RBSE Solutions Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Exercise 10.2

Chapter 10 Area of Triangles and Quadrilaterals Ex 10.2

Question 1. In figure, ABCD is a parallelogram, AE \( \perp \) DC and CF \( \perp \) AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.Answer: First, we find the area of the parallelogram ABCD using the given base AB and its corresponding height AE. Area of parallelogram ABCD = base \( \times \) height
\( \implies \) Area = AB \( \times \) AE
\( \implies \) Area = \( 16 \) cm \( \times \) \( 8 \) cm = \( 128 \) cm\( ^2 \). A key property of parallelograms is that their area remains constant, regardless of which base and corresponding height are used. Now, we use AD as the base and CF as the corresponding height, with the area we just calculated: Area of parallelogram ABCD = AD \( \times \) CF
\( \implies 128 \) cm\( ^2 \) = AD \( \times \) \( 10 \) cm
\( \implies \) AD = \( \frac{128}{10} \) cm
\( \implies \) AD = \( 12.8 \) cm.In simple words: We first found the total area of the shape using one pair of base and height measurements. Then, we used that same area with the other height measurement to calculate the length of the unknown side AD.

🎯 Exam Tip: Remember that the area of a parallelogram can be calculated using any base and its corresponding altitude; this principle is crucial for finding unknown dimensions when different heights are given.

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 (llgm ABCD).Answer: We are given a parallelogram ABCD, and E, F, G, H are the midpoints of its sides AB, BC, CD, and DA, respectively. We need to prove that the area of the quadrilateral EFGH is half the area of parallelogram ABCD. **Step 1: Proving EFGH is a parallelogram.** Let's first consider the diagonal AC. In triangle ABC, E and F are the midpoints of sides AB and BC. By the midpoint theorem, the line segment EF is parallel to AC, and its length is half of AC.
\( \implies \) EF \( \parallel \) AC and EF \( = \frac{1}{2} \) AC ...(i) Similarly, in triangle ADC, H and G are the midpoints of sides AD and DC. By the midpoint theorem, the line segment HG is parallel to AC, and its length is half of AC.
\( \implies \) HG \( \parallel \) AC and HG \( = \frac{1}{2} \) AC ...(ii) From equations (i) and (ii), we can see that EF \( \parallel \) HG and EF \( = \) HG. Since one pair of opposite sides (EF and HG) are both parallel and equal in length, EFGH is a parallelogram. **Step 2: Proving Area(EFGH) = \( \frac{1}{2} \) Area(ABCD).** Now, let's draw a line connecting H and F. This line HF is a diagonal for EFGH. Consider the parallelogram HABF. Triangle HEF and parallelogram HABF share the same base HF and lie between the same parallel lines HF and AB. Therefore, the area of triangle HEF is half the area of parallelogram HABF.
\( \implies \) ar ( \( \triangle \)HEF) = \( \frac{1}{2} \) ar (llgm HABF) ...(iii) Similarly, consider the parallelogram HFCD. Triangle HGF and parallelogram HFCD share the same base HF and lie between the same parallel lines HF and DC. Therefore, the area of triangle HGF is half the area of parallelogram HFCD.
\( \implies \) ar ( \( \triangle \)HGF) = \( \frac{1}{2} \) ar (llgm HFCD) ...(iv) Adding equations (iii) and (iv): ar ( \( \triangle \)HEF) + ar ( \( \triangle \)HGF) = \( \frac{1}{2} \) ar (llgm HABF) + \( \frac{1}{2} \) ar (llgm HFCD) The sum of the areas of triangle HEF and triangle HGF gives the area of parallelogram EFGH.
\( \implies \) ar (llgm EFGH) = \( \frac{1}{2} \) [ar (llgm HABF) + ar (llgm HFCD)] Since parallelograms HABF and HFCD together form the larger parallelogram ABCD, their combined area is the area of ABCD.
\( \implies \) ar (llgm EFGH) = \( \frac{1}{2} \) ar (llgm ABCD). Thus, the area of the parallelogram formed by joining the midpoints of the sides of a parallelogram is half the area of the original parallelogram.In simple words: We first showed that EFGH is also a parallelogram using a rule about midpoints. Then, by drawing a line inside and splitting the big parallelogram into two smaller ones, we found that the areas of the triangles HEF and HGF are half of their respective smaller parallelograms. Adding these triangle areas gives EFGH, which is half the area of the whole big parallelogram ABCD.

🎯 Exam Tip: This proof relies on the midpoint theorem and the property that a triangle and a parallelogram on the same base and between the same parallels have areas in a 1:2 ratio. Clearly state each step of the proof, including any constructions.

Question 3. P and Q are points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that: ar ( \( \triangle \)APB) = ar (ABQC).Answer: We are given a parallelogram ABCD, with point P on side DC and point Q on side AD. We need to show that the area of triangle APB is equal to the area of quadrilateral ABQC. **Part 1: Area of triangle APB** Triangle APB and parallelogram ABCD share the same base AB. They are also located between the same parallel lines AB and DC. According to a key theorem, when a triangle and a parallelogram share the same base and lie between the same parallel lines, the area of the triangle is exactly half the area of the parallelogram.
\( \implies \) ar ( \( \triangle \)APB) = \( \frac{1}{2} \) ar (llgm ABCD) ...(i) **Part 2: Area of quadrilateral ABQC** The quadrilateral ABQC and parallelogram ABCD share the same base BC. They are also located between the same parallel lines BC and AD. Following a similar property for figures on the same base and between the same parallels, the area of quadrilateral ABQC is half the area of parallelogram ABCD. The figure ABQC is a trapezium with parallel sides AQ and BC.
\( \implies \) ar (ABQC) = \( \frac{1}{2} \) ar (llgm ABCD) ...(ii) From equations (i) and (ii), since both ar ( \( \triangle \)APB) and ar (ABQC) are equal to half the area of parallelogram ABCD, we can conclude:
\( \implies \) ar ( \( \triangle \)APB) = ar (ABQC).In simple words: We used a rule that says if a shape shares a base with a parallelogram and is between the same parallel lines, its area can be half the parallelogram's area. We applied this rule to triangle APB and also to the shape ABQC, and since both came out to be half the area of the big parallelogram, they must be equal to each other.

🎯 Exam Tip: Mastering the theorem about areas of figures on the same base and between the same parallels is crucial. Always identify the common base and the parallel lines correctly for each figure involved.

Question 4. (i) ar ( \( \triangle \)APB) + ar (APCD) = \( \frac{1}{2} \) ar (llgm ABCD)
(ii) ar ( \( \triangle \)APD) + ar (APBC) = ar ( \( \triangle \)APB) + ar ( \( \triangle \)PCD).Answer: Let ABCD be a parallelogram and P be any point inside it. We need to prove two statements. **Construction:** Through point P, draw a line segment EF parallel to AB (and also to DC), with E on AD and F on BC. This line EF divides the parallelogram ABCD into two smaller parallelograms, ABFE and EFCD. **(i) To prove: ar ( \( \triangle \)APB) + ar (APCD) = \( \frac{1}{2} \) ar (llgm ABCD)** Consider triangle APB and parallelogram ABFE. They share the same base AB and lie between the same parallel lines AB and EF. Thus, according to the theorem:
\( \implies \) ar ( \( \triangle \)APB) = \( \frac{1}{2} \) ar (llgm ABFE) ...(1) Similarly, consider triangle DPC and parallelogram EFCD. They share the same base DC and lie between the same parallel lines EF and DC. (Here, `APCD` in the question refers to `triangle DPC`.)
\( \implies \) ar ( \( \triangle \)DPC) = \( \frac{1}{2} \) ar (llgm EFCD) ...(2) Adding equations (1) and (2): ar ( \( \triangle \)APB) + ar ( \( \triangle \)DPC) = \( \frac{1}{2} \) ar (llgm ABFE) + \( \frac{1}{2} \) ar (llgm EFCD)
\( \implies \) ar ( \( \triangle \)APB) + ar ( \( \triangle \)DPC) = \( \frac{1}{2} \) [ar (llgm ABFE) + ar (llgm EFCD)] Since parallelogram ABFE and parallelogram EFCD together form the entire parallelogram ABCD:
\( \implies \) ar ( \( \triangle \)APB) + ar ( \( \triangle \)DPC) = \( \frac{1}{2} \) ar (llgm ABCD) ...(3) This proves the first part. **(ii) To prove: ar ( \( \triangle \)APD) + ar (APBC) = ar ( \( \triangle \)APB) + ar ( \( \triangle \)PCD)** We know that the total area of the parallelogram ABCD is the sum of the areas of the four triangles formed by joining P to its vertices: ar (llgm ABCD) = ar ( \( \triangle \)APB) + ar ( \( \triangle \)PBC) + ar ( \( \triangle \)PCD) + ar ( \( \triangle \)PDA). From part (i) [equation (3)], we already established that: ar ( \( \triangle \)APB) + ar ( \( \triangle \)PCD) = \( \frac{1}{2} \) ar (llgm ABCD). Substituting this into the total area equation: ar (llgm ABCD) = \( \frac{1}{2} \) ar (llgm ABCD) + ar ( \( \triangle \)PBC) + ar ( \( \triangle \)PDA) Subtracting \( \frac{1}{2} \) ar (llgm ABCD) from both sides: ar ( \( \triangle \)PBC) + ar ( \( \triangle \)PDA) = ar (llgm ABCD) - \( \frac{1}{2} \) ar (llgm ABCD)
\( \implies \) ar ( \( \triangle \)PBC) + ar ( \( \triangle \)PDA) = \( \frac{1}{2} \) ar (llgm ABCD) ...(4) Now we have two important results: From (3): ar ( \( \triangle \)APB) + ar ( \( \triangle \)PCD) = \( \frac{1}{2} \) ar (llgm ABCD) From (4): ar ( \( \triangle \)APD) + ar ( \( \triangle \)PBC) = \( \frac{1}{2} \) ar (llgm ABCD) Therefore, by equating these two results: ar ( \( \triangle \)APB) + ar ( \( \triangle \)PCD) = ar ( \( \triangle \)APD) + ar ( \( \triangle \)PBC). This proves the second part.In simple words: For the first part, we split the parallelogram into two smaller parallelograms using a line through P. We then showed that the areas of two opposite triangles (APB and DPC) are half of these smaller parallelograms, which sum up to half the main parallelogram. For the second part, we used the idea that the total area of the parallelogram is the sum of all four triangles formed by P. Since we know the sum of two opposite triangles (APB and PCD) is half the total area, the sum of the other two opposite triangles (APD and PBC) must be the remaining half, proving they are equal.

🎯 Exam Tip: When a point P is inside a parallelogram, drawing a line through P parallel to two sides is a common construction technique. Remember that the sum of the areas of the two triangles formed on opposite bases (like APB and DPC) is always half the area of the parallelogram.

Question 5. In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = \( \frac{1}{2} \) ar (PQRS)Answer: We are given two parallelograms, PQRS and ABRS, sharing a common side SR. Point X is located on the side BR. **(i) To prove: ar (PQRS) = ar (ABRS)** We know a fundamental property of parallelograms: if two parallelograms share the same base and are located between the same pair of parallel lines, then their areas are equal. In this case, parallelograms PQRS and ABRS both share the common base SR. They also lie between the same parallel lines SR and PB (the line containing P, Q, A, and B). Therefore, their areas are equal.
\( \implies \) ar (PQRS) = ar (ABRS). **(ii) To prove: ar ( \( \triangle \)AXS) = \( \frac{1}{2} \) ar (PQRS)** Consider triangle AXS and parallelogram ABRS. They share the same base AS. Also, they lie between the same parallel lines AS and BR (because ABRS is a parallelogram, so its opposite sides AS and BR are parallel). According to another key theorem, if a triangle and a parallelogram are on the same base and between the same parallel lines, the area of the triangle is half the area of the parallelogram.
\( \implies \) ar ( \( \triangle \)AXS) = \( \frac{1}{2} \) ar (ABRS). From part (i), we already proved that ar (ABRS) = ar (PQRS). Substituting this into the equation above:
\( \implies \) ar ( \( \triangle \)AXS) = \( \frac{1}{2} \) ar (PQRS). This completes the proof.In simple words: First, we showed that the two parallelograms have the same area because they sit on the same bottom line and reach up to the same top line. Then, for the second part, we looked at a triangle inside one of these parallelograms. Since that triangle shares a side with the parallelogram and stretches to the opposite parallel side, its area is half that parallelogram's area. Because both parallelograms have the same area, the triangle's area is also half the area of the other parallelogram.

🎯 Exam Tip: Always clearly identify the common base and the pair of parallel lines when using theorems about areas of parallelograms and triangles. For compound proofs, ensure each sub-proof flows logically from previous results or established theorems.

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 fields 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:**1. Number of parts the field is divided into:** When point A on side RS is joined to points P and Q, the parallelogram field PQRS is divided into three distinct parts. **2. Shapes of these parts:** The three parts are all triangles:
(i) \( \triangle \)APS
(ii) \( \triangle \)APQ
(iii) \( \triangle \)AQR **3. How the farmer should sow wheat and pulses in equal portions:** Consider \( \triangle \)APQ and the parallelogram PQRS. They both share the same base PQ and lie between the same parallel lines PQ and RS. According to the area theorem for triangles and parallelograms, the area of \( \triangle \)APQ is exactly half the area of the parallelogram PQRS.
\( \implies \) ar ( \( \triangle \)APQ) = \( \frac{1}{2} \) ar (llgm PQRS) This also means that the combined area of the other two triangles, \( \triangle \)APS and \( \triangle \)AQR, must be the remaining half of the parallelogram's area: ar ( \( \triangle \)APS) + ar ( \( \triangle \)AQR) = ar (llgm PQRS) - ar ( \( \triangle \)APQ)
\( \implies \) ar ( \( \triangle \)APS) + ar ( \( \triangle \)AQR) = ar (llgm PQRS) - \( \frac{1}{2} \) ar (llgm PQRS)
\( \implies \) ar ( \( \triangle \)APS) + ar ( \( \triangle \)AQR) = \( \frac{1}{2} \) ar (llgm PQRS) So, to sow wheat and pulses in equal portions, the farmer can sow one crop (e.g., wheat) in the region of \( \triangle \)APQ and sow the other crop (e.g., pulses) in the combined region of \( \triangle \)APS and \( \triangle \)AQR (or vice-versa). This way, each crop will occupy exactly half the total field area.In simple words: The field splits into three triangles. One triangle (APQ) takes up exactly half the total field area because it shares its bottom side with the parallelogram and stretches to the opposite top line. The other two triangles (APS and AQR) together make up the other half. So, the farmer can plant wheat in the big middle triangle and pulses in the two side triangles combined, or the other way around, to get equal amounts.

🎯 Exam Tip: For problems involving areas within parallelograms, remember that any triangle sharing a base with the parallelogram and having its third vertex on the opposite parallel line will have half the area of the parallelogram.