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NCERT Exemplar · Class 9 Mathematics Heron's Formula

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EXERCISE 12.4 1–8 (part 4 of 4)

  1. Exercise 1

    How much paper of each shade is needed to make a kite given in Fig. 12.4\displaystyle 12.4, in which ABCD is a square with diagonal 44\displaystyle 44 cm. NCERT_Question_Class9_Maths_Exemplar_Ch12_Ex12-4_Q1

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    Diagonals of square \(\displaystyle ABCD\) are equal and bisect each other at right angles. \[\text{Area of square } ABCD = \frac{1}{2}\,(44)^2 = 968 \text{ cm}^2 \] They split it into four congruent triangles at \(\displaystyle O\): \[\text{ar}(\triangle AOB) = \text{ar}(\triangle BOC) = \text{ar}(\triangle COD) = \text{ar}(\triangle DOA) = \frac{968}{4} = 242 \text{ cm}^2 \] NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-4_Q1 Yellow is regions I and II, two of these triangles: \[\text{Yellow} = 2 \times 242 = 484 \text{ cm}^2 \] Red is region IV, one triangle: \[\text{Red} = 242 \text{ cm}^2 \] Green is region III plus the tail \(\displaystyle \triangle CEF\), where \(\displaystyle CE=CF=20\) cm and \(\displaystyle EF=14\) cm: \[s = \frac{20+20+14}{2} = 27 \text{ cm} \] \[\text{ar}(\triangle CEF) = \sqrt{27 \times 7 \times 7 \times 13} = 21\sqrt{39} \approx 131.14 \text{ cm}^2 \] \[\text{Green} = 242 + 131.14 = 373.14 \text{ cm}^2 \] Answer: Yellow \(\displaystyle 484 \text{ cm}^2\), Red \(\displaystyle 242 \text{ cm}^2\), Green \(\displaystyle \approx 373.14 \text{ cm}^2\).NCERT prints: Green: $\displaystyle 373.04$ m² — an arithmetic misprint ($\displaystyle 242$ + $\displaystyle 131.14$ = $\displaystyle 373.14$) and a unit misprint, since the diagonal is $\displaystyle 44$ cm, not m.

  2. Exercise 2

    The perimeter of a triangle is 50\displaystyle 50 cm. One side of a triangle is 4\displaystyle 4 cm longer than the smaller side and the third side is 6\displaystyle 6 cm less than twice the smaller side. Find the area of the triangle.

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    NCERT’s answer
    $\displaystyle 20 \sqrt{30} \mathrm{~cm}^{2}$
    Let the smallest side be \(\displaystyle x\) cm; the other sides are \(\displaystyle x+4\) and \(\displaystyle 2x-6\) cm. \[x + (x+4) + (2x-6) = 50 \] \[4x - 2 = 50 \quad\Rightarrow\quad x = 13 \] The sides are \(\displaystyle 13, 17, 20\) cm. \[s = \frac{13+17+20}{2} = 25 \text{ cm} \] \[\text{Area} = \sqrt{25 \times 12 \times 8 \times 5} = \sqrt{12000} = 20\sqrt{30} \approx 109.54 \text{ cm}^2 \] Answer: \(\displaystyle 20\sqrt{30} \approx 109.54 \text{ cm}^2\).

  3. Exercise 3

    The area of a trapezium is 475 cm2\displaystyle 475 \mathrm{~cm}^{2} and the height is 19\displaystyle 19 cm. Find the lengths of its two parallel sides if one side is 4\displaystyle 4 cm greater than the other.

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    NCERT’s answer
    $\displaystyle 23$ cm, $\displaystyle 27$ cm
    \[\text{Area} = \frac{1}{2}(a+b)h \quad\Rightarrow\quad 475 = \frac{1}{2}(a+b)(19) \] \[a+b = \frac{950}{19} = 50 \text{ cm} \] With \(\displaystyle a = b+4\): \[b + (b+4) = 50 \quad\Rightarrow\quad b = 23, \quad a = 27 \] Answer: the parallel sides are \(\displaystyle 23\) cm and \(\displaystyle 27\) cm.

  4. Exercise 4

    A rectangular plot is given for constructing a house, having a measurement of 40\displaystyle 40 m long and 15\displaystyle 15 m in the front. According to the laws, a minimum of 3\displaystyle 3 m , wide space should be left in the front and back each and 2\displaystyle 2 m wide space on each of other sides. Find the largest area where house can be constructed.

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    The margins leave \(\displaystyle 3\) m off the \(\displaystyle 40\) m length at each end and \(\displaystyle 2\) m off the \(\displaystyle 15\) m width at each side. \[\text{Length available} = 40 - 3 - 3 = 34 \text{ m} \] \[\text{Width available} = 15 - 2 - 2 = 11 \text{ m} \] NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-4_Q4 \[\text{Largest area} = 34 \times 11 = 374 \text{ m}^2 \] Answer: \(\displaystyle 374 \text{ m}^2\).NCERT prints: $\displaystyle 374$ cm² — a misprint: the plot's dimensions are given in metres, so the area is $\displaystyle 374$ m², not cm².

  5. Exercise 5

    A field is in the shape of a trapezium having parallel sides 90\displaystyle 90 m and 30\displaystyle 30 m. These sides meet the third side at right angles. The length of the fourth side is 100\displaystyle 100 m . If it costs Rs 4\displaystyle 4 to plough 1 m2\displaystyle 1 \mathrm{~m}^{2} of the field, find the total cost of ploughing the field.

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    NCERT’s answer
    Rs $\displaystyle 19200$
    Draw \(\displaystyle CE \perp BD\), meeting it at \(\displaystyle E\). Since \(\displaystyle AC \parallel BD\) and both \(\displaystyle AB, CE \perp BD\), \(\displaystyle ABEC\) is a rectangle, so \(\displaystyle CE = AB\) and \(\displaystyle ED = BD - AC = 90 - 30 = 60\) m. NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-4_Q5 \[CD^2 = CE^2 + ED^2 \quad \text{(Pythagoras in } \triangle CED\text{)} \] \[100^2 = CE^2 + 60^2 \quad\Rightarrow\quad CE = 80 \text{ m} \] \[\text{Area} = \frac{1}{2}(BD+AC)\times CE = \frac{1}{2}(90+30)(80) = 4800 \text{ m}^2 \] \[\text{Cost} = 4800 \times 4 = \text{Rs } 19200 \] Answer: Rs \(\displaystyle 19200\).

  6. Exercise 6

    In Fig. 12.5\displaystyle 12.5, ΔABC\displaystyle \Delta \mathrm{ABC} has sides AB=7.5 cm,AC=6.5 cm\displaystyle \mathrm{AB}=7.5 \mathrm{~cm}, \mathrm{AC}=6.5 \mathrm{~cm} and BC=7 cm\displaystyle \mathrm{BC}=7 \mathrm{~cm}. On base BC a parallelogram DBCE of same area as that of ΔABC\displaystyle \Delta \mathrm{ABC} is constructed. Find the height DF of the parallelogram. NCERT_Question_Class9_Maths_Exemplar_Ch12_Ex12-4_Q6

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    NCERT’s answer
    $\displaystyle 3$ cm
    \[s = \frac{7.5+7+6.5}{2} = 10.5 \text{ cm} \] \[\text{ar}(\triangle ABC) = \sqrt{10.5 \times 3.5 \times 4 \times 3} = \sqrt{441} = 21 \text{ cm}^2 \] NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-4_Q6 Parallelogram \(\displaystyle DBCE\) stands on the same base \(\displaystyle BC\) and has the same area as \(\displaystyle \triangle ABC\): \[BC \times DF = \text{ar}(\triangle ABC) \] \[7 \times DF = 21 \quad\Rightarrow\quad DF = 3 \text{ cm} \] Answer: \(\displaystyle DF = 3\) cm.

  7. Exercise 7

    The dimensions of a rectangle ABCD are 51\displaystyle 51 cm × 25\displaystyle 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9\displaystyle 9 : 8\displaystyle 8, is cut off from the rectangle as shown in the Fig. 12.6. If the area of the trapezium PQCD is 56\displaystyle \frac{5}{6} th part of the area of the rectangle, find the lengths QC and PD. NCERT_Question_Class9_Maths_Exemplar_Ch12_Ex12-4_Q7

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    NCERT’s answer
    $\displaystyle 45$ cm, $\displaystyle 40$ cm
    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-4_Q7 \[\text{ar}(ABCD) = AB \times BC = 25 \times 51 = 1275 \text{ cm}^2 \] \[QC = 9x, \quad PD = 8x \quad (\text{ratio } 9:8) \] Height of the trapezium is \(\displaystyle AB\), since \(\displaystyle AB \perp AD\) and \(\displaystyle AB \perp BC\). \[\text{ar}(PQCD) = \frac{1}{2}(QC+PD)\times AB = \frac{1}{2}(17x)(25) = \frac{425x}{2} \] \[\frac{425x}{2} = \frac{5}{6}\times 1275 = 1062.5 \] \[x = 5 \] \[QC = 9x = 45 \text{ cm}, \quad PD = 8x = 40 \text{ cm} \] Answer: \(\displaystyle QC = 45\) cm, \(\displaystyle PD = 40\) cm.

  8. Exercise 8

    A design is made on a rectangular tile of dimensions 50\displaystyle 50 cm × 70\displaystyle 70 cm as shown in Fig. 12.7. The design shows 8\displaystyle 8 triangles, each of sides 26\displaystyle 26 cm, 17\displaystyle 17 cm and 25\displaystyle 25 cm. Find the total area of the design and the remaining area of the tile. NCERT_Question_Class9_Maths_Exemplar_Ch12_Ex12-4_Q8

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    NCERT’s answer
    $\displaystyle 1632 \mathrm{~cm}^{2}, 1868 \mathrm{~cm}^{2}$
    NCERT_Solution_Class9_Maths_Exemplar_Ch12_Ex12-4_Q8 \[s = \frac{26+17+25}{2} = 34 \text{ cm} \] \[\text{ar(one triangle)} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{34\times 8\times 17\times 9} = \sqrt{41616} = 204 \text{ cm}^2 \] \[\text{ar(design)} = 8\times 204 = 1632 \text{ cm}^2 \] \[\text{ar(tile)} = 50\times 70 = 3500 \text{ cm}^2 \] \[\text{remaining area} = 3500 - 1632 = 1868 \text{ cm}^2 \] Answer: design area \(\displaystyle =1632\) cm\(\displaystyle ^2\); remaining area \(\displaystyle =1868\) cm\(\displaystyle ^2\).