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NCERT Exemplar · Class 9 Mathematics Heron's Formula

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EXERCISE 12.3 1–10 (part 3 of 4)

  1. Exercise 1

    Find the cost of laying grass in a triangular field of sides 50\displaystyle 50 m, 65\displaystyle 65 m and 65\displaystyle 65 m at the rate of Rs 7\displaystyle 7 per m2\displaystyle \mathrm{m}^{2}.

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    NCERT’s answer
    Rs $\displaystyle 10500$
    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q1 \[s = \frac{50+65+65}{2} = 90 \ \mathrm{m} \] \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{90 \times 40 \times 25 \times 25} = 1500 \ \mathrm{m}^{2} \] \[\text{Cost} = 1500 \times 7 = \text{Rs } 10500 \] Answer: Rs $\displaystyle 10$,500.

  2. Exercise 2

    The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13\displaystyle 13 m, 14\displaystyle 14 m and 15\displaystyle 15 m. The advertisements yield an earning of Rs 2000\displaystyle 2000 per m2\displaystyle \mathrm{m}^{2} a year. A company hired one of its walls for 6\displaystyle 6 months. How much rent did it pay?

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    NCERT’s answer
    Rs $\displaystyle 84,000$
    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q2 \[s = \frac{13+14+15}{2} = 21 \ \mathrm{m} \] \[\text{Area} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84 \ \mathrm{m}^{2} \] \[\text{Rent} = 84 \times 2000 \times \frac{6}{12} = \text{Rs } 84000 \] Answer: Rs $\displaystyle 84$,000.

  3. Exercise 3

    From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14\displaystyle 14 cm, 10\displaystyle 10 cm and 6\displaystyle 6 cm. Find the area of the triangle.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q3 \[\text{Area} = \tfrac{1}{2}a \cdot PM + \tfrac{1}{2}a \cdot PN + \tfrac{1}{2}a \cdot PQ = \tfrac{1}{2}a(14+10+6) = 15a \] \[\text{Area} = \frac{\sqrt3}{4}a^{2} \quad \text{(equilateral triangle)} \] \[\frac{\sqrt3}{4}a^{2} = 15a \implies a = \frac{60}{\sqrt3} = 20\sqrt3 \ \mathrm{cm} \] \[\text{Area} = 15a = 15 \times 20\sqrt3 = 300\sqrt3 \ \mathrm{cm}^{2} \approx 519.6 \ \mathrm{cm}^{2} \] Answer: \(\displaystyle 300\sqrt3 \ \mathrm{cm}^{2} \approx 519.6 \ \mathrm{cm}^{2}\).NCERT prints: $\displaystyle 300$√$\displaystyle 3$ cm — a unit misprint: an area cannot be in cm; the computed $\displaystyle 300$√$\displaystyle 3$ cm² is what matches.

  4. Exercise 4

    The perimeter of an isosceles triangle is 32\displaystyle 32 cm . The ratio of the equal side to its base is 3\displaystyle 3 : 2. Find the area of the triangle.

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    NCERT’s answer
    $\displaystyle 32 \sqrt{2} \mathrm{~cm}^{2}$
    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q4 \[3x+3x+2x = 32 \implies x = 4 \] \[\text{Equal side} = 12 \ \mathrm{cm}, \quad \text{Base} = 8 \ \mathrm{cm} \] \[s = \frac{12+12+8}{2} = 16 \ \mathrm{cm} \] \[\text{Area} = \sqrt{16 \times 4 \times 4 \times 8} = \sqrt{2048} = 32\sqrt2 \ \mathrm{cm}^{2} \approx 45.3 \ \mathrm{cm}^{2} \] Answer: \(\displaystyle 32\sqrt2 \ \mathrm{cm}^{2} \approx 45.3 \ \mathrm{cm}^{2}\).

  5. Exercise 5

    Find the area of a parallelogram given in Fig. 12.2. Also find the length of the altitude from vertex A on the side DC. NCERT_Question_Class9_Maths_Exemplar_Ch12_Ex12-3_Q5

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    NCERT’s answer
    $\displaystyle 180 \mathrm{~cm}^{2}$
    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q5 \[\text{In } \triangle BCD: \quad s = \frac{BC+CD+BD}{2} = \frac{17+12+25}{2} = 27\ \mathrm{cm} \] \[\text{Area}(\triangle BCD) = \sqrt{s(s-BC)(s-CD)(s-BD)} = \sqrt{27 \times 10 \times 15 \times 2} = 90\ \mathrm{cm}^2 \] \[\text{Area}(ABCD) = 2 \times \text{Area}(\triangle BCD) = 180\ \mathrm{cm}^2 \quad \text{(a diagonal bisects a parallelogram into two congruent triangles)} \] \[\text{Area}(ABCD) = DC \times AM \implies 180 = 12 \times AM \implies AM = 15\ \mathrm{cm} \] Answer: area \(\displaystyle =180\ \mathrm{cm}^2\); altitude from \(\displaystyle A\) on \(\displaystyle DC\) \(\displaystyle =15\ \mathrm{cm}\).

  6. Exercise 6

    A field in the form of a parallelogram has sides 60\displaystyle 60 m and 40\displaystyle 40 m and one of its diagonals is 80\displaystyle 80 m long. Find the area of the parallelogram.

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    NCERT’s answer
    $\displaystyle 600 \sqrt{15} \mathrm{~m}^{2}$
    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q6 Diagonal AC splits the parallelogram into congruent triangles ABC and ACD, each with sides $\displaystyle 60$ m, $\displaystyle 40$ m, $\displaystyle 80$ m. \[s = \frac{60+40+80}{2} = 90 \ \mathrm{m} \] \[\text{Area}(\triangle ABC) = \sqrt{90 \times 30 \times 50 \times 10} = \sqrt{1350000} = 300\sqrt{15} \ \mathrm{m}^{2} \] \[\text{Area}(ABCD) = 2 \times 300\sqrt{15} = 600\sqrt{15} \ \mathrm{m}^{2} \approx 2323.8 \ \mathrm{m}^{2} \] Answer: \(\displaystyle 600\sqrt{15} \ \mathrm{m}^{2} \approx 2323.8 \ \mathrm{m}^{2}\).

  7. Exercise 7

    The perimeter of a triangular field is 420\displaystyle 420 m and its sides are in the ratio 6\displaystyle 6 : 7\displaystyle 7 : 8. Find the area of the triangular field.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 2100 \sqrt{15} \mathrm{~m}^{2}$
    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q7 \[6x+7x+8x = 420 \implies x = 20 \] \[\text{Sides} = 120, 140, 160 \ \mathrm{m} \] \[s = \frac{120+140+160}{2} = 210 \ \mathrm{m} \] \[\text{Area} = \sqrt{210 \times 90 \times 70 \times 50} = \sqrt{66150000} = 2100\sqrt{15} \ \mathrm{m}^{2} \approx 8133.3 \ \mathrm{m}^{2} \] Answer: \(\displaystyle 2100\sqrt{15} \ \mathrm{m}^{2} \approx 8133.3 \ \mathrm{m}^{2}\).

  8. Exercise 8

    The sides of a quadrilateral ABCD are 6\displaystyle 6 cm, 8\displaystyle 8 cm, 12\displaystyle 12 cm and 14\displaystyle 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 24(\sqrt{6}+1) \mathrm{cm}^{2}$
    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q8 Diagonal AC divides ABCD into right triangle ABC and triangle ACD. \[AC^{2} = AB^{2}+BC^{2} = 6^{2}+8^{2} = 100 \implies AC = 10 \ \mathrm{cm} \quad \text{(Pythagoras, } \angle B = 90^\circ\text{)} \] \[\text{Area}(\triangle ABC) = \tfrac{1}{2} \times 6 \times 8 = 24 \ \mathrm{cm}^{2} \] \[s = \frac{12+14+10}{2} = 18 \ \mathrm{cm} \] \[\text{Area}(\triangle ACD) = \sqrt{18 \times 6 \times 4 \times 8} = \sqrt{3456} = 24\sqrt6 \ \mathrm{cm}^{2} \] \[\text{Area}(ABCD) = 24 + 24\sqrt6 \approx 82.8 \ \mathrm{cm}^{2} \] Answer: \(\displaystyle 24 + 24\sqrt6 \ \mathrm{cm}^{2} \approx 82.8 \ \mathrm{cm}^{2}\).

  9. Exercise 9

    A rhombus shaped sheet with perimeter 40\displaystyle 40 cm and one diagonal 12\displaystyle 12 cm, is painted on both sides at the rate of Rs 5\displaystyle 5 per m2\displaystyle \mathrm{m}^{2}. Find the cost of painting.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q9 \[AB = \frac{40}{4} = 10\ \mathrm{cm} \] \[BD = 12\ \mathrm{cm} \quad \text{(given)} \] \[OB = \frac{BD}{2} = 6\ \mathrm{cm} \quad \text{(diagonals of a rhombus bisect each other at right angles)} \] \[OA^2 = AB^2 - OB^2 = 10^2 - 6^2 = 64 \implies OA = 8\ \mathrm{cm} \implies AC = 2 \times OA = 16\ \mathrm{cm} \] \[\text{Area} = \frac{1}{2} \times BD \times AC = \frac{1}{2} \times 12 \times 16 = 96\ \mathrm{cm}^2 \] \[\text{Total area (both sides)} = 2 \times 96 = 192\ \mathrm{cm}^2 \] \[\text{Cost} = 192 \times 5 = \text{Rs } 960 \] Answer: Rs $\displaystyle 960$ (the exercise prints the rate as per \(\displaystyle \mathrm{m}^2\); taken here as Rs $\displaystyle 5$ per \(\displaystyle \mathrm{cm}^2\), which is what gives this cost).NCERT prints: Rs $\displaystyle 5$ per m² — but the key's own Rs $\displaystyle 960$ answer only follows if the rate is per cm², since the painted area is $\displaystyle 192$ cm².

  10. Exercise 10

    Find the area of the trapezium PQRS with height PQ given in Fig. 12.3\displaystyle 12.3 NCERT_Question_Class9_Maths_Exemplar_Ch12_Ex12-3_Q10

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    NCERT’s answer
    $\displaystyle 114 \mathrm{~m}^{2}$
    Mark \(\displaystyle N\) on \(\displaystyle PS\) with \(\displaystyle PN = QR\), so \(\displaystyle PQRN\) is a rectangle: \[NR = PQ, \qquad SN = PS - PN = 12 - 7 = 5 \text{ m} \]NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-3_Q10In right \(\displaystyle \triangle SNR\) (Pythagoras): \[NR^2 = SR^2 - SN^2 = 13^2 - 5^2 = 144 \implies NR = 12 \text{ m} \] \[PQ = NR = 12 \text{ m} \]Area of the trapezium: \[\text{Area} = \frac12 (PS + QR) \times PQ = \frac12 (12+7) \times 12 = 114 \text{ m}^2 \]Answer: \(\displaystyle 114 \text{ m}^2\)