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NCERT Class 9 Mathematics Exercise Set 6.2 Solutions: Measuring Space: Perimeter and Area

Step-by-step working for every question of Exercise Set 6.2 in NCERT's Class 9 Mathematics book, chapter “Measuring Space: Perimeter and Area”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 6.2 Q1

   Fig. 6.31: rectangle ABCD with AB = 10 cm and BC = 8 cm, E a point on BC. Find the area of triangle ADE.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Given rectangle ABCD has side AB = 10 cm and side BC = 8 cm.
   2. In a rectangle, opposite sides are equal, so side AD is equal to side BC, which is 8 cm.
   3. Point E lies on the side BC, so the perpendicular distance from E to the opposite side AD is equal to the length of side AB, which is 10 cm.
   4. This distance of 10 cm acts as the height of triangle ADE with respect to the base AD.
   5. The area of triangle ADE is calculated as half of the base AD multiplied by the height.
   6. Area = 1/2 × 8 cm × 10 cm.
   7. Area = 40 cm².

   Answer: 40 cm²

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2. Ex 6.2 Q2

   A trapezium has parallel sides 40 cm and 20 cm and both non-parallel sides 26 cm. Find its area.

   1. Let the trapezium be ABCD with AB ∥ CD, where AB = 40 cm and CD = 20 cm.
   2. Since the non-parallel sides are equal (AD = BC = 26 cm), the trapezium is isosceles.
   3. Drop perpendiculars from D and C to AB, meeting AB at E and F respectively.
   4. Then EF = CD = 20 cm, and AE = FB = (40 − 20) ÷ 2 = 10 cm.
   5. In right triangle ADE, DE² + AE² = AD², so DE² + 10² = 26².
   6. DE² = 676 − 100 = 576, so DE = √576 = 24 cm.
   7. The height of the trapezium is 24 cm.
   8. Area = ½ × (sum of parallel sides) × height = ½ × (40 + 20) × 24.
   9. Area = ½ × 60 × 24 = 720 cm².

   Answer: 24 cm, 720 cm²

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3. Ex 6.2 Q3

   A triangle has two sides 8 cm and 11 cm and perimeter 32 cm. Find its area.

   1. Let the third side of the triangle be x cm.
   2. The perimeter is the sum of all sides, so 8 + 11 + x = 32.
   3. Solving for x gives x = 32 − 19 = 13 cm.
   4. The sides of the triangle are 8 cm, 11 cm, and 13 cm.
   5. Calculate the semi-perimeter s = 32 ÷ 2 = 16 cm.
   6. Use Heron's formula: Area = √(s(s−a)(s−b)(s−c)).
   7. Substitute the values: Area = √(16(16−8)(16−11)(16−13)).
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   8. Simplify the terms inside the square root: Area = √(16 × 8 × 5 × 3).
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   9. Multiply the numbers: 16 × 8 = 128 and 5 × 3 = 15, so Area = √(128 × 15).
   10. Calculate the product: 128 × 15 = 1920.
   11. Simplify √1920 by factoring out perfect squares: 1920 = 64 × 30.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   12. Area = √64 × √30 = 8√30 cm².
   13. Approximate the value using √30 ≈ 5.477: 8 × 5.477 ≈ 43.82 cm².

   Answer: 13 cm, 8√30 cm², 43.82

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4. Ex 6.2 Q4

   A triangular plot has sides in the ratio 3 : 5 : 7 and perimeter 300 m. Find its area.

   1. Let the sides be 3k, 5k and 7k. Then 3k + 5k + 7k = 15k = 300, so k = 20.
   2. The sides are 60 m, 100 m and 140 m.
   3. Semi-perimeter s = 300 ÷ 2 = 150 m, so s − a = 90, s − b = 50 and s − c = 10.
   4. By Heron's formula, area = √(150 × 90 × 50 × 10) = √6750000.
   5. Since 1500² = 2250000 and 6750000 = 2250000 × 3, area = 1500√3 m².
   6. With √3 ≈ 1.7320508, area ≈ 2598.08 m².

   Answer: Sides 60 m, 100 m, 140 m; area = 1500√3 m² ≈ 2598.08 m²

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5. Ex 6.2 Q5

   A rhombus of area 128 cm² has one diagonal twice the other. Find the shorter diagonal.

   1. Let the length of the shorter diagonal be x cm.
   2. Since one diagonal is twice the other, the longer diagonal is 2x cm.
   3. The area of a rhombus is given by the formula: Area = (1/2) × d₁ × d₂.
   4. Substitute the given area and diagonal expressions into the formula: 128 = (1/2) × x × 2x.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   5. Simplify the equation to get 128 = x².
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   6. Take the square root of both sides to find x = √128.
      why?

      If two non-negative quantities are equal, their square roots are equal. A square has two roots, + and −; a length or a count cannot be negative, so the negative root is dropped there.

   7. Simplify the radical: √128 = √(64 × 2) = 8√2.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   8. Calculate the approximate decimal value: 8 × 1.414 ≈ 11.31.

   Answer: 8√2 cm, 11.31

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6. Ex 6.2 Q6

   ABCD is a parallelogram, P and Q any points on AB. Find area(△PCD) : area(△QCD).

   1. Consider the triangle PCD and the parallelogram ABCD.
   2. Triangle PCD and parallelogram ABCD share the same base CD.
   3. Since P lies on AB and AB is parallel to CD, the perpendicular height from P to CD is equal to the height of the parallelogram.
   4. The area of a triangle is half the product of its base and height, so area(△PCD) = 1/2 × area(parallelogram ABCD).
   5. Consider the triangle QCD and the parallelogram ABCD.
   6. Triangle QCD and parallelogram ABCD share the same base CD.
   7. Since Q lies on AB and AB is parallel to CD, the perpendicular height from Q to CD is equal to the height of the parallelogram.
   8. The area of a triangle is half the product of its base and height, so area(△QCD) = 1/2 × area(parallelogram ABCD).
   9. Comparing the two results, area(△PCD) is equal to area(△QCD).
   10. Therefore, the ratio of their areas is 1 : 1.
       why?

       The areas of two similar triangles are in the ratio of the squares of their corresponding sides.

   Answer: 1 : 1

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7. Ex 6.2 Q7

   O is any point on diagonal PR of parallelogram PQRS. Prove area(△PSO) = area(△PQO).

   1. The diagonal PR splits parallelogram PQRS into triangles PQR and PSR, which are congruent (PQ = RS, QR = SP, PR common), so they have equal areas.
   2. These two triangles have the same base PR, so their heights to PR are equal: the perpendicular from Q to line PR has the same length h as the perpendicular from S.
   3. Triangles PQO and PSO have the same base PO, which lies on line PR.
   4. Their heights to this base are the perpendiculars from Q and from S to line PR, both of length h.
   5. So area(△PQO) = ½ × PO × h = area(△PSO).

   Answer: area(△PSO) = area(△PQO)

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8. Ex 6.2 Q8

   Prove that joining the midpoints of the sides of a 4-gon in order gives a parallelogram whose area is half that of the 4-gon.

   1. Let ABCD be the 4-gon (convex, as usually drawn) and P, Q, R, S the midpoints of AB, BC, CD, DA.
   2. PQRS is a parallelogram: in △ABC the midpoint theorem gives PQ ∥ AC and PQ = ½AC, and in △ADC it gives SR ∥ AC and SR = ½AC, so PQ and SR are equal and parallel.
   3. Corner triangle at B: Q is the midpoint of BC, so PQ is a median of △PBC and area(△BPQ) = ½ area(△BPC).
   4. P is the midpoint of AB, so CP is a median of △ABC and area(△BPC) = ½ area(△ABC). Hence area(△BPQ) = ¼ area(△ABC).
   5. In the same way area(△DRS) = ¼ area(△DAC). Adding, area(△BPQ) + area(△DRS) = ¼ (area(△ABC) + area(△ADC)) = ¼ area(ABCD).
   6. Using diagonal BD instead, area(△APS) + area(△CQR) = ¼ (area(△ABD) + area(△CBD)) = ¼ area(ABCD).
   7. So the four corner triangles together have area ½ area(ABCD).
   8. PQRS is what remains of ABCD after the four corner triangles are removed, so area(PQRS) = area(ABCD) − ½ area(ABCD) = ½ area(ABCD).

   Answer: area(PQRS) = ½ area(ABCD)

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9. Ex 6.2 Q9

   In △ABC, D is the midpoint of BC and P any point on the median AD (Fig. 6.32). Show area(△ABP) = area(△ACP).

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Since D is the midpoint of BC, BD = DC.
   2. In △ABC, AD is a median, so area(△ABD) = area(△ACD).
   3. In △PBC, PD is a median (since D is midpoint of BC), so area(△PBD) = area(△PCD).
   4. Subtract the area of △PBD from the area of △ABD: area(△ABD) − area(△PBD) = area(△ABP).
   5. Subtract the area of △PCD from the area of △ACD: area(△ACD) − area(△PCD) = area(△ACP).
   6. Since area(△ABD) = area(△ACD) and area(△PBD) = area(△PCD), their differences are equal.
   7. Therefore, area(△ABP) = area(△ACP).

   Answer: area(△ABP) = area(△ACP)

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10. Ex 6.2 Q10

    P is any point inside square ABCD (Fig. 6.33). Ratio of area(△PAB) + area(△PCD) to area(△PBC) + area(△PDA)?

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Let the side of square ABCD be a, so its total area is a².
    2. Let the perpendicular distance from P to side AB be h₁ and to side CD be h₂.
    3. Since P is inside the square, the sum of these distances equals the side length: h₁ + h₂ = a.
    4. The area of triangle PAB is (1/2) × base × height = (1/2) × a × h₁.
    5. The area of triangle PCD is (1/2) × base × height = (1/2) × a × h₂.
    6. The sum of the areas of triangles PAB and PCD is (1/2)a(h₁ + h₂).
    7. Substituting h₁ + h₂ = a, this sum becomes (1/2)a(a) = a²/2.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    8. Similarly, let the distance from P to BC be k₁ and to DA be k₂, so k₁ + k₂ = a.
    9. The sum of the areas of triangles PBC and PDA is (1/2)a(k₁ + k₂) = (1/2)a(a) = a²/2.
    10. The ratio of the two sums is (a²/2) : (a²/2), which simplifies to 1 : 1.
        why?

        Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    Answer: 1 : 1

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11. Ex 6.2 Q11

    In △ABC, D is the midpoint of AB, P any point on BC, Q on AB with CQ ∥ PD (Fig. 6.34). Prove area(△BPQ) = ½ area(△ABC).

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Q and D both lie on AB, with D between B and Q as in the figure, and CQ ∥ PD.
    2. Triangles PDQ and PDC have the same base PD, and Q and C lie on one line parallel to PD, so their heights to PD are equal: area(△PDQ) = area(△PDC).
    3. Since D lies on BQ, area(△BPQ) = area(△BPD) + area(△PDQ).
    4. So area(△BPQ) = area(△BPD) + area(△PDC) = area(△BDC), because P lies on BC.
    5. D is the midpoint of AB, so CD is a median of △ABC and area(△BDC) = ½ area(△ABC).
    6. Therefore area(△BPQ) = ½ area(△ABC).

    Answer: area(△BPQ) = ½ area(△ABC)

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