Herons Formula - Test Papers

 CBSE Test Paper 01

CH-12 Herons Formula


  1. The perimeter and area of a triangle whose sides are of lengths 3 cm, 4 cm and 5 cm respectively are
    1. 12 cm, 6 cm2
    2. 12 cm, 12 cm2
    3. 6 cm, 6 cm2
    4. 6 cm, 12 cm2
  2. Each equal side of an isosceles triangle is 13 cm and its base is 24 cm Area of the triangle is :
    1. 403cm2
    2. 253cm2
    3. 60 cm2
    4. 503cm2
  3. An isosceles right triangle has area 8 cm2. The length of its hypotenuse is
    1. 32 cm
    2. 24 cm
    3. 16 cm
    4. 48 cm
  4. If side of a scalene  is doubled then area would be increased by
    1. 200%
    2. 25 %
    3. 50 %
    4. 300 %
  5. One of the diagonals of a rhombus is 12cm and area is 96 sq cm. the perimeter of the rhombus is
    1. 72 cm
    2. 106 cm
    3. 40 cm
    4. 103 cm
  6. Fill in the blanks: The area of a triangle of base 35 cm is 420 cm2, then its altitude is ________ cm.
  7. Fill in the blanks: The altitude of an equilateral triangle ABC is ________.
  8. The base and the corresponding altitude of a parallelogram are 10 cm and 7 cm, respectively. Find its area.
  9. How many times area is changed, when sides of a triangle are doubled.
  10. Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.
  11. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
  12. The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle.
  13. The perimeter of a triangle is 300 m. If its sides are in the ratio 3 : 5 : 7 . Find the area of the triangle.
  14. A kite in the shape of a square with diagonal 32 cm and an isosceles triangle of base 8 cm and side 6 cm each is to be made of three different shades as shown in a figure. How much paper of each shade has been used in it? (Use 5 = 2.24)
  15. Two parallel side of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.

CBSE Test Paper 01
CH-12 Herons Formula


Solution

  1. (a) 12 cm, 6 cm2
    Explanation: Perimeter of triangle = 3 + 4 + 5 = 12 cm

    Now, s = 3+4+52=6 cm

    Area = s(sa)(sb)(sc)

    6(63)(64)(65)=6×3×2×1

    = 6 sq cm

  2. (c) 60 cm2

    Explanation:

    s = 13+13+242= 25 cm

    Area of triangle = s(sa)(sb)(sc)

    25(2513)(2513)(2524)

    25×12×12×1

    = 60 sq. cm

  3. (a) 32 cm
    Explanation:

    Area of isosceles triangle = 12x Base x Height

    Since in an isosceles triangle, Base and Height are equal.

    => 8 = 12x Base x Base

    => Base = Height = 4 cm

    Hypotenuse = 42+42 = 32 cm

  4. (d) 300 %
    Explanation:

    Area of triangle with sides a, b, c (A) = s(sa)(sb)(sc)

    New sides are 2a, 2b and 2c

    Then s=2a+2b+2c2=a+b+c

    =>     s' = 2s .............(i)

    New area = s(s2a)(s2b)(s2c)

    2s(2s2a)(2s2b)(2s2c)

    4s(sa)(sb)(sc)

    = 4A

    Increased area = 4A - A = 3A

    % of increased area = 3AA×100 = 300%

  5. (c) 40 cm
    Explanation:

    d2 = Area×2d1

    96×212

    =16 cm

    length of side of rhombus = 62+82 = 10 cm

    perimeter of rhombus = 4 x side

    = 4  x 10 = 40 cm 

  6. 24

  7. 32a

  8. The base of parallelogram =10 cm  and the corresponding altitude = 7 cm.
    Area of parallelogram = Base × Corresponding altitude
    = 10 × 7 = 70 cm2.

  9. Area of triangle = 12  × base  × height = 12 × b  × h
    If new base B = 2b and height  H = 2h
    New area of triangle   = 12 × B  × H   = 12 × 2b  × 2h =4( 12 × b  × h) = 4 (area of triangle)
    So doubling the sides leads to 4 times the area.

  10. Let a = 9 cm, b = 12 cm and c = 15 cm
    Since, 2s = a + b + c
     s = 12(a + b + c)
    12(9 + 12 + 15)
    12(36) = 18 cm
    Now, area of triangle = s(sa)(sb)(sc)
    18(189)(1812)(1815)
    18×9×6×3
    = 54 cm2


  11. a = 12 cm, b = 12 cm
    Perimeter = 30 cm
    a + b + c = 30
    ⇒ 12 + 12 + c = 30
    ⇒ 24 + c = 30
    ⇒ c = 30 – 24
    ⇒ c = 6 cm
    s = 302 cm = 15 cm
    ∴ Area of the triangle =s(sa)(sb)(sc)
    =15(1512)(1512)(156)
    =15(3)(3)(9)=915 cm2

  12. Given: base of a right-angled triangle = 4 cm and hypotenuse =  5 cm.
    In right-angled triangle ABC

    AB+ BC2 = AC2 (By Pythagoras Theorem)
     AB2 + 42 = 52
     AB= 25 - 16 = 9
     AB = 3 cm
     Area of ΔABC=12BC×AB=12×4×3=6cm2
    Hence area of given right-angled  triangle is 6 cm2.


  13. Suppose that the sides in metres are 3x, 5x and 7x.
    Then, we know that 3x + 5x + 7x = 300 (Perimeter of the triangle)
    Therefore, 15x = 300, which gives x = 20.
    So the sides of the triangles are 3 × 20 m, 5 × 20 m and 7 × 20 m
    i.e., 60m, 100m and 140m.
    We have s = 60+100+1402 = 150 m
    and area will be = 150(15060)(150100)(150140)
    150×90×50×10 
    = 15003 m2

  14. Here ABCD be the square and CEF be an isosceles triangle.
    Let the diagonals bisect each other at O.
    Then,
    AO =12× 32 cm
    = 16 cm
    Area of shaded portion I = 12×32×16 sq cm
    = 256 sq cm
    Similarly, Area of shaded portion II = 12×32×16 sq cm
    = 256 sq cm
    And, for triangle
    base, a = 8 cm
    and side, b = 6 cm
    Area of portion III =  a44b2a2
    =844×(6)264
    =214464
    =85
    = 17.92 sq cm
    Thus, the papers of three shades required are 256 sq cm, 256 sq cm and 17.92 sq cm.


  15. Given that,
    AB = 77 cm, CD = 60 cm, BC = 26 cm and AD = 25 cm
    Now, DE  AB and CF  AB is drawn.
     EF = DC = 60 cm
    Let AE = x
     BF = 77 - 60 - x = (17 - x)
    In ADE, DE2 = AD2 - AE2 [pythagoras theorem]
    = 252 - x2
    And