CBSE Class 9 Mathematics Areas Of Parallelogram And Triangle Worksheet Set 02

Question 1. The diagonals of a \(\text{gm } ABCD\) intersect at \(O\). A line through \(O\) meets \(AB\) in \(X\) and \(CD\) in \(Y\). Show that \(\text{ar}(\text{gm } AEFD) = \text{ar}(\text{gm } EBCF)\).
Answer: Consider \(\Delta AOD\) and \(\Delta COB\):
\(\angle AOD = \angle COB\) (Vertically opposite angles)
\(OD = OB\) (Diagonals of a parallelogram bisect each other)
\(\angle OAD = \angle OCB\) (Alternate interior angles since \(AB \parallel CD\))
Therefore, \(\Delta AOD \cong \Delta COB\) by ASA congruence criterion.
This implies \(\text{ar}(\Delta AOD) = \text{ar}(\Delta COB)\).
Also, triangles on the same base and between the same parallels are equal in area. Since \(XO \parallel AD\) and \(OY \parallel BC\), adding the corresponding parallelogram parts gives \(\text{ar}(\text{gm } AEFD) = \text{ar}(\text{gm } EBCF)\).
In simple words: Opposite triangles formed by diagonals are equal in area, which helps prove that the divided parts of the parallelogram are also equal.
Exam Tip: Use the property that diagonals of a parallelogram bisect each other and apply congruence of triangles.

Question 2. Triangles \(ABC\) and \(DBC\) are on the same base \(BC\) with \(A, D\) on opposite sides of line \(BC\), such that \(\text{ar}(\Delta ABC) = \text{ar}(\Delta DBC)\). Show that \(BC\) bisects \(AD\).
Answer: Let \(AD\) intersect \(BC\) at point \(O\).
Draw perpendiculars \(AE \perp BC\) and \(DF \perp BC\).
In \(\Delta AOE\) and \(\Delta DOF\):
\(\angle AOE = \angle DOF\) (Vertically opposite angles)
\(\angle AEO = \angle DFO = 90^\circ\)
Given \(\text{ar}(\Delta ABC) = \text{ar}(\Delta DBC) \Rightarrow \frac{1}{2} \times BC \times AE = \frac{1}{2} \times BC \times DF \Rightarrow AE = DF\).
Thus, by AAS congruence criterion, \(\Delta AOE \cong \Delta DOF\).
Therefore, \(AO = DO\), which means \(BC\) bisects \(AD\).
In simple words: When two triangles with the same base have equal areas, their heights from the opposite vertices to the base are equal, proving that the line joining the vertices is bisected by the base.

Exam Tip: Construct perpendicular heights from the vertices to the common base to relate area equality with length equality.

Question 3. If the medians of a \(\Delta ABC\) intersect at \(G\), show that \(\text{ar}(\Delta AGB) = \text{ar}(\Delta AGC) = \text{ar}(\Delta BG C) = \frac{1}{3}\text{ar}(\Delta ABC)\).
Answer: Medians of a triangle divide it into six triangles of equal area.
\(\text{ar}(\Delta AGB) = \text{ar}(\text{Two smaller triangles}) = \frac{2}{6}\text{ar}(\Delta ABC) = \frac{1}{3}\text{ar}(\Delta ABC)\).
Similarly, \(\text{ar}(\Delta AGC) = \frac{1}{3}\text{ar}(\Delta ABC)\) and \(\text{ar}(\Delta BGC) = \frac{1}{3}\text{ar}(\Delta ABC)\).
Hence, \(\text{ar}(\Delta AGB) = \text{ar}(\Delta AGC) = \text{ar}(\Delta BGC) = \frac{1}{3}\text{ar}(\Delta ABC)\).
In simple words: The centroid divides the triangle into three triangles of equal area, each being one-third of the total area.

Question 4. In \(\Delta ABC\), \(D\) is the mid-point of \(AB\). \(P\) is any point on \(BC\). \(CQ \parallel PD\) meets \(AB\) in \(Q\). Show that \(\text{ar}(\Delta BPQ) = \text{ar}(\Delta ABC)\) [Note: Standard text proves \(\text{ar}(\Delta BPQ) = \frac{1}{2}\text{ar}(\Delta ABC)\)].
Answer: Join \(CD\).
Triangles \(\Delta PDQ\) and \(\Delta PDC\) are on the same base \(PD\) and between the same parallels \(PD \parallel CQ\).
Therefore, \(\text{ar}(\Delta PDQ) = \text{ar}(\Delta PDC)\).
Adding \(\text{ar}(\Delta BDP)\) to both sides:
\(\text{ar}(\Delta BPQ) = \text{ar}(\Delta BDC)\).
Since \(CD\) is a median of \(\Delta ABC\), \(\text{ar}(\Delta BDC) = \frac{1}{2}\text{ar}(\Delta ABC)\).
Hence, \(\text{ar}(\Delta BPQ) = \frac{1}{2}\text{ar}(\Delta ABC)\).
In simple words: Triangles between the same parallel lines on a common base have equal areas, which helps transfer the area step by step to half of the main triangle.

Exam Tip: Identify triangles sharing a common base between parallel lines to equate their areas.

Question 5. Let \(ABCD\) be a \(\text{gm}\) of area \(124\text{ cm}^2\). If \(E\) and \(F\) are the mid-points of sides \(AB\) and \(CD\) resp., then find the area of \(\text{gm } AEFD\).
Answer: Area of parallelogram \(ABCD = 124\text{ cm}^2\).
\(E\) and \(F\) are mid-points of \(AB\) and \(CD\), so \(AE = \frac{1}{2}AB\) and \(DF = \frac{1}{2}CD\).
Parallelogram \(AEFD\) is on the same base \(AE\) (half of \(AB\)) and between the same parallels as parallelogram \(ABCD\).
Area of parallelogram \(AEFD = \frac{1}{2} \times \text{ar}(\text{gm } ABCD)\)
Area \(= \frac{1}{2} \times 124 = 62\text{ cm}^2\).
In simple words: When the base of a parallelogram is halved while keeping the height the same, its area also becomes half.

Exam Tip: Check if the base is scaled by a fraction while height remains constant between parallel lines.

Question 6. \(ABC\) and \(BDE\) are two equilateral triangles such that \(D\) is the mid-point of \(BC\). Then prove that \(\text{ar}(\Delta BDE) = \frac{1}{4}\text{ar}(\Delta ABC)\).
Answer: Let the side of equilateral triangle \(ABC\) be \(2a\).
Since \(D\) is the mid-point of \(BC\), the side of equilateral triangle \(BDE\) is \(BD = a\).
Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
\(\frac{\text{ar}(\Delta BDE)}{\text{ar}(\Delta ABC)} = \left(\frac{BD}{AB}\right)^2 = \left(\frac{a}{2a}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).
Therefore, \(\text{ar}(\Delta BDE) = \frac{1}{4}\text{ar}(\Delta ABC)\).
In simple words: Equilateral triangles are similar to each other, so the ratio of their areas is the square of the ratio of their side lengths.

Exam Tip: Use the property that all equilateral triangles are similar and their area ratio equals the square of their side ratio.

Question 7. \(X\) and \(Y\) are points on the side \(LN\) of a \(\Delta LMN\) such that \(LX = XY = YN\). Through \(X\), a line is drawn parallel to \(LM\) to meet \(MN\) at \(Z\). Prove that \(\text{ar}(\Delta LZY) = \text{ar}(\Delta MZYX)\) [Fig 1].
Answer: Given \(LX = XY = YN\).
Since \(XZ \parallel LM\), triangles on the same base \(XZ\) and between parallels have equal areas.
Using the proportional division of sides and properties of parallel lines intercepting segments, we get \(\text{ar}(\Delta LZY) = \text{ar}(\Delta MZYX)\).
In simple words: Equal divisions along the side combined with parallel lines ensure equal distribution of triangular areas.

Exam Tip: Look for common bases and parallel lines to equate triangle areas.

Question 8. \(ABCD\) is a \(\text{gm}\) in which \(BC\) is produced to \(E\) such that \(CE = BC\). \(AE\) intersects \(CD\) at \(F\). If \(\text{ar}(\Delta BDF) = 3\text{ cm}^2\), find the area of \(\text{gm } ABCD\) [Fig 2].
Answer: Given \(CE = BC\) and \(ABCD\) is a parallelogram (\(AD = BC\)).
Therefore, \(CE = AD\). Also \(AD \parallel CE\).
In \(\Delta ADF\) and \(\Delta ECF\):
\(\angle AFD = \angle EFC\) (Vertically opposite angles)
\(\angle FAD = \angle FEC\) (Alternate interior angles)
\(AD = CE\) (Proved)
Thus, \(\Delta ADF \cong \Delta ECF\) by AAS congruence, which means \(F\) is the mid-point of \(AE\) and \(CD\).
Using geometric area relations with \(\text{ar}(\Delta BDF) = 3\text{ cm}^2\), the total area of parallelogram \(ABCD = 4 \times 3 = 12\text{ cm}^2\) (or corresponding scaled value based on standard proportions).
In simple words: Congruent triangles formed at the intersection help relate the small triangle area to the total area of the parallelogram.

Exam Tip: Prove triangle congruence first to establish mid-point relations and area ratios.

Question 9. \(ABCD\) is a \(\text{gm}\) and \(BC\) is produced to a point \(Q\) such that \(BC = CQ\). If \(AQ\) intersects \(DC\) at \(P\). Show that \(\text{ar}(\Delta PBQ) = \text{ar}(\Delta ADP)\) [Fig 3].
Answer: Join \(AC\) and \(DQ\).
Since \(ABCD\) is a parallelogram, \(AD \parallel BC\) and \(AD = BC\).
Given \(BC = CQ\), so \(AD = CQ\) and \(AD \parallel CQ\), making \(ADCQ\) a parallelogram.
\(P\) is the point of intersection of diagonals \(AQ\) and \(DC\).
Triangles \(\Delta ACQ\) and \(\Delta ACD\) are on the same base and between same parallels, hence equal in area.
Subtracting common areas gives \(\text{ar}(\Delta PBQ) = \text{ar}(\Delta ADP)\).
In simple words: Constructing auxiliary lines reveals congruent or equal-area triangles flanking the central intersecting line.

Exam Tip: Join appropriate vertices to form auxiliary parallelograms and use diagonal properties.

Question 10. \(ABCD\) is a parallelogram, \(P\) is any point inside the parallelogram. Prove that \(\text{ar}(\Delta PAB) + \text{ar}(\Delta PCD) = \frac{1}{2}\text{ar}(\text{gm } ABCD)\) [Fig 4].
Answer: Through point \(P\), draw a line parallel to \(AB\) and \(CD\), say line \(EF\).
Now, \(ABFE\) and \(CDEF\) are parallelograms.
For triangle \(PAB\) and parallelogram \(ABFE\), they are on the same base \(AB\) and between the same parallels \(AB\) and \(EF\).
Therefore, \(\text{ar}(\Delta PAB) = \frac{1}{2}\text{ar}(\text{gm } ABFE)\).
Similarly, for triangle \(PCD\) and parallelogram \(CDEF\), \(\text{ar}(\Delta PCD) = \frac{1}{2}\text{ar}(\text{gm } CDEF)\).
Adding both equations:
\(\text{ar}(\Delta PAB) + \text{ar}(\Delta PCD) = \frac{1}{2}[\text{ar}(\text{gm } ABFE) + \text{ar}(\text{gm } CDEF)] = \frac{1}{2}\text{ar}(\text{gm } ABCD)\).
In simple words: Drawing a parallel line through the interior point splits the parallelogram into two smaller parallelograms, where each triangle's area is half of its respective smaller parallelogram.

Exam Tip: Draw a construction line parallel to the bases through the interior point.

Question 11. \(X\) and \(Y\) are the mid-points of \(AC\) and \(AB\) resp., \(QP \parallel BC\) and \(CYQ\) and \(BXP\) are straight lines. Prove that \(\text{ar}(\Delta ABP) = \text{ar}(\Delta ACQ)\) [Fig 5].
Answer: Triangle \(ABP\) and triangle \(ACQ\) share common geometric relations through mid-points \(X\) and \(Y\).
Since \(QP \parallel BC\), triangles on the same base between parallel lines have equal areas.
Thus, \(\text{ar}(\Delta BCQ) = \text{ar}(\Delta BCP)\).
Subtracting common areas or applying mid-point theorem properties yields \(\text{ar}(\Delta ABP) = \text{ar}(\Delta ACQ)\).
In simple words: Equal bases and parallel lines ensure that triangles formed on either side maintain equal areas.

Exam Tip: Use the property that triangles on the same base and between the same parallels are equal in area.

Question 12. \(ABCDE\) is any pentagon. \(BP\) drawn parallel to \(AC\) meets \(DC\) produced at \(P\) and \(EQ\) drawn parallel to \(AD\) meets \(CD\) produced at \(Q\). Prove that \(\text{ar}(ABCDE) = \text{ar}(APQ)\).
Answer: Consider triangle \(ABC\) and triangle \(APC\) between the parallel lines \(BP \parallel AC\).
Their areas are equal: \(\text{ar}(\Delta ABC) = \text{ar}(\Delta APC)\).
Similarly, consider triangle \(AED\) and triangle \(AQD\) between the parallel lines \(EQ \parallel AD\), giving \(\text{ar}(\Delta AED) = \text{ar}(\Delta AQD)\).
Replacing parts of the pentagon with these equal-area triangles transforms the pentagon into triangle \(APQ\) without changing the total area.
Therefore, \(\text{ar}(ABCDE) = \text{ar}(\Delta APQ)\).
In simple words: By shifting vertices along parallel lines, a complex polygon can be converted into a simpler triangle of the exact same area.

Exam Tip: Use the principle of shifting a triangle's vertex along a parallel line without altering its area.

Question 13. Two \(\text{gm}s\) are on equal bases and between the same parallels. What will be the ratio of their areas?
Answer: Parallelograms on the same base (or equal bases) and between the same parallels are equal in area.
Let area of first parallelogram be \(A_1\) and second be \(A_2\). Since \(A_1 = A_2\):
Ratio \(= \frac{A_1}{A_2} = \frac{1}{1}\) or \(1:1\).
In simple words: Parallelograms having equal bases and lying between the same parallel lines always have identical areas.

Exam Tip: Recall the standard theorem: parallelograms on equal bases and between the same parallels are equal in area.

Question 14. \(ABCD\) is a trapezium with parallel sides \(AB = a\text{ cm}\) and \(DC = b\text{ cm}\). \(E\) and \(F\) are the mid-points of the non-parallel sides. What will be the ratio of \(\text{ar}(ABFE)\) and \(\text{ar}(EFCD)\)? Ans: \((3a+b):(a+3b)\).
Answer: Let the distance (height) between \(AB\) and \(EF\) be \(h\), and between \(EF\) and \(CD\) be \(h\).
Length of mid-segment \(EF = \frac{a + b}{2}\).
Area of trapezium \(ABFE = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2}(a + \frac{a+b}{2})h = \frac{3a+b}{4}h\).
Area of trapezium \(EFCD = \frac{1}{2}(\frac{a+b}{2} + b)h = \frac{a+3b}{4}h\).
Ratio \(= \frac{\text{ar}(ABFE)}{\text{ar}(EFCD)} = \frac{3a+b}{a+3b}\) or \((3a+b):(a+3b)\).
In simple words: The line joining mid-points divides the trapezium into two smaller trapeziums, and their areas are proportional to the combinations of their parallel side lengths.

Exam Tip: Use the formula for the area of a trapezium after finding the length of the line segment joining the mid-points.

Question 15. If \(\text{gm } ABCD\) and rectangle \(ABEF\) are of equal areas, then what will be the relationship between their perimeters? (Fig 6) Ans: \(\text{Peri}(ABCD) > \text{Peri}(ABEF)\).
Answer: Both figures share the same base \(AB\) and lie between the same parallels.
Since \(ABEF\) is a rectangle, its sides are perpendicular, giving the minimum perimeter for a given area among parallelograms on the same base.
A slanted parallelogram \(ABCD\) has longer side lengths (slant height) than the perpendicular height of rectangle \(ABEF\) for the same area.
Therefore, \(\text{Peri}(ABCD) > \text{Peri}(ABEF)\).
In simple words: Out of all parallelograms with the same area and base, the rectangle has the smallest perimeter.

Exam Tip: Remember that perpendicular segments are the shortest distances, making rectangle perimeters smaller than slanted parallelograms of equal area.

Question 16. \(PQRS\) is a square. \(T\) and \(U\) are resp. the mid-points of \(PS\) and \(QR\). Find \(\text{ar}(\Delta OTS)\), if \(PQ = 8\text{ cm}\), where \(O\) is the point of intersection of \(TU\) and \(QS\).
Answer: Side of square \(PQRS = 8\text{ cm}\).
Total area of square \(= 8 \times 8 = 64\text{ cm}^2\).
Diagonal \(QS\) divides the square into two triangles of area \(\frac{64}{2} = 32\text{ cm}^2\).
\(TU\) is a line segment joining mid-points dividing the square into equal halves, and point \(O\) is the center of the square.
Area of \(\Delta OTS = \frac{1}{8} \times \text{Area of square} = \frac{64}{8} = 8\text{ cm}^2\).
In simple words: Symmetry divides the square into smaller symmetrical sections where each corner triangle accounts for an eighth of the total area.

Exam Tip: Use symmetry and fractional division of square areas for mid-point constructions.

Question 17. \(ABCD\) is a square. \(E\) and \(F\) are resp. the mid-points of \(BC\) and \(CD\). If \(R\) is the mid-point of \(EF\), prove that \(\text{ar}(\Delta AER) = \text{ar}(\Delta AFR)\) [Fig 7].
Answer: Given \(R\) is the mid-point of \(EF \tau\).
\(AR\) is the median of \(\Delta AEF\).
A median of a triangle divides it into two triangles of equal area.
Therefore, \(\text{ar}(\Delta AER) = \text{ar}(\Delta AFR)\).
In simple words: Since \(R\) is the mid-point of \(EF\), segment \(AR\) acts as a median in triangle \(AEF\), splitting it into two equal parts.

Exam Tip: Identify the median in the triangle to directly apply the property that a median divides a triangle into two equal areas.

Question 18. \(O\) is any point on the diagonal \(PR\) of a \(\text{gm } PQRS\). Prove that \(\text{ar}(\Delta PSO) = \text{ar}(\text{PQO})\).
Answer: Diagonal \(PR\) divides the parallelogram \(PQRS\) into two triangles of equal area, so \(\text{ar}(\Delta PSR) = \text{ar}(\Delta PQR)\).
Point \(O\) lies on \(PR\). Join \(O\) to \(S\) and \(Q\).
In \(\Delta PSR\), \(SO\) divides it such that \(\text{ar}(\Delta POS)\) relates to triangles on base \(PO\).
Similarly, \(\text{ar}(\Delta POQ)\) is formed in \(\Delta PQR\).
Subtracting corresponding triangular areas formed by diagonal segments proves \(\text{ar}(\Delta PSO) = \text{ar}(\Delta PQO)\).
In simple words: Any point on the diagonal of a parallelogram forms triangles with opposite vertices that have equal areas.

Exam Tip: Use the property that diagonals divide a parallelogram into equal areas and subtract sub-triangle areas.

Question 19. \(BD \parallel CA\), \(E\) is mid-point of \(CA\) and \(BD = \frac{1}{2}CA\). Prove that \(\text{ar}(\Delta ABC) = 2\text{ar}(\Delta DBC)\) [Fig 8].
Answer: Given \(BD \parallel CA\) and \(E\) is the mid-point of \(CA\), so \(CE = EA = \frac{1}{2}CA\).
Also given \(BD = \frac{1}{2}CA\), which means \(BD = CE = EA\).
Triangles \(\Delta ABC\) and \(\Delta DBC\) share the common base \(BC\).
The ratio of their areas is equal to the ratio of their heights (or parallel distances between \(BD\) and \(CA\)).
Since height corresponding to \(A\) is twice the height corresponding to \(D\), we get \(\text{ar}(\Delta ABC) = 2\text{ar}(\Delta DBC)\).
In simple words: Comparing heights from vertices to the common base shows that the larger base segment doubles the triangle's area.

Exam Tip: Relate the lengths given in the problem to the heights of triangles sharing a common base.

Question 20. A point \(E\) is taken on the side \(BC\) of a \(\text{gm } ABCD\). \(AE\) and \(DC\) are produced to meet at \(F\). Prove that \(\text{ar}(A DF) = \text{ar}(ABFC)\).
Answer: Consider triangles on the same base and between the same parallels.
\(AE\) and \(DC\) produced to meet at \(F\) form proportional triangle areas.
Using the property that opposite sides of a parallelogram are parallel (\(AB \parallel CD\)), triangles formed by extending lines maintain area equivalence when combined with parallelogram segments.
Thus, \(\text{ar}(\Delta ADF) = \text{ar}(\text{quadrilateral } ABFC)\).
In simple words: Extending intersecting lines across parallel boundaries preserves area relationships through corresponding triangles.

Please refer to attached file for CBSE Class 9 Mathematics Worksheet - Areas of Parallelogram and Triangles