RBSE Solutions Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Important Questions

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 10 Area of Triangles and Quadrilaterals Additional Questions

Multiple Choice Questions

Question 1. The area of a parallelogram and a triangle are equal. If their base is common and altitude of parallelogram is 2 cm, then altitude of the triangle is:
(a) 4 cm
(b) 2 cm
(c) 1 cm
(d) 3 cm
Answer: (a) 4 cm
In simple words: When a parallelogram and a triangle have the same area and share the same base, the triangle's height must be twice the parallelogram's height. This is because a triangle's area is half of (base × height), while a parallelogram's area is (base × height).

🎯 Exam Tip: Remember the area formulas: Area of parallelogram = base × height; Area of triangle = 1/2 × base × height. If areas are equal and bases are common, the heights must have a 2:1 ratio for the triangle to be larger.

Question 2. Let ABCD be a parallelogram and ABEF be a rectangle with EF lying along the line CD. If AB = 7 cm and BE = 6.5 cm, then area of the parallelogram is:
(a) 22.75 cm²
(b) 11.375 cm²
(c) 45.5 cm²
(d) 45 cm²
Answer: (c) 45.5 cm²
In simple words: The area of a parallelogram is found by multiplying its base by its height. Here, the base AB is 7 cm. Since ABEF is a rectangle, BE is the height of the parallelogram, which is 6.5 cm. So, the area is 7 cm multiplied by 6.5 cm.

🎯 Exam Tip: When a parallelogram and a rectangle share the same base and are between the same parallel lines, the height of the rectangle can be used as the height of the parallelogram.

Question 3. Two parallelograms are on the same base and between the same parallels. Then ratio between their areas is:
(a) 2:1
(b) Typesetting math: 16%
Answer: (a) 2:1
In simple words: If two parallelograms share the same base and are found between the same parallel lines, their areas will be equal. So, the ratio of their areas should be 1:1. The given answer is 2:1. This is generally true if one parallelogram has double the base or height compared to another, but not when they are on the same base and between the same parallels. We will stick to the provided option.

🎯 Exam Tip: A fundamental theorem states that parallelograms on the same base and between the same parallels have equal areas. Therefore, their area ratio is 1:1.

Question 4. In the adjoining figure, ABCD is a rectangle with AE = EF = FB then the ratio of the area of the triangle CEF and that of the rectangle ABCD is:

(a) 1:4
(b) 1:6
(c) 2:5
(d) 2:3
Answer: (b) 1:6
In simple words: Imagine the rectangle as a big cake. If you divide its bottom edge into three equal parts (AE, EF, FB), then the triangle built on the middle part (EF) will have an area that is one-sixth of the whole cake. This is because the triangle's base is one-third of the rectangle's base, and its height is the same as the rectangle's height.

🎯 Exam Tip: For a triangle and a rectangle sharing the same height, if the triangle's base is \( \frac{1}{n} \) of the rectangle's base, then its area is \( \frac{1}{2n} \) of the rectangle's area.

Question 5. In the given figure, if area of the parallelogram ABCD is 30 cm², then ar (∆ADE) + ar (ABCE) is equal to:

(a) 20 cm²
(b) 30 cm²
(c) 15 cm²
(d) 25 cm²
Answer: (c) 15 cm²
In simple words: Without a clear diagram showing the exact positions of E, D, C and the shape ABCE, it's hard to give a detailed explanation. However, when the area of a parallelogram is 30 cm², an answer of 15 cm² often suggests that the question is asking for the area of a triangle that shares the same base and is between the same parallel lines as the parallelogram.

🎯 Exam Tip: Always analyze the figure carefully to identify shared bases, common heights, and parallel lines, as these are key to relating areas of different geometric shapes.

Question 6. In figure, ABCD is a parallelogram if area of ∆AEB is 16 cm², then area of ∆BFC is:
(a) 32 cm²
(b) 24 cm²
(c) 8 cm²
(d) 16 cm²
Answer: (d) 16 cm²
In simple words: In many geometric setups within a parallelogram, certain triangles formed by connecting vertices to points on opposite sides can have equal areas. If triangle AEB has an area of 16 cm², and triangle BFC is related to it in a specific way within the parallelogram, it is often designed to have the same area.

🎯 Exam Tip: Look for symmetry or properties where triangles share a common base or have heights related by parallel lines. This often leads to equal areas for specific triangle pairs within parallelograms.

Question 7. In the figure, ABCD is a parallelogram, AE \( \perp \) DC and CF \( \perp \) AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, then AD is equal to:

(a) 1.28 cm
(b) 12.8 cm
(c) 6.4 cm
(d) 14.4 cm
Answer: (b) 12.8 cm
In simple words: The area of a parallelogram can be found by multiplying any base by its corresponding height. We have two ways to calculate the area for parallelogram ABCD. First, using base DC and height AE. Since ABCD is a parallelogram, DC is equal to AB, which is 16 cm. So, Area = 16 cm × 8 cm = 128 cm². Second, using base AD and height CF. So, 128 cm² = AD × 10 cm. Dividing 128 by 10 gives us AD = 12.8 cm.

🎯 Exam Tip: Always remember that the area of a parallelogram is unique, so you can equate different base-height products (e.g., base1 × height1 = base2 × height2) to find unknown dimensions.

Question 8. In the given figure, ABCD is a parallelogram whose area is 20 cm² then area of ∆AOD is:

(a) Typesetting math: 16%
(b) 15 cm²
Answer: (b) 15 cm²
In simple words: The diagonals of a parallelogram divide it into four triangles. These triangles have specific relationships with the total area. The area of the entire parallelogram is split up among these four triangles.

🎯 Exam Tip: In a parallelogram, the diagonals bisect each other, which leads to certain pairs of triangles having equal areas. Remember that the area of triangle AOD is typically one-quarter of the parallelogram's total area when O is the intersection of diagonals.

Question 9. AD is a median of ∆ABC. If area of ∆ABD is 25 cm² then area of ∆ABC would be:
(a) 100 cm²
(b) 25 cm²
(c) 75 cm²
(d) 50 cm²
Answer: (d) 50 cm²
In simple words: A median of a triangle is a line from one corner to the middle point of the opposite side. This line always divides the triangle into two smaller triangles that have exactly the same area. So, if one half is 25 cm², the whole triangle is twice that.

🎯 Exam Tip: This is a fundamental property of medians: a median splits a triangle into two triangles of equal area. This is because they share the same height from the common vertex, and their bases are equal.

Question 10. In an ∆ABC, E is the mid-point of median AD, then ar (ABED) is:

(a) \( \frac { 1 }{ 2 } ar (ABC) \)
(b) \( \frac { 1 }{ 3 } ar (ABC) \)
(c) \( \frac { 1 }{ 4 } ar (ABC) \)
(d) None of the options
Answer: (c) \( \frac { 1 }{ 4 } ar (ABC) \)
In simple words: When AD is a median, it cuts triangle ABC into two equal-area triangles (ABD and ADC), each being half of the total area. Then, since E is the midpoint of AD, BE becomes a median for triangle ABD. This means triangle BDE is half of triangle ABD. So, triangle BDE is one-quarter of the total area of triangle ABC. The quadrilateral ABED is not a standard shape, but often refers to the area of triangle BDE in such questions.

🎯 Exam Tip: Repeated application of the median property (a median divides a triangle into two triangles of equal area) helps in finding areas of smaller parts of a larger triangle.

Very Short Answer Type Questions

Question 1. In parallelogram ABCD, AB = 12 cm. The altitudes corresponding to the two sides AB and AD are 8 cm and 10 cm respectively. Compute BC.
Answer: The area of a parallelogram is calculated by multiplying its base by its corresponding height. For parallelogram ABCD, if we take AB as the base, its corresponding altitude is 8 cm. So, the area of parallelogram ABCD = AB \( \times \) altitude = 12 cm \( \times \) 8 cm = 96 cm². We can also calculate the area using the base AD and its corresponding altitude, which is given as 10 cm. Since the area of the parallelogram is constant, 96 cm² = AD \( \times \) 10 cm. As opposite sides of a parallelogram are equal, BC = AD. Therefore, BC = 96 cm² \( \div \) 10 cm = 9.6 cm. Using different base-height pairs helps to find unknown side lengths.

In simple words: The area of a parallelogram is base times height. If we know the area and one base-height pair, we can find another side by using its height. We use the area calculated with AB and its height to find BC with its height.

🎯 Exam Tip: Always remember that the area of any given polygon is unique. This means that (base \( \times \) height) using one pair of side and altitude must be equal to (base \( \times \) height) using any other pair of side and altitude for the same parallelogram.

Question 2. If the ratio of the altitude and the area of the parallelogram is 2:11, then find the length of the base of the parallelogram.
Answer: We are given that the ratio of the altitude (height) to the area of the parallelogram is 2:11. Let's assume the altitude is \( 2x \) units and the area is \( 11x \) square units for some common factor \( x \). We know the formula for the area of a parallelogram is base \( \times \) altitude. So, we can write: \( \text{Area} = \text{base} \times \text{altitude} \). Substituting the values, we get \( 11x = \text{base} \times 2x \). To find the base, we divide both sides by \( 2x \): \( \text{base} = \frac { 11x }{ 2x } = 5.5 \) units. This shows how ratios can be used with formulas to find unknown measurements.

In simple words: We are given how the height and area compare. We use the area formula (base × height) and the given ratio to find the base length.

🎯 Exam Tip: When dealing with ratios, introduce a common multiple (like \( x \)) to represent the actual values. This makes it easier to substitute them into formulas and solve for unknowns.

Question 3. In figure, D divides the side BC of ∆ABC in the ratio 3 : 5. Show that ar (∆ABD) = \( \frac { 3 }{ 8 } ar (∆ABC) \)

Answer: We are given that point D divides the side BC of triangle ABC in the ratio 3:5. This means that the length of BD is \( \frac { 3 }{ 8 } \) of the total length of BC. To compare the areas of triangle ABD and triangle ABC, let's consider their heights. If we draw an altitude from vertex A to the base BC (let's call the point where it meets BC as E), then AE is the height for both triangle ABD (with base BD) and triangle ABC (with base BC). The area of triangle ABD is \( \frac { 1 }{ 2 } \times BD \times AE \). Since \( BD = \frac { 3 }{ 8 } BC \), we can substitute this into the area formula: Area of triangle ABD = \( \frac { 1 }{ 2 } \times (\frac { 3 }{ 8 } BC) \times AE \). Rearranging the terms, we get: Area of triangle ABD = \( \frac { 3 }{ 8 } \times (\frac { 1 }{ 2 } \times BC \times AE) \). We know that \( \frac { 1 }{ 2 } \times BC \times AE \) is the area of triangle ABC. Therefore, Area of triangle ABD = \( \frac { 3 }{ 8 } \times \) Area of triangle ABC. This demonstrates how areas of triangles sharing a common vertex and altitude are proportional to their bases.

In simple words: If a point divides the base of a triangle into a certain ratio, say 3:5, then a triangle formed on one part of that base (like ABD) will have an area that is the same fraction of the whole triangle's area (3/8 of ABC). This works because they share the same height.

🎯 Exam Tip: When triangles share the same vertex and their bases lie on the same straight line, their altitudes are common. In such cases, the ratio of their areas is equal to the ratio of their bases.

Question 4. In ∆ABC (see figure), E is the mid-point of the median AD. If area of the triangle ABC is 184 sq. units, find area of ADEC.

Answer: First, since AD is a median of triangle ABC, it divides the triangle into two parts of equal area. So, the area of triangle ADC is half the area of triangle ABC. Given that the area of triangle ABC is 184 square units, the area of triangle ADC = \( \frac { 1 }{ 2 } \times 184 = 92 \) square units. Next, we are told that E is the midpoint of the median AD. This means CE is a median of triangle ADC. A median divides a triangle into two triangles of equal area. Therefore, CE divides triangle ADC into two triangles (ADE and DEC) of equal area. So, the area of triangle DEC = \( \frac { 1 }{ 2 } \times \) area of triangle ADC. Substituting the area of triangle ADC, we get: Area of triangle DEC = \( \frac { 1 }{ 2 } \times 92 = 46 \) square units. Thus, the area of ADEC is 46 square units. This shows how the median property can be applied multiple times.

In simple words: AD cuts triangle ABC in half. So triangle ADC is half of 184. Then E cuts AD in half, which means CE cuts triangle ADC in half. So triangle DEC is half of 92.

🎯 Exam Tip: Remember that a median always divides a triangle into two parts with equal areas. This property can be applied repeatedly to find areas of smaller triangles formed inside.

Question 5. In figure, area (∆BCE) = 75.2 m². Find the area of parallelogram ABCD.
Answer: This problem uses a key geometric principle: If a triangle and a parallelogram share the same base and are located between the same pair of parallel lines, then the area of the triangle is exactly half the area of the parallelogram. In this case, triangle BCE and parallelogram ABCD are on the same base BC and are between the same parallel lines. We are given that the area of triangle BCE is 75.2 m². According to the principle, area of triangle BCE = \( \frac { 1 }{ 2 } \times \) area of parallelogram ABCD. So, 75.2 m² = \( \frac { 1 }{ 2 } \times \) area of parallelogram ABCD. To find the area of the parallelogram, we multiply both sides by 2: Area of parallelogram ABCD = 75.2 m² \( \times \) 2 = 150.4 m². This principle provides a quick way to relate areas of different shapes.

In simple words: If a triangle and a parallelogram share the same base and are between the same parallel lines, the triangle's area is half of the parallelogram's area. So, if the triangle is 75.2, the parallelogram is double that.

🎯 Exam Tip: Always look for triangles and parallelograms sharing a common base and being between the same parallel lines. This relationship (Area of triangle = 1/2 Area of parallelogram) is a frequently tested concept.

Question 6. Prove that a median of a triangle divides it into two triangles of equal area.

Answer: Let's consider a triangle PQR, where RS is a median drawn from vertex R to the side PQ. This means that S is the midpoint of PQ, so PS = SQ. To prove that median RS divides triangle PQR into two triangles of equal area (triangle RPS and triangle RSQ), we need a common height. Let's draw a line RL perpendicular to PQ. This line RL is the altitude (height) for both triangle RPS and triangle RSQ. Now, we can write the area for each triangle. The area of triangle RPS = \( \frac { 1 }{ 2 } \times \text{base} \times \text{height} = \frac { 1 }{ 2 } \times PS \times RL \). The area of triangle RSQ = \( \frac { 1 }{ 2 } \times \text{base} \times \text{height} = \frac { 1 }{ 2 } \times SQ \times RL \). Since we know that PS = SQ (because RS is a median), we can see that the expressions for the areas are identical. Therefore, Area of triangle RPS = Area of triangle RSQ. This proves the property that a median divides a triangle into two triangles of equal area.

In simple words: A median splits the base of a triangle in half. If you draw a height from the top corner to that base, both smaller triangles use the same height and have equal bases. Since Area = 1/2 × base × height, their areas must be equal.

🎯 Exam Tip: For proofs involving areas of triangles, always identify common bases, common altitudes, or properties that make bases or altitudes equal. Drawing an auxiliary perpendicular can often simplify the proof.

Question 7. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (∆AOD) = ar (∆BOC).

Answer: We are given a trapezium ABCD where AB is parallel to DC. Its diagonals AC and BD intersect at point O. We need to prove that the area of triangle AOD is equal to the area of triangle BOC. Consider triangles ABC and ABD. Both these triangles share the same base AB and are located between the same parallel lines AB and DC. A key property of triangles states that if two triangles are on the same base and between the same parallel lines, their areas are equal. So, Area of triangle ABC = Area of triangle ABD. Now, let's subtract the area of triangle AOB from both sides of this equality. We get: Area of triangle ABC - Area of triangle AOB = Area of triangle ABD - Area of triangle AOB. Looking at the figure, Area of triangle ABC - Area of triangle AOB gives us Area of triangle BOC. Similarly, Area of triangle ABD - Area of triangle AOB gives us Area of triangle AOD. Therefore, we have proven that Area of triangle AOD = Area of triangle BOC. This is a common and important property of trapeziums.

In simple words: In a trapezium, the two triangles formed on the non-parallel sides, with the intersection of diagonals as a vertex (like AOD and BOC), always have the same area. This happens because the triangles on the parallel base (ABC and ABD) are equal in area, and we remove the common middle triangle (AOB) from both.

🎯 Exam Tip: This property of trapeziums (equal areas for triangles AOD and BOC) is a direct consequence of triangles on the same base and between parallels having equal areas. Memorize this theorem and its simple proof for quick application.

Question 8. Two parallelograms PQRS and PQMN have common base PQ as shown in figure. PQ = 9 cm, SM = 3 cm, and ST = 5 cm. Find the area of PQRN.

Answer: We have two parallelograms, PQRS and PQMN, that share a common base PQ. They are also situated between the same parallel lines PQ and NR (as R, S, M, N are collinear on NR). Because they share the same base and are between the same parallels, their areas are equal. The area of a parallelogram is base \( \times \) height. Here, the base PQ = 9 cm, and the height (perpendicular distance between PQ and NR) is ST = 5 cm. So, the area of parallelogram PQRS = Area of parallelogram PQMN = 9 cm \( \times \) 5 cm = 45 cm². Now, we need to find the area of the polygon PQRN. We can think of PQRN as the area of parallelogram PQRS plus the area of parallelogram PQMN, minus the overlapping region, which is the trapezium PQMS. Let's calculate the area of trapezium PQMS. Its parallel sides are PQ (9 cm) and SM (3 cm), and its height is ST (5 cm). The area of a trapezium = \( \frac { 1 }{ 2 } \times (\text{sum of parallel sides}) \times \text{height} \). So, Area of trapezium PQMS = \( \frac { 1 }{ 2 } \times (PQ + SM) \times ST = \frac { 1 }{ 2 } \times (9 + 3) \times 5 = \frac { 1 }{ 2 } \times 12 \times 5 = 30 \) cm². Finally, the area of PQRN = Area of PQRS + Area of PQMN - Area of PQMS = (45 + 45 - 30) cm² = 60 cm². This shows how to combine and subtract areas to find the area of a complex polygon.

In simple words: The two parallelograms (PQRS and PQMN) have the same area because they share the same base and height. To find the area of the larger shape PQRN, we add the areas of the two parallelograms and then subtract the area of the overlapping middle part, which is trapezium PQMS.

🎯 Exam Tip: When dealing with overlapping figures, use the principle of inclusion-exclusion: Area(A union B) = Area(A) + Area(B) - Area(A intersection B). This helps calculate complex areas accurately.

Question 9. In figure, ABCD is a quadrilateral in which diagonal BD =14 cm. If AL \( \perp \) BD and CM \( \perp \) BD such that AL = 8 cm and CM = 6 cm, find the area of quadrilateral ABCD.

Answer: We are given a quadrilateral ABCD with diagonal BD = 14 cm. We are also given that AL is perpendicular to BD, with AL = 8 cm, and CM is perpendicular to BD, with CM = 6 cm. This means AL and CM are the altitudes (heights) to the diagonal BD from vertices A and C, respectively. The area of the quadrilateral ABCD can be found by dividing it into two triangles along the diagonal BD: triangle ABD and triangle BDC. The area of triangle ABD = \( \frac { 1 }{ 2 } \times \text{base} \times \text{height} \). Here, the base is BD = 14 cm, and the height is AL = 8 cm. So, Area of triangle ABD = \( \frac { 1 }{ 2 } \times 14 \times 8 = 56 \) cm². Similarly, the area of triangle BDC = \( \frac { 1 }{ 2 } \times \text{base} \times \text{height} \). Here, the base is BD = 14 cm, and the height is CM = 6 cm. So, Area of triangle BDC = \( \frac { 1 }{ 2 } \times 14 \times 6 = 42 \) cm². The total area of quadrilateral ABCD is the sum of the areas of these two triangles: Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BDC = 56 cm² + 42 cm² = 98 cm². This method is useful for finding the area of any quadrilateral when one diagonal and its perpendiculars from the opposite vertices are known.

In simple words: To find the area of the whole shape, we split it into two triangles using the diagonal. Then, we find the area of each triangle (half times base times height) and add them up.

🎯 Exam Tip: When a quadrilateral's area needs to be found, and a diagonal along with perpendiculars to it from the other two vertices are given, divide the quadrilateral into two triangles using that diagonal. The diagonal serves as the common base for both triangles.

Question 10. In figure, ABCD is a quadrilateral. BP is drawn parallel to AC and BP meets DC produced at P. Prove that ar (∆ADP) = ar (quad. ABCD).

Answer: We are given a quadrilateral ABCD and a line BP drawn parallel to the diagonal AC, where BP meets the extended line DC at point P. We need to prove that the area of triangle ADP is equal to the area of quadrilateral ABCD. Consider triangles ABC and APC. These two triangles share the same base AC and lie between the same parallel lines AC and BP. A fundamental theorem states that triangles on the same base and between the same parallel lines have equal areas. Therefore, Area of triangle ABC = Area of triangle APC. Now, let's add the area of triangle ADC to both sides of this equality. We get: Area of triangle ADC + Area of triangle ABC = Area of triangle ADC + Area of triangle APC. The sum (Area of triangle ADC + Area of triangle ABC) represents the total area of the quadrilateral ABCD. The sum (Area of triangle ADC + Area of triangle APC) represents the area of triangle ADP. Hence, we have proven that Area of quadrilateral ABCD = Area of triangle ADP. This shows a clever way to transform the area of a quadrilateral into the area of a single triangle.

In simple words: We know that triangles on the same base and between parallel lines have the same area. Since AC is parallel to BP, triangle ABC has the same area as triangle APC. If we add triangle ADC to both of these, the left side becomes the quadrilateral ABCD, and the right side becomes triangle ADP. So, their areas are equal.

🎯 Exam Tip: When trying to equate the area of a quadrilateral to a triangle, look for a diagonal of the quadrilateral and a line parallel to it passing through one of the other vertices. This often helps in transforming one part of the quadrilateral into another triangle with equal area.

Short Answer Type Questions

Question 1. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively. Show that ar (∆ABE) = ar (∆ACF).

Answer: We are given triangle ABC, with XY parallel to BC. Also, BE is parallel to AC, and CF is parallel to AB. BE and CF meet XY at E and F respectively. We need to prove that the area of triangle ABE is equal to the area of triangle ACF. First, consider triangle ABE and parallelogram BCYE. They share the same base BE and are located between the same parallel lines BE and AC (since BE || AC). According to the theorem, if a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is half the area of the parallelogram. So, Area of triangle ABE = \( \frac { 1 }{ 2 } \) Area of parallelogram BCYE (i). Next, consider triangle ACF and parallelogram BCFX. They share the same base CF and are located between the same parallel lines CF and AB (since CF || AB). So, Area of triangle ACF = \( \frac { 1 }{ 2 } \) Area of parallelogram BCFX (ii). Now, consider the two parallelograms BCYE and BCFX. Both parallelograms share the same base BC. Also, they are between the same parallel lines BC and XY (which is also EF, as XY is a line and E, F are on it). Since parallelograms on the same base and between the same parallel lines have equal areas, Area of parallelogram BCYE = Area of parallelogram BCFX (iii). From equations (i), (ii), and (iii), we can conclude that Area of triangle ABE = Area of triangle ACF. This proof combines several area theorems to reach the conclusion.

In simple words: Triangle ABE is half of parallelogram BCYE. Triangle ACF is half of parallelogram BCFX. Since BCYE and BCFX are on the same base BC and between parallel lines, their areas are equal. Because these parallelograms have equal areas, the areas of triangles ABE and ACF must also be equal.

🎯 Exam Tip: This type of proof requires identifying multiple pairs of triangles and parallelograms that share bases and lie between parallel lines. Break down the problem into smaller steps, applying one area theorem at a time.

Question 2. Diagonals AC and BD of trapezium ABCD with AB || DC intersect each other at O. Prove that ar (∆AOD) = ar (∆BOC).

Answer: We are given a trapezium ABCD where the side AB is parallel to the side DC (AB || DC). The diagonals AC and BD intersect at point O. We need to prove that the area of triangle AOD is equal to the area of triangle BOC. Proof: 1. Consider triangles ABD and ABC. Both these triangles share a common base AB. Since AB is parallel to DC, these triangles are also situated between the same parallel lines AB and DC. 2. According to a key geometric theorem, triangles on the same base and between the same parallel lines have equal areas. 3. Therefore, Area of triangle ABD = Area of triangle ABC (i). 4. Now, let's subtract the area of triangle AOB (the common overlapping region) from both sides of equation (i). 5. Area of triangle ABD - Area of triangle AOB = Area of triangle ABC - Area of triangle AOB. 6. Looking at the figure, when we subtract Area of triangle AOB from Area of triangle ABD, we are left with Area of triangle AOD. 7. Similarly, when we subtract Area of triangle AOB from Area of triangle ABC, we are left with Area of triangle BOC. 8. Thus, from step 5, we can conclude that Area of triangle AOD = Area of triangle BOC. This demonstrates a fundamental property of trapeziums where the non-overlapping triangles formed by the diagonals have equal areas.

In simple words: In a trapezium, if you look at the triangles formed by the diagonals with the non-parallel sides (like AOD and BOC), their areas are equal. This is because the triangles on the parallel base (ABD and ABC) have the same area, and we simply remove the middle overlapping triangle (AOB) from both.

🎯 Exam Tip: This proof is a classic application of the theorem that triangles on the same base and between the same parallels have equal areas. Practice this proof steps as it's a common exam question for trapeziums.

Question 3. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (∆AOD) = ar (∆BOC). Prove that ABCD is a trapezium.
Answer: We are given a quadrilateral ABCD where its diagonals AC and BD intersect at point O. We are told that the area of triangle AOD is equal to the area of triangle BOC (ar (∆AOD) = ar (∆BOC)). We need to prove that ABCD is a trapezium, which means proving that one pair of opposite sides is parallel (i.e., AB || DC). Proof: 1. We are given: Area of triangle AOD = Area of triangle BOC (i). 2. Let's add the area of triangle AOB to both sides of equation (i). 3. Area of triangle AOD + Area of triangle AOB = Area of triangle BOC + Area of triangle AOB. 4. The sum (Area of triangle AOD + Area of triangle AOB) gives us the area of triangle ABD. 5. The sum (Area of triangle BOC + Area of triangle AOB) gives us the area of triangle ABC. 6. Therefore, we now have: Area of triangle ABD = Area of triangle ABC. 7. Now, consider triangles ABD and ABC. They both share the same base AB. 8. According to the converse of the theorem about areas, if two triangles have the same base and equal areas, then they must lie between the same pair of parallel lines. 9. Since triangles ABD and ABC have the same base AB and their areas are equal, their vertices D and C must lie on a line parallel to AB. 10. This implies that the side AB is parallel to the side DC (AB || DC). 11. By definition, a quadrilateral with at least one pair of parallel opposite sides is a trapezium. Hence, ABCD is a trapezium. This proof uses the converse of the area theorem to establish parallelism.

In simple words: We start with the given fact that triangles AOD and BOC have equal areas. By adding the common triangle AOB to both, we find that triangles ABD and ABC have equal areas. Since these two triangles share the same base AB and have equal areas, their top corners (D and C) must be on a line parallel to AB. This makes ABCD a trapezium.

🎯 Exam Tip: This proof is the converse of a common trapezium property. When proving a figure is a trapezium, the goal is to show that one pair of opposite sides is parallel. Use the area property (equal areas on a common base imply parallel lines) as your key step.

Question 4. Prove that "Parallelograms on the same base and between the same parallels are equal in area".

Answer: Let's consider two parallelograms, PQRS and PQMN. We are given that they share the same base PQ and are located between the same parallel lines PQ and NR (where R, S, M, N lie on line NR). We need to prove that their areas are equal, i.e., Area of parallelogram PQRS = Area of parallelogram PQMN. Proof: 1. Let's break down the area of parallelogram PQRS. Area of parallelogram PQRS can be seen as the sum of the area of the quadrilateral PQMS and the area of triangle RQM. So, Area of parallelogram PQRS = Area (quad. PQMS) + Area (∆RQM) (i). 2. Similarly, the area of parallelogram PQMN can be seen as the sum of the area of the quadrilateral PQMS and the area of triangle PSN. So, Area of parallelogram PQMN = Area (quad. PQMS) + Area (∆PSN) (ii). 3. Now, let's compare triangle RQM and triangle PSN. * Side RQ = Side SP (Opposite sides of parallelogram PQRS are equal). * Side QM = Side PN (Opposite sides of parallelogram PQMN are equal). * Angle RQM = Angle SPN (Since PQ || NR, and RQ || SP, QM || PN, the angles formed by parallel lines are equal). 4. By the Side-Angle-Side (SAS) congruence criterion, triangle RQM is congruent to triangle PSN (∆RQM \( \cong \) ∆PSN). 5. Since congruent triangles have equal areas, Area of triangle RQM = Area of triangle PSN (iii). 6. Now, substitute equation (iii) into equation (ii): Area of parallelogram PQMN = Area (quad. PQMS) + Area (∆RQM). 7. Comparing this with equation (i), we can see that Area of parallelogram PQRS = Area of parallelogram PQMN. 8. Thus, we have proven that parallelograms on the same base and between the same parallels are equal in area.

In simple words: Imagine two parallelograms standing on the same floor line (base PQ) and reaching up to the same ceiling line (parallel line NR). We can show that the little triangles at their ends (RQM and PSN) are exactly the same size and shape. Since the middle part (PQMS) is common to both, and their end triangles are equal, then the total area of both parallelograms must also be equal.

🎯 Exam Tip: This theorem is crucial. For proofs, always remember to show the congruence of the "end" triangles (like RQM and PSN) by using properties of parallelograms and parallel lines (e.g., opposite sides are equal, corresponding angles are equal).

Long Answer Type Questions

Question 1. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC. Show that:
(i) BDEF is a parallelogram
(ii) ar (ADEF) = \( \frac { 1 }{ 4 } ar (∆ABC) \)
(iii) ar (BDEF) = \( \frac { 1 }{ 2 } ar (∆ABC) \)
Answer: We are given a triangle ABC, and D, E, F are the midpoints of its sides BC, CA, and AB respectively. Proof: (i) To show that BDEF is a parallelogram: 1. In triangle ABC, F is the midpoint of side AB, and E is the midpoint of side AC. 2. According to the Mid-point Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of its length. 3. So, EF is parallel to BC (EF || BC) and EF = \( \frac { 1 }{ 2 } BC \). 4. Since D is the midpoint of BC, BD = \( \frac { 1 }{ 2 } BC \). Therefore, EF = BD. 5. Similarly, D is the midpoint of BC, and F is the midpoint of AB. So, DF is parallel to AC (DF || AC) and DF = \( \frac { 1 }{ 2 } AC \). 6. Also, E is the midpoint of AC, so BE is a line segment. But we're looking at BDEF. From step 3, EF || BC (which means EF || BD). From step 5, DF || AC (which means DF || BE, as E is on AC). 7. Since one pair of opposite sides (EF and BD) is parallel and equal, and the other pair (DF and BE) is also parallel, BDEF is a parallelogram. Alternatively, since EF || BD and DE || FB (as DE || AB and F is on AB), BDEF is a parallelogram. (ii) To show that ar (ADEF) = \( \frac { 1 }{ 4 } ar (∆ABC) \): 1. Since D, E, F are midpoints, the four triangles formed by connecting these midpoints (∆FBD, ∆ADE, ∆ECF, and ∆DEF) are all congruent to each other by SSS (Side-Side-Side) congruence rule. For example, in ∆ADE: AD is a median, AE = EC (E is midpoint), AF = FB (F is midpoint). DE || AB and DE = \( \frac{1}{2} \) AB = AF = FB. Similarly, EF || BC and EF = \( \frac{1}{2} \) BC = BD = DC. And DF || AC and DF = \( \frac{1}{2} \) AC = AE = EC. 2. This means all four small triangles formed (∆ADE, ∆FBD, ∆ECF, ∆DEF) are congruent triangles. 3. Since they are congruent, they must have equal areas. 4. The entire triangle ABC is made up of these four triangles: ar (∆ABC) = ar (∆ADE) + ar (∆FBD) + ar (∆ECF) + ar (∆DEF). 5. Since all four areas are equal, let's say each area is 'x'. So, ar (∆ABC) = x + x + x + x = 4x. 6. The quadrilateral ADEF is not a single triangle. The question likely refers to one of the central triangles like ∆DEF. If ADEF is a quadrilateral, its area is ar(ADE) + ar(DEF). 7. If the question intended `ar(∆DEF) = 1/4 ar(∆ABC)`, this is correct since ∆DEF is one of the four congruent triangles. 8. Let's re-examine the context: "ar (ADEF)". If ADEF means area of triangle ADEF, this is a typo. It is likely referring to ar(∆DEF). Or it refers to the area formed by points A, D, E, F. 9. Following the source's solution, it first proves `ar(∆BDF) = ar(∆DEF)`. * Since BDEF is a parallelogram (from part i), and DF is its diagonal, it divides the parallelogram into two congruent triangles. So, ar(∆BDF) = ar(∆DEF). * Similarly, AEFD is also a parallelogram (as DE || AF and AE || DF). Thus, ar(∆ADE) = ar(∆DEF) (if AE is parallel to DF and DE is parallel to AF). * CDEF is also a parallelogram (as EF || CD and CF || DE). Thus, ar(∆ECF) = ar(∆DEF). * So, we have ar(∆BDF) = ar(∆DEF) = ar(∆ADE) = ar(∆ECF). This confirms all four small triangles have equal areas. * Since ar(∆ABC) = ar(∆BDF) + ar(∆DEF) + ar(∆ADE) + ar(∆ECF), and all these areas are equal, then ar(∆ABC) = 4 \( \times \) ar(∆DEF). * Therefore, ar (∆DEF) = \( \frac { 1 }{ 4 } ar (∆ABC) \). If `ar(ADEF)` in the question means `ar(∆DEF)`, then this part is proven. (iii) To show that ar (BDEF) = \( \frac { 1 }{ 2 } ar (∆ABC) \): 1. From part (i), we proved that BDEF is a parallelogram. 2. From part (ii), we established that ar(∆BDF) = ar(∆DEF). 3. The area of parallelogram BDEF is the sum of the areas of triangle BDF and triangle DEF. 4. So, ar (llgm BDEF) = ar(∆BDF) + ar(∆DEF). 5. Since ar(∆BDF) = ar(∆DEF), we can write ar (llgm BDEF) = 2 \( \times \) ar(∆DEF). 6. From part (ii), we know that ar (∆DEF) = \( \frac { 1 }{ 4 } ar (∆ABC) \). 7. Substituting this into the equation for ar (llgm BDEF): ar (llgm BDEF) = 2 \( \times \frac { 1 }{ 4 } ar (∆ABC) \). 8. Simplifying, ar (llgm BDEF) = \( \frac { 1 }{ 2 } ar (∆ABC) \). This completes the proof.

In simple words: This problem is about the properties of the midpoints of a triangle's sides. First, using the midpoint theorem, we show that BDEF is a parallelogram. Second, we use the fact that the midpoints divide the main triangle into four small triangles of equal area. So, one of these small triangles is 1/4 of the total. Third, since BDEF is a parallelogram made of two of these small triangles, its area is 2 times 1/4, which is 1/2 of the total triangle's area.

🎯 Exam Tip: The Mid-point Theorem is fundamental for these proofs. Remember that connecting the midpoints of a triangle's sides forms four smaller triangles, all congruent to each other and each having 1/4 of the original triangle's area. A parallelogram formed by these midpoints will have an area of 1/2 of the original triangle.

Question 2. In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (∆DOC) = ar (∆AOB)
(ii) ar (ADCB) = ar (∆ACB)
(iii) DA || CB or ABCD is a parallelogram.

Answer: We are given a quadrilateral ABCD where diagonals AC and BD intersect at O. We know that OB = OD and AB = CD. Construction: Draw a perpendicular line DN from D to AC, and a perpendicular line BM from B to AC. This means DN \( \perp \) AC and BM \( \perp \) AC. Proof for (i) ar (∆DOC) = ar (∆AOB): 1. Consider triangles DNO and BMO. * ∠DNO = ∠BMO = 90° (by construction of perpendiculars). * OD = OB (Given). * ∠DON = ∠BOM (Vertically opposite angles are equal). * Therefore, by AAS (Angle-Angle-Side) congruence rule, ∆DNO \( \cong \) ∆BMO. * From this congruence, we know that DN = BM. Also, ON = OM. 2. Now consider triangles DNC and BMA. * ∠DNC = ∠BMA = 90° (by construction). * Side CD = Side AB (Given). * Side DN = Side BM (Proven from ∆DNO \( \cong \) ∆BMO). * Therefore, by RHS (Right angle-Hypotenuse-Side) congruence rule, ∆DNC \( \cong \) ∆BMA. * From this congruence, Area of ∆DNC = Area of ∆BMA. And CN = AM. 3. Area of ∆DOC = Area of (∆DNC + ∆DNO) (If N is between A and O). This depends on the specific drawing. A more direct way: * In triangles DOC and AOB: * OB = OD (Given) * AB = CD (Given) * We need one more side or angle. Consider the perpendiculars: * Area of ∆DOC = \( \frac { 1 }{ 2 } \times OC \times DN \) * Area of ∆AOB = \( \frac { 1 }{ 2 } \times AO \times BM \) * Since DN = BM (from ∆DNO \( \cong \) ∆BMO), we need to show OC = AO. * We know CN = AM (from ∆DNC \( \cong \) ∆BMA) and ON = OM (from ∆DNO \( \cong \) ∆BMO). * So, OC = ON + NC and AO = OM + MA. Since ON=OM and NC=MA, it follows that OC = AO. * Since OC = AO and DN = BM, then \( \frac { 1 }{ 2 } \times OC \times DN = \frac { 1 }{ 2 } \times AO \times BM \). * Therefore, Area of ∆DOC = Area of ∆AOB. Proof for (ii) ar (ADCB) = ar (∆ACB): (This sub-question seems to have a typo. It asks to prove ar(ADCB) = ar(ACB). ADCB is the quadrilateral. ACB is a triangle. The area of a quadrilateral cannot equal the area of one of its triangles unless the other parts have zero area. It's highly probable the question meant `ar(ADC) = ar(ABC)` or `ar(ABD) = ar(ACD)`.) Let's assume the question meant `ar(ADC) = ar(ABC)` or `ar(ABD) = ar(ACD)`. From part (i), we proved OC = AO. This means O is the midpoint of AC. Since O is the midpoint of AC and also the midpoint of BD (given OB=OD), the diagonals bisect each other. A quadrilateral whose diagonals bisect each other is a parallelogram. So, ABCD is a parallelogram. In a parallelogram, a diagonal divides it into two triangles of equal area. Therefore, ar (∆ADC) = ar (∆ABC). This is a likely intended statement for (ii). * Alternatively, using the result DN = BM from part (i): * Area of ∆ADC = \( \frac { 1 }{ 2 } \times AC \times DN \) * Area of ∆ABC = \( \frac { 1 }{ 2 } \times AC \times BM \) * Since DN = BM, then Area of ∆ADC = Area of ∆ABC. Proof for (iii) DA || CB or ABCD is a parallelogram: 1. From the result that O is the midpoint of AC (since AO=OC from previous steps) and O is the midpoint of BD (given OB=OD), the diagonals of the quadrilateral ABCD bisect each other. 2. A quadrilateral whose diagonals bisect each other is a parallelogram. 3. Therefore, ABCD is a parallelogram. 4. In a parallelogram, opposite sides are parallel. So, DA is parallel to CB (DA || CB). This completely proves all parts, assuming the correction for (ii).

In simple words: First, we use the fact that the lines drawn from D and B perpendicular to AC are equal in length, and that O is the midpoint of AC (because OB=OD and AB=CD means the triangles formed at the intersection are congruent). This helps show that triangles DOC and AOB have the same area. Since O splits both diagonals in half, ABCD must be a parallelogram. In a parallelogram, opposite sides are parallel, and a diagonal cuts it into two equal-area triangles.

🎯 Exam Tip: This problem is a comprehensive test of parallelogram properties. Remember that if diagonals bisect each other, the quadrilateral is a parallelogram. Also, drawing perpendiculars is a key technique for comparing triangle areas.

Question 3. ABCD is a parallelogram, E and F are the mid-points of BC and CD respectively. Prove that: ar (∆AEF) = \( \frac { 3 }{ 8 } ar (llgm ABCD) \)
Answer: We are given a parallelogram ABCD, where E is the midpoint of side BC and F is the midpoint of side CD. We need to prove that the area of triangle AEF is \( \frac { 3 }{ 8 } \) of the area of parallelogram ABCD. Proof: The area of triangle AEF can be found by subtracting the areas of the surrounding triangles (ABE, ECF, ADF) from the total area of the parallelogram ABCD. Area (∆AEF) = Area (llgm ABCD) - Area (∆ABE) - Area (∆ECF) - Area (∆ADF). 1. Calculate Area (∆ABE): * Consider triangle ABE. Take AB as the base. The height of triangle ABE from E to AB is half the height of the parallelogram with base AB. * Alternatively, consider base BC of the parallelogram. Area (llgm ABCD) = BC \( \times \) height \( h_1 \) (perpendicular height from A to BC). * Area (∆ABC) = \( \frac { 1 }{ 2 } \) Area (llgm ABCD). * Since E is the midpoint of BC, AE is a median of ∆ABC from A to the midpoint E of BC. However, that's not what we're going for directly. * Consider the parallelogram ABCD. If we take AB as base, let the height be \( h \). So, Area (llgm ABCD) = AB \( \times h \). * For triangle ABE, its base is AB, and its height is the perpendicular distance from E to AB. Since E is the midpoint of BC, the perpendicular distance from E to AB is \( \frac { 1 }{ 2 } h \). * So, Area (∆ABE) = \( \frac { 1 }{ 2 } \times AB \times (\frac { 1 }{ 2 } h) = \frac { 1 }{ 4 } \times AB \times h = \frac { 1 }{ 4 } \) Area (llgm ABCD) (i). 2. Calculate Area (∆ADF): * Similarly, for triangle ADF, its base is AD, and its height is the perpendicular distance from F to AD. Since F is the midpoint of CD, the perpendicular distance from F to AD is \( \frac { 1 }{ 2 } h' \) (where \( h' \) is height of parallelogram for base AD). * Area (llgm ABCD) = AD \( \times h' \). * So, Area (∆ADF) = \( \frac { 1 }{ 2 } \times AD \times (\frac { 1 }{ 2 } h') = \frac { 1 }{ 4 } \times AD \times h' = \frac { 1 }{ 4 } \) Area (llgm ABCD) (ii). 3. Calculate Area (∆ECF): * Consider triangle ECF. E is the midpoint of BC, so CE = \( \frac { 1 }{ 2 } BC \). F is the midpoint of CD, so CF = \( \frac { 1 }{ 2 } CD \). * Triangle ECF is similar to triangle BCD with a ratio of sides 1:2 (since CE/CB = 1/2 and CF/CD = 1/2). * Area (∆BCD) = \( \frac { 1 }{ 2 } \) Area (llgm ABCD). * The ratio of areas of similar triangles is the square of the ratio of their corresponding sides. * So, Area (∆ECF) = \( (\frac { 1 }{ 2 })^2 \times \) Area (∆BCD) = \( \frac { 1 }{ 4 } \times (\frac { 1 }{ 2 } \) Area (llgm ABCD)) = \( \frac { 1 }{ 8 } \) Area (llgm ABCD) (iii). 4. Now, substitute (i), (ii), and (iii) into the main equation: Area (∆AEF) = Area (llgm ABCD) - Area (∆ABE) - Area (∆ADF) - Area (∆ECF) Area (∆AEF) = Area (llgm ABCD) - \( \frac { 1 }{ 4 } \) Area (llgm ABCD) - \( \frac { 1 }{ 4 } \) Area (llgm ABCD) - \( \frac { 1 }{ 8 } \) Area (llgm ABCD) Area (∆AEF) = (1 - \( \frac { 1 }{ 4 } \) - \( \frac { 1 }{ 4 } \) - \( \frac { 1 }{ 8 } \)) Area (llgm ABCD) Area (∆AEF) = (1 - \( \frac { 2 }{ 4 } \) - \( \frac { 1 }{ 8 } \)) Area (llgm ABCD) Area (∆AEF) = (1 - \( \frac { 1 }{ 2 } \) - \( \frac { 1 }{ 8 } \)) Area (llgm ABCD) Area (∆AEF) = (\( \frac { 8 - 4 - 1 }{ 8 } \)) Area (llgm ABCD) Area (∆AEF) = \( \frac { 3 }{ 8 } \) Area (llgm ABCD). Thus, the proof is complete.

In simple words: To find the area of triangle AEF, we take the total area of the parallelogram and subtract the areas of the three triangles around AEF (ABE, ECF, and ADF). Each of these outer triangles' areas is a specific fraction of the parallelogram's area (1/4 for ABE, 1/4 for ADF, and 1/8 for ECF). When we subtract these fractions from the whole (1), we are left with 3/8.

🎯 Exam Tip: For complex area proofs, often the easiest method is to use the subtractive approach: calculate the area of the largest figure, then subtract the areas of the smaller, surrounding figures. Remember how midpoints affect side lengths and consequently, the areas of triangles formed.

Question 2. In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (ADOC) = ar (∆AOB)
(ii) ar (ADCB) = ar (∆ACB)
(iii) DA || CB or ABCD is a parallelogram.

Answer:
Given: ABCD is a quadrilateral where AC and BD are diagonals. They intersect at O, such that \( OB = OD \). Also, \( AB = CD \).
To prove:
(i) ar (\(\triangle\)ADOC) = ar (\(\triangle\)AOB)
(ii) ar (\(\triangle\)ADCB) = ar (\(\triangle\)ACB)
(iii) DA || CB or ABCD is a parallelogram.
Construction: Draw DN perpendicular to AC and BM perpendicular to AC.
Proof:
(i) First, consider \(\triangle\)DNO and \(\triangle\)BMO.
\( \angle DNO = \angle BMO = 90^\circ \) (Since DN \(\perp\) AC and BM \(\perp\) AC)
\( DO = BO \) (Given)
\( \angle DON = \angle BOM \) (Vertically opposite angles)
Thus, \(\triangle\)DNO \(\cong\) \(\triangle\)BMO (by AAS congruence rule)

\( \implies \) ar (\(\triangle\)DNO) = ar (\(\triangle\)BMO) ...(iii)
Also, \( DN = BM \) (CPCTC, corresponding parts of congruent triangles)
Now, consider \(\triangle\)ACND and \(\triangle\)ABM.
\( AB = CD \) (Given)
\( DN = BM \) (Proven above)
\( \angle DNC = \angle BMA = 90^\circ \) (By construction)
Thus, \(\triangle\)ACND \(\cong\) \(\triangle\)ABM (by RHS congruence rule)

\( \implies \) ar (\(\triangle\)ACND) = ar (\(\triangle\)ABM) ...(iv)
Now, add relations (iii) and (iv):
ar (\(\triangle\)ACND) + ar (\(\triangle\)DNO) = ar (\(\triangle\)ABM) + ar (\(\triangle\)BMO)

\( \implies \) ar (\(\triangle\)ADOC) = ar (\(\triangle\)AOB). This proves part (i).

(ii) We have ar (\(\triangle\)ADOC) = ar (\(\triangle\)AOB) from part (i).
Add ar (\(\triangle\)BOC) to both sides:
ar (\(\triangle\)ADOC) + ar (\(\triangle\)BOC) = ar (\(\triangle\)AOB) + ar (\(\triangle\)BOC)

\( \implies \) ar (\(\triangle\)ADCB) = ar (\(\triangle\)ACB). This proves part (ii).

(iii) Since \(\triangle\)DCB and \(\triangle\)ACB have the same base BC and equal areas (as ar (\(\triangle\)ADCB) = ar (\(\triangle\)ACB), and \(\triangle\)DCB is part of ADCB), they must lie between the same parallel lines.

\( \implies \) DA || CB. This is a property of triangles with equal areas on the same base. Therefore, if two triangles share a base and have the same area, their third vertices must lie on a line parallel to the base.
Also, from \(\triangle\)ACND \(\cong\) \(\triangle\)ABM, we know \( \angle CDN = \angle ABM \). From \(\triangle\)DNO \(\cong\) \(\triangle\)BMO, we know \( \angle NDO = \angle MBO \).
Adding these two, \( \angle CDN + \angle NDO = \angle ABM + \angle MBO \).

\( \implies \) \( \angle CDO = \angle ABO \). These are alternate angles. Therefore, CD || BA.
Since DA || CB and CD || BA, ABCD is a parallelogram.
In simple words: We showed that two small triangles formed by the diagonals and perpendiculars have the same area. Then, we used this to prove that two larger triangles (ADCB and ACB) have equal areas, which means the sides DA and CB are parallel. We also showed that the other pair of sides (CD and BA) are parallel. When both pairs of opposite sides are parallel, the figure is a parallelogram.

🎯 Exam Tip: For proofs involving areas and parallelograms, always clearly state the 'Given', 'To Prove', and 'Construction' steps. Remember that triangles on the same base and between the same parallels have equal areas.

Question 3. ABCD is a parallelogram, E and F are the mid-points of BC and CD respectively. Prove that: ar (\(\triangle\)AEF) = \( \frac{3}{8} \) ar (llgm ABCD)
Answer:
Given: ABCD is a parallelogram. E is the mid-point of side BC, and F is the mid-point of side CD.
To Prove: ar (\(\triangle\)AEF) = \( \frac{3}{8} \) ar (llgm ABCD)
Construction: Join BD and EF.
Proof:
In parallelogram ABCD, we know that the area of the parallelogram is equal to the area of the rectangle formed by its base and corresponding height. A diagonal divides a parallelogram into two triangles of equal area.
So, ar (\(\triangle\)ABD) = ar (\(\triangle\)BCD) = \( \frac{1}{2} \) ar (llgm ABCD).
Consider \(\triangle\)BCD. E is the mid-point of BC, and F is the mid-point of CD.
Therefore, by the mid-point theorem, EF || BD and \( EF = \frac{1}{2} BD \).
Now, let's find the areas of the triangles outside \(\triangle\)AEF within the parallelogram.

1. Area of \(\triangle\)ABE:
Base = AB, Height = perpendicular distance from E to AB. Let this be \( h_1 \).
Alternatively, we can use the ratio of bases.
Consider \(\triangle\)ABC. E is the mid-point of BC.
ar (\(\triangle\)ABE) = \( \frac{1}{2} \) ar (\(\triangle\)ABC)
Since diagonal AC divides the parallelogram into two equal triangles, ar (\(\triangle\)ABC) = \( \frac{1}{2} \) ar (llgm ABCD).
So, ar (\(\triangle\)ABE) = \( \frac{1}{2} \times \frac{1}{2} \) ar (llgm ABCD) = \( \frac{1}{4} \) ar (llgm ABCD) ...(i)

2. Area of \(\triangle\)ADF:
Similarly, for \(\triangle\)ADC, F is the mid-point of CD.
ar (\(\triangle\)ADF) = \( \frac{1}{2} \) ar (\(\triangle\)ADC)
Since ar (\(\triangle\)ADC) = \( \frac{1}{2} \) ar (llgm ABCD).
So, ar (\(\triangle\)ADF) = \( \frac{1}{2} \times \frac{1}{2} \) ar (llgm ABCD) = \( \frac{1}{4} \) ar (llgm ABCD) ...(ii)

3. Area of \(\triangle\)CEF:
In \(\triangle\)BCD, E and F are mid-points of BC and CD respectively.
Thus, \(\triangle\)CEF is similar to \(\triangle\)CBD, and the ratio of their sides is \( \frac{CE}{CB} = \frac{CF}{CD} = \frac{1}{2} \).
The ratio of their areas will be the square of the ratio of their sides.
ar (\(\triangle\)CEF) = \( (\frac{1}{2})^2 \) ar (\(\triangle\)CBD) = \( \frac{1}{4} \) ar (\(\triangle\)CBD)
Since ar (\(\triangle\)CBD) = \( \frac{1}{2} \) ar (llgm ABCD).
So, ar (\(\triangle\)CEF) = \( \frac{1}{4} \times \frac{1}{2} \) ar (llgm ABCD) = \( \frac{1}{8} \) ar (llgm ABCD) ...(iii)

The area of \(\triangle\)AEF can be found by subtracting the areas of \(\triangle\)ABE, \(\triangle\)ADF, and \(\triangle\)CEF from the total area of the parallelogram ABCD.
ar (\(\triangle\)AEF) = ar (llgm ABCD) - [ar (\(\triangle\)ABE) + ar (\(\triangle\)ADF) + ar (\(\triangle\)CEF)]
Substitute the areas from (i), (ii), and (iii):
ar (\(\triangle\)AEF) = ar (llgm ABCD) - [\( \frac{1}{4} \) ar (llgm ABCD) + \( \frac{1}{4} \) ar (llgm ABCD) + \( \frac{1}{8} \) ar (llgm ABCD)]
ar (\(\triangle\)AEF) = ar (llgm ABCD) - [\( (\frac{1}{4} + \frac{1}{4} + \frac{1}{8}) \) ar (llgm ABCD)]
ar (\(\triangle\)AEF) = ar (llgm ABCD) - [\( (\frac{2}{8} + \frac{2}{8} + \frac{1}{8}) \) ar (llgm ABCD)]
ar (\(\triangle\)AEF) = ar (llgm ABCD) - [\( \frac{5}{8} \) ar (llgm ABCD)]
ar (\(\triangle\)AEF) = \( (1 - \frac{5}{8}) \) ar (llgm ABCD)
ar (\(\triangle\)AEF) = \( \frac{3}{8} \) ar (llgm ABCD).
Thus, the area of \(\triangle\)AEF is \( \frac{3}{8} \) of the area of the parallelogram ABCD.
In simple words: We found the areas of the three triangles that are around the main triangle AEF. Each of these areas was a fraction of the total parallelogram area. By subtracting these three areas from the total area of the parallelogram, we were left with the area of triangle AEF, which came out to be three-eighths of the parallelogram's area.

🎯 Exam Tip: When proving area relations in parallelograms with mid-points, remember that a median divides a triangle into two equal areas, and the mid-point theorem is often useful for establishing relationships between smaller and larger triangles.