The sides of a quadrangular field, taken in order are 29 m, 36 m, 7m
and 24 m respectively. The angle contained by the last

2 thoughts on “The sides of a quadrangular field, taken in order are 29 m, 36 m, 7m<br /> and 24 m respectively. The angle contained by the last”

1. ➡️ The sides of the quadrangular field taken in order are 26,27,7,24m respectively. The angle contained by the last 2 sides is a right angle. … So, the diagonal of a right angled triangle will be the third side of the normal triangle. So, calculate the third side using Pythagoras theorem as √AD2+CD2=AC.

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2. Answer:

   • Area of the quadrangular field is 444 m².

   Step-by-step explanation:

   Given that:

   • The sides of a quadrangular field, taken in order are 29 m, 36 m, 7 m and 24 m respectively.
   • The angle contained by the last two sides is a right angle.

   To Find:

   • The area of the field.

   Solution:

   As the sides are provided in order. Therefore, the length of the last two sides are 7 m and 24 m respectively.

   According to the question the last two sides form a right angle (90°).

   Finding diagonal of the field:

   As the last two sides of the field form a right angle. Including diagonal the it would form a right angle triangle. Therefore, we can find the diagonal using the Pythagoras theorem.

   As we know that:

   In a right angle triangle using Pythagoras theorem,

   [tex]\sf{\circ\;(Hypotenuse)^2=(Base)^2+(Height)^2}[/tex]

   Here,

   • Hypotenuse = Diagonal of the field
   • Base = 24 m
   • Height = 7 m

   Let us assume:

   • The diagonal of the field be d.

   Substituting the values,

   [tex]\sf{\longmapsto (d)^2=(24)^2+(7)^2}[/tex]

   [tex]\sf{\longmapsto (d)^2=576+49}[/tex]

   [tex]\sf{\longmapsto (d)^2=625}[/tex]

   [tex]\sf{\longmapsto d=\sqrt{625}}[/tex]

   [tex]\sf{\longmapsto d=25}[/tex]

   Hence,

   • Diagonal of the park = 25 m

   As, the field is divided into two triangles.

   Now, we need to find the areas of both the triangles and add them to get the total area of the field.

   Finding area of 1st triangle:

   • 1st side = 29 m
   • 2nd side = 36 m
   • 3rd side = 25 m

   As we know that:

   In a triangle using Heron’s formula:

   [tex]\sf{\circ\;Area\;of\;the\;triangle=\sqrt{s(s-a)(s-b)(s-c)}\;sq.\;units}[/tex]

   Where,

   • s = Semi-perimeter
   • a = Length of 1st side
   • b = Length of 2nd side
   • c = Length of 3rd side

   Finding semi-perimeter of the triangle:

   Semi-perimeter = Perimeter/2

   = (29 + 36 + 25) m/2

   = 90 m/2

   = 45 m

   Substituting the values,

   [tex]\sf{Area=\sqrt{45(45-29)(45-36)(45-25)}}[/tex]

   [tex]\sf{=\sqrt{45\times16\times9\times20}}[/tex]

   [tex]\sf{=\sqrt{129600}}[/tex]

   [tex]\sf{=360}[/tex]

   Hence, area of the 1st triangle is 360 m².

   Finding area of 2nd triangle:

   • 1st side = 7 m
   • 2nd side = 24 m
   • 3rd side = 25 m

   Finding semi-perimeter of the triangle:

   Semi-perimeter = Perimeter/2

   = (7 + 24 + 25) m/2

   = 56 m/2

   = 28 m

   Finding area of the 2nd triangle using Heron’s formula:

   [tex]\sf{Area=\sqrt{28(28-7)(28-24)(28-25)}}[/tex]

   [tex]\sf{=\sqrt{28\times21\times4\times3}}[/tex]

   [tex]\sf{=\sqrt{7056}}[/tex]

   [tex]\sf{=84}[/tex]

   Hence, area of the 2nd triangle is 84 m².

   Finding area of the quadrangular field:

   Ar. of the field = (Ar. of 1st Δ) + (Ar. of 2nd Δ)

   = (360 + 84) m²

   = 444 m²

   Hence, area of the field = 444 m²

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