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NCERT Exemplar · Class 9 Mathematics Heron's Formula

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EXERCISE 12.1 1–9 (part 1 of 4)

  1. Exercise 1

    An isosceles right triangle has area 8 cm2\displaystyle 8 \mathrm{~cm}^{2}. The length of its hypotenuse is (A) 32 cm\displaystyle \sqrt{32} \mathrm{~cm} (B) 16 cm\displaystyle \sqrt{16} \mathrm{~cm} (C) 48 cm\displaystyle \sqrt{48} \mathrm{~cm} (D) 24 cm\displaystyle \sqrt{24} \mathrm{~cm}

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    NCERT’s answer
    (A)
    (A) \(\displaystyle \sqrt{32}\ \text{cm}\)Legs \(\displaystyle a\) equal (isosceles right triangle). \[\frac{1}{2}a^2 = 8\ \text{cm}^2 \quad \Rightarrow \quad a^2 = 16 \] \[\text{hypotenuse}^2 = a^2+a^2 = 2a^2 = 32 \] \[\text{hypotenuse} = \sqrt{32}\ \text{cm} \]

  2. Exercise 2

    The perimeter of an equilateral triangle is 60\displaystyle 60 m . The area is (A) 103 m2\displaystyle 10 \sqrt{3} \mathrm{~m}^{2} (B) 153 m2\displaystyle 15 \sqrt{3} \mathrm{~m}^{2} (C) 203 m2\displaystyle 20 \sqrt{3} \mathrm{~m}^{2} (D) 1003 m2\displaystyle 100 \sqrt{3} \mathrm{~m}^{2}

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 100\sqrt{3}\ \text{m}^2\)Equilateral triangle, perimeter $\displaystyle 60$ m. \[a = \frac{60}{3} = 20\ \text{m} \] \[\text{Area} = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}(20)^2 = 100\sqrt{3}\ \text{m}^2 \]

  3. Exercise 3

    The sides of a triangle are 56\displaystyle 56 cm, 60\displaystyle 60 cm and 52\displaystyle 52 cm long. Then the area of the triangle is (A) 1322 cm2\displaystyle 1322 \mathrm{~cm}^{2} (B) 1311 cm2\displaystyle 1311 \mathrm{~cm}^{2} (C) 1344 cm2\displaystyle 1344 \mathrm{~cm}^{2} (D) 1392 cm2\displaystyle 1392 \mathrm{~cm}^{2}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 1344\ \text{cm}^2\)Heron's formula for sides $\displaystyle 56$, $\displaystyle 60$, $\displaystyle 52$ cm. \[s = \frac{56+60+52}{2} = 84\ \text{cm} \] \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{84\times28\times24\times32} \] \[\text{Area} = \sqrt{1806336} = 1344\ \text{cm}^2 \]

  4. Exercise 4

    The area of an equilateral triangle with side 23 cm\displaystyle 2 \sqrt{3} \mathrm{~cm} is (A) 5.196 cm2\displaystyle 5.196 \mathrm{~cm}^{2} (B) 0.866 cm2\displaystyle 0.866 \mathrm{~cm}^{2} (C) 3.496 cm2\displaystyle 3.496 \mathrm{~cm}^{2} (D) 1.732 cm2\displaystyle 1.732 \mathrm{~cm}^{2}

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 5.196\ \text{cm}^2\)Equilateral triangle, side \(\displaystyle 2\sqrt{3}\) cm. \[\text{Area} = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}\left(2\sqrt{3}\right)^2 = \frac{\sqrt{3}}{4}(12) \] \[\text{Area} = 3\sqrt{3} \approx 5.196\ \text{cm}^2 \]

  5. Exercise 5

    The length of each side of an equilateral triangle having an area of 93 cm2\displaystyle 9 \sqrt{3} \mathrm{~cm}^{2} is (A) 8\displaystyle 8 cm (B) 36\displaystyle 36 cm (C) 4\displaystyle 4 cm (D) 6\displaystyle 6 cm

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 6\ \text{cm}\)Equilateral triangle, area \(\displaystyle 9\sqrt{3}\ \text{cm}^2\). \[\frac{\sqrt{3}}{4}a^2 = 9\sqrt{3} \quad \Rightarrow \quad a^2 = 36 \] \[a = 6\ \text{cm} \]

  6. Exercise 6

    If the area of an equilateral triangle is 163 cm2\displaystyle 16 \sqrt{3} \mathrm{~cm}^{2}, then the perimeter of the triangle is (A) 48\displaystyle 48 cm (B) 24\displaystyle 24 cm (C) 12\displaystyle 12 cm (D) 36\displaystyle 36 cm

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 24\ \text{cm}\)Equilateral triangle, area \(\displaystyle 16\sqrt{3}\ \text{cm}^2\). \[\frac{\sqrt{3}}{4}a^2 = 16\sqrt{3} \quad \Rightarrow \quad a^2 = 64,\ a = 8\ \text{cm} \] \[\text{Perimeter} = 3a = 24\ \text{cm} \]

  7. Exercise 7

    The sides of a triangle are 35\displaystyle 35 cm, 54\displaystyle 54 cm and 61\displaystyle 61 cm, respectively. The length of its longest altitude (A) 165 cm\displaystyle 16 \sqrt{5} \mathrm{~cm} (B) 105 cm\displaystyle 10 \sqrt{5} \mathrm{~cm} (C) 245 cm\displaystyle 24 \sqrt{5} \mathrm{~cm} (D) 28\displaystyle 28 cm

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 24\sqrt{5}\ \text{cm}\)Heron's formula for sides $\displaystyle 35$, $\displaystyle 54$, $\displaystyle 61$ cm. \[s = \frac{35+54+61}{2} = 75\ \text{cm} \] \[\text{Area} = \sqrt{75\times40\times21\times14} = 420\sqrt{5}\ \text{cm}^2 \] Longest altitude falls on the shortest side, $\displaystyle 35$ cm. \[h = \frac{2\times\text{Area}}{35} = \frac{840\sqrt{5}}{35} = 24\sqrt{5}\ \text{cm} \]

  8. Exercise 8

    The area of an isosceles triangle having base 2\displaystyle 2 cm and the length of one of the equal sides 4\displaystyle 4 cm, is (A) 15 cm2\displaystyle \sqrt{15} \mathrm{~cm}^{2} (B) 152 cm2\displaystyle \sqrt{\frac{15}{2}} \mathrm{~cm}^{2} (C) 215 cm2\displaystyle 2 \sqrt{15} \mathrm{~cm}^{2} (D) 415 cm2\displaystyle 4 \sqrt{15} \mathrm{~cm}^{2}

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    NCERT’s answer
    (A)
    (A) \(\displaystyle \sqrt{15}\ \text{cm}^2\)Heron's formula for sides $\displaystyle 4$, $\displaystyle 4$, $\displaystyle 2$ cm. \[s = \frac{4+4+2}{2} = 5\ \text{cm} \] \[\text{Area} = \sqrt{5\times1\times1\times3} = \sqrt{15}\ \text{cm}^2 \]

  9. Exercise 9

    The edges of a triangular board are 6\displaystyle 6 cm, 8\displaystyle 8 cm and 10\displaystyle 10 cm. The cost of painting it at the rate of 9\displaystyle 9 paise per cm2\displaystyle \mathrm{cm}^{2} is (A) Rs 2.00\displaystyle 2.00 (B) Rs 2.16\displaystyle 2.16 (C) Rs 2.48\displaystyle 2.48 (D) Rs 3.00\displaystyle 3.00

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    NCERT’s answer
    (B)
    (B) Rs $\displaystyle 2.16$Sides $\displaystyle 6$, $\displaystyle 8$, $\displaystyle 10$ cm form a right triangle. \[6^2+8^2 = 10^2 \] \[\text{Area} = \frac{1}{2}\times6\times8 = 24\ \text{cm}^2 \] \[\text{Cost} = 24\times9 = 216\ \text{paise} = \text{Rs } 2.16 \]