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  1. Asked: In:

    A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

    parth joshi

    Draw CF ∥ AD and CG ⊥AB. In quadrilateral ADCF, CF ∥ AD [∵ By construction] CD ∥ AF [∵ ABCD is a trapezium] Therefore, ADCF is a parallelogram. So, AD = CF = 13 m and CD = AF = 10 m [∵ Opposite sides of a parallelogram] Therefore, BF = AB - AF = 25 - 10 = 15 m Here, the Sides of triangle are a = 13Read more

    Draw CF ∥ AD and CG ⊥AB.
    In quadrilateral ADCF,
    CF ∥ AD [∵ By construction]
    CD ∥ AF [∵ ABCD is a trapezium]
    Therefore, ADCF is a parallelogram. So,
    AD = CF = 13 m and CD = AF = 10 m [∵ Opposite sides of a parallelogram]
    Therefore, BF = AB – AF = 25 – 10 = 15 m
    Here, the Sides of triangle are a = 13 m, b = 14 m and c = 15 m.
    So, the semi- perimeter of triangle
    S = a + b + c/2 = 13 + 14 + 15/2 = 42/2 = 21 m.
    Therefore, using Heron’s formula, area of triangle BCF = √s(s -a)(s -b)(s -c)
    = √21(21 -13)(21 -14)(21 – 15)
    = √21(8)(7)(6)
    = √7056
    = 84 m²
    But, the area of triangle BCF = 1/2 BF × CG
    So, 1/2 × BF × CG = 84
    ⇒ 1/2 × 15 × CG = 84
    ⇒ CG = (84×2/15) = 11.2 m
    Therefore, area of trapezium ABCD = 1/2 × (AB + CD) × CG
    = 1/2 × (25 + 10) × 11.2
    = 35 × 5.6
    196 m²
    Hence, the area of the field is 196 m².

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  2. Asked: In:

    A park, in the shape of a quadrilateral ABCD, has angle C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

    parth joshi

    In quadrilateral ABCD, join BD. In triangle BDC, by Pythagoras theorem BD² BC² + CD² ⇒ BD² = 12² + 5² = 144 + 25 = 169 ⇒ BD = 13 cm Area of triangle BDC = 1/2 × BC × DC = 1/2 × 12 × 5 = 30 cm² Here, the sides of triangle ABD are a = 9 cm, b = 8 cm and c = 13 cm. So, the semi-perimeter of triangle ABRead more

    In quadrilateral ABCD, join BD.
    In triangle BDC, by Pythagoras theorem
    BD² BC² + CD²
    ⇒ BD² = 12² + 5² = 144 + 25 = 169
    ⇒ BD = 13 cm
    Area of triangle BDC = 1/2 × BC × DC = 1/2 × 12 × 5 = 30 cm²
    Here, the sides of triangle ABD are a = 9 cm, b = 8 cm and c = 13 cm.
    So, the semi-perimeter of triangle ABD s = ABD s = a+b+c/2 = 9+8+13/2 = 30/2 = 15 cm
    Therefore, using Heron’s formula area of triangle ABD = √s(s -a)(s -b)(s -c)
    = √15(15 -9)(15 -8)(15 -13)
    = √15(6)(7)(2) = √1260 = 35.5 cm²(approx.)
    Total area of park = 30 + 3535 = 65.5 cm²
    Hence, total area of park is 65.5 cm².

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  3. Asked: In:

    Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

    parth joshi

    Join diagonal AC of quadrilateral ABCD. Here, the sides of triangle ABC are a = 3 cm, b = 4 cn and c = 5 cm. So, the semi-perimeter of triangle S = a+b+c/2 = 3 + 4 + 5/2 = 12/2 = 6 cm Therefore, using Heron's formula area of triangle = √s(s -a)(s -b)(s -c) = √6(6 -3)(6 -4)(6 -5) = √6(3)(2)(1) = √36Read more

    Join diagonal AC of quadrilateral ABCD.
    Here, the sides of triangle ABC are a = 3 cm, b = 4 cn and c = 5 cm.
    So, the semi-perimeter of triangle S = a+b+c/2 = 3 + 4 + 5/2 = 12/2 = 6 cm
    Therefore, using Heron’s formula area of triangle = √s(s -a)(s -b)(s -c)
    = √6(6 -3)(6 -4)(6 -5) = √6(3)(2)(1) = √36 = 6 cm²
    And the sides of triangle ACD are a’ = 4 cm, b’ = 5 cn and c’ = 5 cm.
    So, the semi-perimeter of triangle S’ = a’ + b’ + c’/2 = 4 + 5 + 5/2 = 14/2 = 7 cm
    Therefore, using Heron’s formula area of triangle = √s'(s’ – a’)(s’ -b’)(s’ -c’)
    = √7(7 – 4)(7 – 5)(7 – 5)
    = √7(3)(2)(2)
    = 2√21 = 9.2 cm²(approx.)
    Total area of quadrilateral = Area of triangle ABC + Area of triangle ACD
    ⇒ Total area of quadrilateral ABCD = 6 + 9.2 = 15.2 cm²
    Hence, the area of quadrilateral ABCD is 15.2 cm².

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  4. Asked: In:

    A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

    parth joshi

    Here, the side of triangle are a, a and units. So, the semi-perimeter of triangle is given by s = a+a+a/2 = 3a/2 Therefore, using Heron's formula, the area of triangle = √s(s - a) (s - b) (s - c) = √((3a/2(3a/2) - a)(3a/2) - a)) = √((3a/2(3a - 2a/2)(3a - 2a/2)(3a - 2a/2)) = √3a/2(a/2)(a/2)(a/2) = (aRead more

    Here, the side of triangle are a, a and units.
    So, the semi-perimeter of triangle is given by s = a+a+a/2 = 3a/2
    Therefore, using Heron’s formula, the area of triangle = √s(s – a) (s – b) (s – c)
    = √((3a/2(3a/2) – a)(3a/2) – a)) = √((3a/2(3a – 2a/2)(3a – 2a/2)(3a – 2a/2))
    = √3a/2(a/2)(a/2)(a/2) = (a²/4)√3
    Perimeter of equilaterral triangle = 3a
    According to question, 3a = 180 cm ⇒ a = 180/3 = 60 cm
    Therefore, the area of triangle = (a²/4)√3 = ((60)²/4))√3 = (3600/4)√3 = 900 √3 cm².

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  6. Asked: In:

    Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

    parth joshi

    Here, the sides of triangle are a = 18 cm, b = 10 cm and primeter is 42 cm. We Know that the parimeter of triangle = a + b + c ⇒ 42 = 18 + 10 + c ⇒ c = 14 cm So, the semi- perimeter of triangle is given by S = a + b + c/2 = 42/2 = 21 cm Therefore, using Heron's formula, the area of triangle = √s(s -Read more

    Here, the sides of triangle are a = 18 cm, b = 10 cm and primeter is 42 cm.
    We Know that the parimeter of triangle = a + b + c
    ⇒ 42 = 18 + 10 + c
    ⇒ c = 14 cm
    So, the semi- perimeter of triangle is given by
    S = a + b + c/2 = 42/2 = 21 cm
    Therefore, using Heron’s formula, the area of triangle = √s(s – a)(s – b)(s – c)
    = √21(21 – 18)(21 – 10)(21 – 14)
    = √21(3)(11)(7)
    = √7 × 3 × (3)(11)(7)
    = 7 × 3√11 = 21√11 cm²
    Hence, the area of triangle is 21√11 cm².

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