CBSE Class 9 Mathematics Areas Of Parallelogram And Triangle Worksheet Set 03

Question 1. In the fig. ABCD, DCFE & ABFE are parallelograms. Show that ar(ΔADE) = ar(ΔBCF) [Hint: prove ΔADE ≅ ΔBCF, by SSS rule]
Answer: We compare the corresponding sides of \( \Delta\text{ADE} \) and \( \Delta\text{BCF} \):
1. Since ABCD is a parallelogram, its opposite sides are equal in length:
\( AD = BC \)
2. Since DCFE is a parallelogram, its opposite sides are equal in length:
\( DE = CF \)
3. Since ABFE is a parallelogram, its opposite sides are equal in length:
\( AE = BF \)
By applying the Side-Side-Side (SSS) congruence criterion:
\( \Delta\text{ADE} \cong \Delta\text{BCF} \)
Because congruent geometric figures enclose identical regions, their areas must be equal:
\( \text{ar}(\Delta\text{ADE}) = \text{ar}(\Delta\text{BCF}) \).
Hence proved.
In simple words: All three pairs of sides in the two triangles are opposite sides of parallelograms, which makes them equal. Because the triangles are identical in every way, their areas are the same.

Exam Tip: When asked to prove equality of areas, checking whether the two figures can be proven congruent by SSS or SAS is often the fastest method.

Question 2. In the fig. ABCD is a trapezium with AB || DC. M is the midpoint of BC. Prove that ar(ΔAMD) = 1/2 ar(trap. ABCD) [Hint: Const: Through M draw EMF || AD so that AEFD is a ||gm. prove ΔBME ≅ ΔCMF (by AAS rule) prove ar(trap ABCD) = ar(||gm AEFD) but ar(ΔAMD) = 1/2 ar(||gm AEFD)]
Answer: Construction: Through point M, draw a straight line segment EMF parallel to AD, meeting AB extended at point E and the base DC at point F. This forms quadrilateral AEFD.
Since \( AE \parallel DF \) (as \( AB \parallel DC \)) and \( AD \parallel EF \) by construction, quadrilateral AEFD is a parallelogram.
Proof:
In \( \Delta\text{BME} \) and \( \Delta\text{CMF} \):
1. \( BM = CM \) (since M is given as the midpoint of BC).
2. \( \angle BME = \angle CMF \) (vertically opposite angles).
3. \( \angle MBE = \angle MCF \) (alternate interior angles, since \( AE \parallel DF \) with transversal BC).
By the AAS congruence criterion:
\( \Delta\text{BME} \cong \Delta\text{CMF} \)

\( \implies \text{ar}(\Delta\text{BME}) = \text{ar}(\Delta\text{CMF}) \quad \text{--- (1)} \)
Now, the area of trapezium ABCD can be related to parallelogram AEFD:
\( \text{ar}(\text{trap. } ABCD) = \text{ar}(\text{quad. } AMFD) - \text{ar}(\Delta\text{CMF}) + \text{ar}(\Delta\text{BME}) \)
Using equation (1), the two triangular areas cancel out:
\( \text{ar}(\text{trap. } ABCD) = \text{ar}(\text{parallelogram } AEFD) \quad \text{--- (2)} \)
Next, observe that \( \Delta\text{AMD} \) and parallelogram AEFD share the common base AD and lie between the same parallel lines AD and EF.
Therefore, the area of the triangle is half the area of the parallelogram:
\( \text{ar}(\Delta\text{AMD}) = \frac{1}{2} \text{ar}(\text{parallelogram } AEFD) \)
Substituting from equation (2):
\( \text{ar}(\Delta\text{AMD}) = \frac{1}{2} \text{ar}(\text{trap. } ABCD) \).
Hence proved.
In simple words: Extending the top side turns the trapezium into a parallelogram of equal area because the added corner triangle exactly matches the missing one. The inside triangle shares a base with this parallelogram, so it takes up half the area.

Exam Tip: Be sure to write the construction step explicitly; examiners award a dedicated mark for introducing the parallel line EMF.

Question 3. In the fig. ABFE is a trapezium with AB || EF. C & F are points on the sides AE & BF such that CD || AB. Prove that ar(ΔADE) = ar(ΔBCF) [Hint: prove ar(ΔCDA) = ar(ΔCDB) --- ① ar(ΔCDE) = ar(ΔCDF) --- ② adding ① + ②]
Answer: We are given that \( AB \parallel EF \) and \( CD \parallel AB \), which also means \( CD \parallel EF \).
1. Consider \( \Delta\text{CDA} \) and \( \Delta\text{CDB} \):
Both triangles stand on the identical base CD and lie between the same parallel lines AB and CD.
Therefore, their areas are equal:
\( \text{ar}(\Delta\text{CDA}) = \text{ar}(\Delta\text{CDB}) \quad \text{--- (1)} \)
2. Consider \( \Delta\text{CDE} \) and \( \Delta\text{CDF} \):
Both triangles stand on the identical base CD and lie between the same parallel lines CD and EF.
Therefore, their areas are equal:
\( \text{ar}(\Delta\text{CDE}) = \text{ar}(\Delta\text{CDF}) \quad \text{--- (2)} \)
Adding equation (1) and equation (2):
\( \text{ar}(\Delta\text{CDA}) + \text{ar}(\Delta\text{CDE}) = \text{ar}(\Delta\text{CDB}) + \text{ar}(\Delta\text{CDF}) \)
From the diagram, the combination of \( \Delta\text{CDA} \) and \( \Delta\text{CDE} \) forms \( \Delta\text{ADE} \), while the combination of \( \Delta\text{CDB} \) and \( \Delta\text{CDF} \) forms \( \Delta\text{BCF} \):
\( \text{ar}(\Delta\text{ADE}) = \text{ar}(\Delta\text{BCF}) \).
Hence proved.
In simple words: The top triangles share base CD between the upper parallel lines, and the bottom triangles share base CD between the lower parallel lines. Adding both pairs gives the complete areas of the two large side triangles.

Exam Tip: State the common base and the specific pair of parallel lines clearly for each equation to earn complete justification marks.

Question 4. In the fig. E is any point on the diagonal AC of ||gm ABCD. Prove that ar(ΔABE) = ar(ΔADE) [Hint: In ΔABD, AF is the median => ar(ΔABF) = ar(ΔADF) --- ① again, In ΔBED, EF is the median => ar(ΔBEF) = ar(ΔDEF) --- ② adding ① + ②]
Answer: Construction: Draw diagonal BD intersecting diagonal AC at point F.
In a parallelogram, the diagonals bisect each other. Thus, F is the midpoint of segment BD.
Proof:
1. In \( \Delta\text{ABD} \), segment AF connects vertex A to the midpoint F of the opposite side BD. Therefore, AF is a median.
Since a median divides a triangle into two triangles of equal area:
\( \text{ar}(\Delta\text{ABF}) = \text{ar}(\Delta\text{ADF}) \quad \text{--- (1)} \)
2. In \( \Delta\text{BED} \), segment EF connects vertex E to the midpoint F of base BD. Therefore, EF is a median.
Thus:
\( \text{ar}(\Delta\text{BEF}) = \text{ar}(\Delta\text{DEF}) \quad \text{--- (2)} \)
Adding equation (1) and equation (2):
\( \text{ar}(\Delta\text{ABF}) + \text{ar}(\Delta\text{BEF}) = \text{ar}(\Delta\text{ADF}) + \text{ar}(\Delta\text{DEF}) \)

\( \implies \text{ar}(\Delta\text{ABE}) = \text{ar}(\Delta\text{ADE}) \).
(Note: If point E is chosen between A and F, subtracting equation (2) from equation (1) yields the same equality).
Hence proved.
In simple words: The diagonals cut each other in half, making the diagonal parts medians of the triangles. Since a median splits a triangle into two pieces of equal area, adding or subtracting these pieces gives equal areas on both sides.

Exam Tip: Always cite the property that "a median of a triangle divides it into two triangles of equal area" as your primary justification.

Question 5. A point T is taken on the side PQ of ||gm PQRS such that ST and RQ produced to meet at A. Prove that ΔASQ and ΔATR are equal in area. [Hint: prove ar(ΔTQS) = ar(ΔTQR) add ar(ΔTQA) on both sides]
Answer: In parallelogram PQRS, side PQ is parallel to side SR (\( PQ \parallel SR \)).
1. Consider \( \Delta\text{TQS} \) and \( \Delta\text{TQR} \):
Both triangles stand on the identical base TQ and lie between the same parallel lines PQ and SR.
Therefore, their areas are equal:
\( \text{ar}(\Delta\text{TQS}) = \text{ar}(\Delta\text{TQR}) \quad \text{--- (1)} \)
2. Now, add \( \text{ar}(\Delta\text{TQA}) \) to both sides of equation (1):
\( \text{ar}(\Delta\text{TQS}) + \text{ar}(\Delta\text{TQA}) = \text{ar}(\Delta\text{TQR}) + \text{ar}(\Delta\text{TQA}) \)
Observe the combined shapes from the figure:
- Since S, T, and A are collinear points, \( \Delta\text{TQS} \) and \( \Delta\text{TQA} \) join along side TQ to make \( \Delta\text{ASQ} \):
\( \text{ar}(\Delta\text{TQS}) + \text{ar}(\Delta\text{TQA}) = \text{ar}(\Delta\text{ASQ}) \)
- Since R, Q, and A are collinear points, \( \Delta\text{TQR} \) and \( \Delta\text{TQA} \) join along side TQ to make \( \Delta\text{ATR} \):
\( \text{ar}(\Delta\text{TQR}) + \text{ar}(\Delta\text{TQA}) = \text{ar}(\Delta\text{ATR}) \)
Consequently:
\( \text{ar}(\Delta\text{ASQ}) = \text{ar}(\Delta\text{ATR}) \).
Hence proved.
In simple words: The two inner triangles share base TQ between parallel lines, making their areas match. Adding the common bottom triangle to both creates the two large triangles, so their total areas are also equal.

Exam Tip: Mentioning that points S, T, A are collinear and points R, Q, A are collinear confirms why the two areas cleanly add up to form the larger triangles.

Question 6. In the fig. ABCD and AEFD are two parallelograms. Prove that ar(ΔPEA) = ar(ΔQFD) [Hint: prove ar(||gm ADFE) = ar(||gm ADQP) sub. ar(ADFP) from both sides]
Answer: In quadrilateral PQDA:
1. \( AP \parallel DQ \) (because AP lies along side AB and DQ lies along side CD of parallelogram ABCD, where \( AB \parallel CD \)).
2. \( PQ \parallel AD \) (because PQ lies along side EF of parallelogram AEFD, where \( EF \parallel AD \)).
Because both pairs of opposite sides are parallel, quadrilateral PQDA is a parallelogram.
Now compare parallelogram PQDA and parallelogram AEFD:
Both parallelograms share the common base AD and lie between the same parallel lines AD and line EQ.
Therefore, their areas are equal:
\( \text{ar}(\text{parallelogram } AEFD) = \text{ar}(\text{parallelogram } PQDA) \quad \text{--- (1)} \)
Notice that quadrilateral ADFP is a common region inside both parallelograms.
Subtracting \( \text{ar}(\text{quadrilateral } ADFP) \) from both sides of equation (1):
\( \text{ar}(\text{parallelogram } AEFD) - \text{ar}(\text{quadrilateral } ADFP) = \text{ar}(\text{parallelogram } PQDA) - \text{ar}(\text{quadrilateral } ADFP) \)
From the figure:
- Removing ADFP from parallelogram AEFD leaves \( \Delta\text{PEA} \).
- Removing ADFP from parallelogram PQDA leaves \( \Delta\text{QFD} \).
Therefore:
\( \text{ar}(\Delta\text{PEA}) = \text{ar}(\Delta\text{QFD}) \).
Hence proved.
In simple words: Both large parallelograms share the same base and lie between the same parallel lines, giving them identical total areas. Taking away the shared overlapping shape in the middle leaves the two small end triangles equal in area.

Exam Tip: First proving that PQDA is a parallelogram on the same base AD is the key intermediate step that examiners award marks for.

Question 7. (i) The median of a triangle divides the triangle into two
(a) congruent triangles
(b) isosceles triangles
(c) right triangles
(d) triangles of equal areas
Answer: (d) triangles of equal areas
A median connects a vertex of a triangle to the midpoint of the opposite side. The two resulting triangles share equal base lengths and possess the same perpendicular altitude from the common vertex. Consequently, the median divides the triangle into two triangles of equal area.
In simple words: The median splits the bottom edge into two equal halves while keeping the same height, so both resulting triangles have the exact same area.

Exam Tip: Remember that a median guarantees equal areas, but the two triangles are not necessarily congruent unless the original triangle is isosceles or equilateral.

Question 7. (ii) If a triangle and a parallelogram are on the same base and between the same parallels then the ratio of the area of the triangle to the area of the parallelogram is
(a) 1 : 4
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer: (b) 1 : 2
When a triangle and a parallelogram share the same base \( b \) and lie between the same parallel lines, they have the same perpendicular height \( h \).
The area of the triangle is \( \frac{1}{2}bh \) and the area of the parallelogram is \( bh \).
Therefore, the ratio of their areas is:
\( \frac{\text{Area of triangle}}{\text{Area of parallelogram}} = \frac{\frac{1}{2}bh}{bh} = \frac{1}{2} = 1 : 2 \).
In simple words: A triangle's area formula includes a half, while a parallelogram's formula does not, so the triangle takes up exactly half the space of the parallelogram.

Exam Tip: Be careful with the order in the question: "triangle to parallelogram" is 1 : 2, whereas "parallelogram to triangle" would be 2 : 1.

Question 7. (iii) AD is the median of ΔABC and P is a point of AC such that ar(ΔADP) : ar(ΔABD) = 2 : 3 then ar(ΔPDC) : ar(ΔABC) is
(a) 2 : 5
(b) 1 : 5
(c) 1 : 6
(d) 3 : 5
Answer: (c) 1 : 6
Since AD is the median of \( \Delta\text{ABC} \), it divides the triangle into two triangles of equal area:
\( \text{ar}(\Delta\text{ABD}) = \text{ar}(\Delta\text{ADC}) = \frac{1}{2}\text{ar}(\Delta\text{ABC}) \)
Let \( \text{ar}(\Delta\text{ABD}) = 3x \).
Then \( \text{ar}(\Delta\text{ADC}) = 3x \) and \( \text{ar}(\Delta\text{ABC}) = 6x \).
We are given that \( \text{ar}(\Delta\text{ADP}) : \text{ar}(\Delta\text{ABD}) = 2 : 3 \), which means:
\( \text{ar}(\Delta\text{ADP}) = 2x \)
Because point P lies on AC, \( \Delta\text{ADC} \) is composed of \( \Delta\text{ADP} \) and \( \Delta\text{PDC} \):
\( \text{ar}(\Delta\text{PDC}) = \text{ar}(\Delta\text{ADC}) - \text{ar}(\Delta\text{ADP}) = 3x - 2x = x \)
Now, find the ratio of \( \text{ar}(\Delta\text{PDC}) \) to \( \text{ar}(\Delta\text{ABC}) \):
\( \frac{\text{ar}(\Delta\text{PDC})}{\text{ar}(\Delta\text{ABC})} = \frac{x}{6x} = \frac{1}{6} = 1 : 6 \).
In simple words: The median splits the entire triangle into two halves of 3 units each, making the whole triangle 6 units. If one part of the right half is 2 units, the leftover part is 1 unit, giving a ratio of 1 to 6.

Exam Tip: Assign a variable like \( 3x \) to the median triangle to simplify ratio subtraction without working with cumbersome fractions.