GSEB Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2

Question 1. In figure, ABCD is a parallelogram. Given AB = 16 cm, AE = 8 cm (height corresponding to base AB), and CF = 10 cm (height corresponding to base AD). Find AD.
Answer: For parallelogram ABCD, we can find its area in two different ways.
First, using base AB and its corresponding height (given as AE = 8 cm):
Area of parallelogram ABCD \( = AB \times \text{height}_{\text{AB}} = 16 \, \text{cm} \times 8 \, \text{cm} = 128 \, \text{cm}^2 \)...........(1)
Next, using base AD and its corresponding height (given as CF = 10 cm):
Area of parallelogram ABCD \( = AD \times \text{height}_{\text{AD}} = AD \times 10 \, \text{cm} \)...........(2)
Since the area of the parallelogram remains constant, we can set the two expressions for the area equal to each other.
From (1) and (2), we obtain:
\( AD \times 10 = 128 \)
\( AD = \frac {128}{10} \)
\( AD = 12.8 \, \text{cm} \)
Thus, the length of side AD is 12.8 cm.
In simple words: The area of a parallelogram can be calculated by multiplying any base by its corresponding height. Since the total area stays the same, we can use different base-height pairs. Here, we used base AB with height AE, and then base AD with height CF. By setting these two area calculations equal, we can find the unknown side AD.

Exam Tip: Always remember that the area of a parallelogram can be determined using any side as the base and multiplying it by the altitude (perpendicular height) to that specific base.

Question 2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar(EFGH) = \frac {1}{2}ar (ABCD). OR Prove that the area of the quadrilateral formed by joining the mid-points of the sides of a parallelogram is half the area of the parallelogram.
Answer: Given: A parallelogram ABCD, in which E, F, G, and H are the mid-points of sides AB, BC, CD, and DA in order.
To prove: \( ar(EFGH) = \frac {1}{2} ar(ABCD) \).
Construction: Join HF.
Proof: In parallelogram ABCD, we know that AD || BC and AD = BC (opposite sides of a parallelogram).
Therefore, \( \frac {1}{2} AD || \frac {1}{2}BC \) and \( \frac {1}{2} AD = \frac {1}{2}BC \).
This implies that AH || BF and AH = BF, because H is the midpoint of AD and F is the midpoint of BC.
Consequently, ABFH is a parallelogram (since one pair of opposite sides, AH and BF, are parallel and equal).
Furthermore, \( \triangle EHF \) and parallelogram ABFH share the same base FH and are situated between the same parallel lines AB and FH.
\( \implies ar(\triangle EHF) = \frac {1}{2} ar(||gm ABFH) \)...........(1)
In a similar way, we can prove that FCDH is also a parallelogram, and \( \triangle FGH \) shares the same base FH and is between parallels CD and FH.
\( \implies ar(\triangle FGH) = \frac {1}{2} ar(||gm FCDH) \)...........(2)
Adding equations (1) and (2), we obtain:
\( ar(\triangle EHF) + ar(\triangle FGH) = \frac {1}{2} ar(||gm ABFH) + \frac {1}{2} ar(||gm FCDH) \)
\( \implies ar(EFGH) = \frac {1}{2} [ar(ABFH) + ar(FCDH)] \)
Since parallelogram ABFH and parallelogram FCDH together form the entire parallelogram ABCD,
\( \implies ar(EFGH) = \frac {1}{2} ar(ABCD). \)
In simple words: When you connect the middle points of all sides of a parallelogram, you create a new smaller parallelogram inside. The area of this inside parallelogram is exactly half the area of the original, larger one. This is shown by dividing the main parallelogram into two smaller parallelograms and relating the areas of the internal triangles to them.

Exam Tip: Remember the midpoint theorem for parallelograms and the relationship between the area of a triangle and a parallelogram when they share the same base and are between the same parallel lines.

Question 3. P and Q are any two points lying on the side DC and AD respectively of a parallelogram ABCD. Show that ar(ΔAPB) = ar(ΔBQC).
Answer: Given: ABCD is a parallelogram. P is a point on side DC, and Q is a point on side AD.
To prove: \( ar(\triangle APB) = ar(\triangle BQC) \).
Proof: Triangle APB and parallelogram ABCD share the common base AB and are situated between the same parallel lines AB and DC.
According to the theorem, the area of a triangle is half the area of a parallelogram if they are on the same base and between the same parallels.
Therefore, \( ar(\triangle APB) = \frac {1}{2} ar(||gm ABCD) \)...........(1)
Similarly, triangle BQC and parallelogram ABCD share the common base BC and are situated between the same parallel lines BC and AD.
Thus, \( ar(\triangle BQC) = \frac {1}{2} ar(||gm ABCD) \)...........(2)
From equations (1) and (2), we can conclude that:
\( ar(\triangle APB) = ar(\triangle BQC) \).
In simple words: If a triangle and a parallelogram share the same bottom line and are between the same parallel lines, the triangle's area is half of the parallelogram's area. Since both triangles APB and BQC each have an area equal to half of the parallelogram ABCD, their areas must be equal to each other.

Exam Tip: This question tests a key theorem about areas of triangles and parallelograms sharing a base and parallels. State the theorem clearly in your explanation.

Question 4. In figure, P is a point in the interior of a parallelogram ABCD. Show that: 1. ar(ΔAPB) + ar(ΔPCD) = \frac {1}{2} ar(||gm ABCD) 2. ar(ΔAPD) + ar(ΔPBC) = ar(ΔAPB) + ar(ΔPCD) [Hint: Through P, draw a line parallel to AB]
Answer: Given: P is a point inside parallelogram ABCD.
To prove:
1. \( ar(\triangle APB) + ar(\triangle PCD) = \frac {1}{2} ar(||gm ABCD) \)
2. \( ar(\triangle APD) + ar(\triangle PBC) = ar(\triangle APB) + ar(\triangle PCD) \).
Construction: Through point P, draw a line EF parallel to AB, where E lies on AD and F lies on BC.
Proof:
1. Since EF is parallel to AB (by construction), and AB is parallel to DC (opposite sides of parallelogram ABCD), it follows that EF || AB || DC.
Considering quadrilateral AEFB: EF || AB and AE || BF (since AD || BC). Thus, AEFB is a parallelogram.
Similarly, considering quadrilateral EFCD: EF || DC and ED || FC. Thus, EFCD is a parallelogram.
Now, triangle APB and parallelogram AEFB share the common base AB and are situated between the same parallel lines AB and EF.
Therefore, \( ar(\triangle APB) = \frac {1}{2} ar(||gm AEFB) \)........(3)
Also, triangle PCD and parallelogram EFCD share the common base DC and are situated between the same parallel lines DC and EF.
Therefore, \( ar(\triangle PCD) = \frac {1}{2} ar(||gm EFCD) \)........(4)
Adding equations (3) and (4), we obtain:
\( ar(\triangle APB) + ar(\triangle PCD) = \frac {1}{2} ar(||gm AEFB) + \frac {1}{2} ar(||gm EFCD) \)
\( \implies ar(\triangle APB) + ar(\triangle PCD) = \frac {1}{2} [ar(||gm AEFB) + ar(||gm EFCD)] \)
Since the two smaller parallelograms AEFB and EFCD combine to form the larger parallelogram ABCD,
\( \implies ar(\triangle APB) + ar(\triangle PCD) = \frac {1}{2} ar(||gm ABCD) \)...........(5)
2. Alternative approach for the second part: Through point P, draw a line LM parallel to AD, where L lies on AB and M lies on DC.
Using a similar argument as above, we can demonstrate that:
\( ar(\triangle APD) + ar(\triangle PBC) = \frac {1}{2} ar(||gm ABCD) \)...........(6)
From equations (5) and (6), we can conclude that:
\( ar(\triangle APD) + ar(\triangle PBC) = ar(\triangle APB) + ar(\triangle PCD) \).
In simple words: For a point inside a parallelogram, the sum of areas of two opposite triangles (like APB and PCD) is half the total parallelogram area. Similarly, the sum of areas of the other two opposite triangles (APD and PBC) is also half the parallelogram's area. This means the sum of areas of one pair of opposite triangles equals the sum of areas of the other pair.

Exam Tip: Remember that drawing an auxiliary line through the interior point parallel to one pair of sides is crucial for solving this problem by breaking the main parallelogram into two smaller parallelograms.

Question 5. In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that: 1. ar(||gm PQRS) = ar(||gm ABRS) 2. ar(ΔAXS) = \frac {1}{2} ar(||gm PQRS)
Answer: Given: PQRS and ABRS are parallelograms. X is any point on side BR.
To prove:
1. \( ar(||gm PQRS) = ar(||gm ABRS) \)
2. \( ar(\triangle AXS) = \frac {1}{2} ar(||gm PQRS) \).
Proof:
1. Parallelograms PQRS and ABRS share the common base RS and are situated between the same parallel lines PB and SR.
According to the theorem, two parallelograms on the same base and between the same parallels have equal areas.
Therefore, \( ar(||gm PQRS) = ar(||gm ABRS) \)...........(1)
2. Triangle AXS and parallelogram ABRS share the common base AS and are situated between the same parallel lines AS and BR.
According to the theorem, the area of a triangle is half the area of a parallelogram if they are on the same base and between the same parallels.
Therefore, \( ar(\triangle AXS) = \frac {1}{2} ar(||gm ABRS) \).
From equation (1), we have \( ar(||gm ABRS) = ar(||gm PQRS) \).
Substituting this into the previous equation, we get:
\( ar(\triangle AXS) = \frac {1}{2} ar(||gm PQRS) \).
In simple words: Two parallelograms that stand on the same base and are between the same parallel lines have the same area. Also, a triangle on the same base as a parallelogram and between the same parallel lines has half the area of the parallelogram. So, triangle AXS has half the area of parallelogram ABRS, which in turn has the same area as parallelogram PQRS.

Exam Tip: This problem applies two fundamental theorems about areas of parallelograms and triangles. Clearly state these theorems in your answer.

Question 6. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Answer: The field is divided into three parts: \( \triangle APQ, \triangle APS, \) and \( \triangle AQR \). All of these sections are triangles.
\( \triangle APQ \) and the parallelogram PQRS are situated between the same parallel lines PQ and RS. They also share a common base PQ.
Therefore, the area of \( \triangle APQ = \frac {1}{2} \) the area of parallelogram PQRS.
The remaining two triangles are \( \triangle APS \) and \( \triangle AQR \). The sum of their areas, \( ar(\triangle APS) + ar(\triangle AQR) \), makes up the other half of the parallelogram's area.
To prove this:
The heights of \( \triangle APQ, \triangle APS, \) and \( \triangle AQR \) with respect to their bases (PQ for APQ, and AS and AR for APS and AQR respectively) are the same, let's call it x, because they all lie between the same parallel lines PQ and RS.
Area of \( \triangle APQ = \frac {1}{2} \times \text{base} \times \text{height} = \frac {(PQ)(x)}{2} \)...........(1)
Now, consider the sum of the areas of \( \triangle APS \) and \( \triangle AQR \). Their bases AS and AR are segments on the line RS. The common height for these triangles (from P to RS and Q to RS) is x.
\( ar(\triangle APS) + ar(\triangle AQR) = \frac {1}{2} \times AS \times x + \frac {1}{2} \times AR \times x \)
\( = \frac {1}{2} \times (AS + AR) \times x \)
Since A is a point on the line segment RS, the sum of lengths \( AS + AR \) equals the total length of the side RS.
\( = \frac {(RS)(x)}{2} \)
Since SR = PQ (opposite sides of the parallelogram are equal), we can substitute PQ for SR.
Therefore, \( ar(\triangle APS) + ar(\triangle AQR) = \frac {(PQ)(x)}{2} \)...........(2)
From equations (1) and (2), we observe that \( ar(\triangle APQ) = ar(\triangle APS) + ar(\triangle AQR) \).
The farmer wants to sow wheat and pulses in equal portions. Since the area of \( \triangle APQ \) is equal to the combined area of \( \triangle APS \) and \( \triangle AQR \), the farmer should plant wheat in \( \triangle APQ \) and pulses in the other two triangles \( \triangle APS \) and \( \triangle AQR \), or vice-versa, to ensure equal portions.
In simple words: The field is cut into three triangles: APQ, APS, and AQR. The area of the triangle formed by connecting P and Q to point A on the opposite side (APQ) is exactly half the total area of the parallelogram. The sum of the areas of the other two triangles (APS and AQR) is also half the parallelogram's area. So, the farmer can plant wheat in one half (APQ) and pulses in the other half (APS + AQR), or vice-versa, to divide the crops equally.

Exam Tip: Remember that a triangle on the same base as a parallelogram and between the same parallels has half the area of the parallelogram. This principle helps determine how to divide the field equally.