(90^\circ\)).
Since one angle of the rhombus is \(60^\circ\) (say \(\angle DAB = 60^\circ\)), the diagonal AC bisects angle DAB, so \(\angle OAB = 30^\circ\).
In right-angled triangle AOB (right-angled at O):
\(\cos(30^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AO}{AB} = \frac{AC / 2}{a}\)
Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):
\(\frac{AC / 2}{a} = \frac{\sqrt{3}}{2} \implies AC = a\sqrt{3}\)
Similarly, for angle OAB = \(30^\circ\), \(\sin(30^\circ) = \frac{OB}{AB} = \frac{BD / 2}{a}\)
Since \(\sin(30^\circ) = \frac{1}{2}\):
\(\frac{BD / 2}{a} = \frac{1}{2} \implies BD = a\)
Now, find the ratio of the diagonals AC and BD:
\(\frac{AC}{BD} = \frac{a\sqrt{3}}{a} = \frac{\sqrt{3}}{1}\)
Therefore, the ratio of the lengths of its diagonals is \(\sqrt{3} : 1\).
In simple words: Use trigonometry in the right-angled triangle formed by the intersecting diagonals of the rhombus.

Exam Tip: Recall that diagonals of a rhombus bisect each other at right angles and also bisect the interior angles.

Question 9. Prove that of all the parallelograms of which the sides are given, the rectangle has the greatest area.
Answer: Let the given adjacent sides of the parallelogram be \(a\) and \(b\), and let the included angle between them be \(\theta\).
The area of a parallelogram is given by:
\(\text{Area} = a \times b \times \sin(\theta)\)
For given fixed side lengths \(a\) and \(b\), the area depends entirely on \(\sin(\theta)\).
We know that the maximum value of the sine function for any angle is 1, which occurs when \(\theta = 90^\circ\):
\(\sin(90^\circ) = 1\)
When \(\theta = 90^\circ\), the parallelogram becomes a rectangle.
Therefore, the maximum possible area is \(a \times b\), which is the area of the rectangle.
Hence, of all parallelograms with given sides, the rectangle has the greatest area.
In simple words: Area depends on \(\sin(\theta)\), which reaches its maximum value of 1 when the angle is \(90^\circ\) (making it a rectangle).

Exam Tip: Use the trigonometric formula for area (\(ab\sin\theta\)) and state that \(\sin\theta\) is maximum at \(90^\circ\).

Question 10. Show that diagonal of a parallelogram divide it into 4 triangles of equal area.
Answer: Let ABCD be a parallelogram whose diagonals AC and BD intersect at point O.
We know that the diagonals of a parallelogram bisect each other at point O.
Therefore, AO = OC and BO = OD.
Consider triangle ABC: BO is the median because O is the midpoint of AC. Thus, \(\text{ar}(\Delta AOB) = \text{ar}(\Delta BOC)\).
Similarly, in triangle BCD: CO is the median, so \(\text{ar}(\Delta BOC) = \text{ar}(\Delta COD)\).
In triangle CDA: DO is the median, so \(\text{ar}(\Delta COD) = \text{ar}(\Delta AOD)\).
In triangle DAB: AO is the median, so \(\text{ar}(\Delta AOD) = \text{ar}(\Delta AOB)\).
Combining all these equalities:
\(\text{ar}(\Delta AOB) = \text{ar}(\Delta BOC) = \text{ar}(\Delta COD) = \text{ar}(\Delta AOD)\)
Thus, the diagonals divide the parallelogram into 4 triangles of equal area.
In simple words: Since diagonals bisect each other, each diagonal acts as a median for the triangles formed, splitting the total area into four equal parts.

Exam Tip: Use the property that the diagonals of a parallelogram bisect each other and apply the median theorem four times.

Students must free download and practice these worksheets to gain more marks in exams. CBSE Class 9 Mathematics Worksheet - Areas of Parallelogram and Triangle