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NCERT Exemplar · Class 9 Mathematics Heron's Formula

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EXERCISE 12.2 1–9 (part 2 of 4)

  1. Write True or False and justify your answer:

    Exercise 1

    The area of a triangle with base 4\displaystyle 4 cm and height 6\displaystyle 6 cm is 24 cm2\displaystyle 24 \mathrm{~cm}^{2}.

    Matches the book, not yet reviewed

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

    NCERT’s answer
    False, area of the triangle is $\displaystyle 12 \mathrm{~cm}^{2}$.
    False — the area is \(\displaystyle 12\ \mathrm{cm}^2\), not \(\displaystyle 24\ \mathrm{cm}^2\). \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] \[= \frac{1}{2} \times 4 \times 6 = 12\ \mathrm{cm}^2 \]

  2. Exercise 2

    The area of △ABC\displaystyle \triangle \mathrm{ABC} is 8 cm2\displaystyle 8 \mathrm{~cm}^{2} in which AB=AC=4 cm\displaystyle \mathrm{AB}=\mathrm{AC}=4 \mathrm{~cm} and ∠A=90∘\displaystyle \angle \mathrm{A}=90^{\circ}.

    Matches the book, not yet reviewed

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

    NCERT’s answer
    True, area of the triangle $\displaystyle =\frac{1}{2} \times 4 \times 4=8 \mathrm{~cm}^{2}$
    True — \(\displaystyle AB\) and \(\displaystyle AC\) are perpendicular legs, \(\displaystyle 4\ \mathrm{cm}\) each.NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-2_Q2\[\text{Area} = \frac{1}{2} \times AB \times AC \] \[= \frac{1}{2} \times 4 \times 4 = 8\ \mathrm{cm}^2 \]

  3. Exercise 3

    The area of the isosceles triangle is 5411 cm2\displaystyle \frac{5}{4} \sqrt{11} \mathrm{~cm}^{2}, if the perimeter is 11\displaystyle 11 cm and the base is 5\displaystyle 5 cm.

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

    NCERT’s answer
    True, Each of equal side = $\displaystyle 3$ cm.
    True — equal sides \(\displaystyle 3\ \mathrm{cm}\) each; Heron's formula gives \(\displaystyle \frac{5}{4}\sqrt{11}\ \mathrm{cm}^2\). \[\text{Equal sides} = \frac{11-5}{2} = 3\ \mathrm{cm}\ \text{each} \quad \text{(perimeter 11, base 5)} \] \[s = \frac{11}{2} = 5.5\ \mathrm{cm} \] \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5.5 \times 2.5 \times 2.5 \times 0.5} \] \[= \sqrt{\frac{275}{16}} = \frac{5}{4}\sqrt{11}\ \mathrm{cm}^2 \]

  4. Exercise 4

    The area of the equilateral triangle is 203 cm2\displaystyle 20 \sqrt{3} \mathrm{~cm}^{2} whose each side is 8\displaystyle 8 cm.

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

    NCERT’s answer
    False, area of the triangle $\displaystyle 16 \sqrt{3} \mathrm{~cm}^{2}$.
    False — the area is \(\displaystyle 16\sqrt3\ \mathrm{cm}^2\), not \(\displaystyle 20\sqrt3\ \mathrm{cm}^2\). \[\text{Area} = \frac{\sqrt3}{4} a^2 \quad \text{(equilateral triangle, side } a\text{)} \] \[= \frac{\sqrt3}{4} \times 8^2 = 16\sqrt3\ \mathrm{cm}^2 \]

  5. Exercise 5

    If the side of a rhombus is 10\displaystyle 10 cm and one diagonal is 16\displaystyle 16 cm, the area of the rhombus is 96 cm2\displaystyle 96 \mathrm{~cm}^{2}.

    Matches the book, not yet reviewed

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

    NCERT’s answer
    True, the other diagonal will be $\displaystyle 12$ cm.
    True — the other diagonal is \(\displaystyle 12\ \mathrm{cm}\), so the area is \(\displaystyle 96\ \mathrm{cm}^2\). NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-2_Q5 \[OA^2 + OB^2 = AB^2 \quad \text{(diagonals of a rhombus bisect each other at right angles)} \] \[8^2 + OB^2 = 10^2 \implies OB = 6 \implies BD = 2 \times OB = 12 \] \[\text{Area} = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 16 \times 12 = 96\ \mathrm{cm}^2 \]

  6. Exercise 6

    The base and the corresponding altitude of a parallelogram are 10\displaystyle 10 cm and 3.5\displaystyle 3.5 cm, respectively. The area of the parallelogram is 30 cm2\displaystyle 30 \mathrm{~cm}^{2}.

    Matches the book, not yet reviewed

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

    NCERT’s answer
    False, the area of the parallelogram is $\displaystyle 35 \mathrm{~cm}^{2}$.
    False — the area is \(\displaystyle 10 \times 3.5 = 35\ \mathrm{cm}^2\), not \(\displaystyle 30\ \mathrm{cm}^2\). \[\text{Area} = \text{base} \times \text{height} = 10 \times 3.5 = 35\ \mathrm{cm}^2 \]

  7. Exercise 7

    The area of a regular hexagon of side ' a\displaystyle a ' is the sum of the areas of the five equilateral triangles with side a\displaystyle a.

    Matches the book, not yet reviewed

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

    NCERT’s answer
    False, area is the sum of all the six equilateral triangles.
    False — a regular hexagon splits into six equilateral triangles of side \(\displaystyle a\), not five. NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-2_Q7 \[\text{Area(hexagon)} = 6 \times \frac{\sqrt3}{4} a^2 \quad \text{(six equilateral triangles from the centre)} \]

  8. Exercise 8

    The cost of levelling the ground in the form of a triangle having the sides 51\displaystyle 51 m, 37\displaystyle 37 m and 20\displaystyle 20 m at the rate of Rs 3\displaystyle 3 per m2\displaystyle \mathrm{m}^{2} is Rs 918.

    Matches the book, not yet reviewed

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

    NCERT’s answer
    True, area $\displaystyle =306 \mathrm{~m}^{2}$.
    True — the area is \(\displaystyle 306\ \mathrm{m}^2\), and \(\displaystyle 306 \times 3 = \text{Rs } 918\). \[s = \frac{51+37+20}{2} = 54\ \mathrm{m} \] \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{54 \times 3 \times 17 \times 34} = 306\ \mathrm{m}^2 \] \[\text{Cost} = 306 \times 3 = \text{Rs } 918 \]

  9. Exercise 9

    In a triangle, the sides are given as 11\displaystyle 11 cm, 12\displaystyle 12 cm and 13\displaystyle 13 cm. The length of the altitude is 10.25\displaystyle 10.25 cm corresponding to the side having length 12\displaystyle 12 cm.

    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.

    True — \(\displaystyle h = \sqrt{105} \approx 10.25\ \mathrm{cm}\). \[s = \frac{11+12+13}{2} = 18 \] \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18 \times 7 \times 6 \times 5} = 6\sqrt{105}\ \mathrm{cm}^2 \] \[\text{Area} = \frac{1}{2} \times 12 \times h \implies h = \sqrt{105}\ \mathrm{cm} \approx 10.25\ \mathrm{cm} \]NCERT prints: area = $\displaystyle 12$√$\displaystyle 105$ cm² — a misprint: that value gives h = $\displaystyle 2$√$\displaystyle 105$ ≈ $\displaystyle 20.5$ cm, not the $\displaystyle 10.25$ cm the question itself states.