Question

Length of two sides of a field are 12 m and 9 m. Other two sides are 10 m and 7 m. Angle between first two sides is a right angle. What is area of this field?( √6= 2.45).​

Answer 1

Answer:

Step-by-step explanation:

The sides of a quadrilateral ABCD are 6cm,8cm,12cm and 14cm (taken in order).

i.e. ∠ABC=90  

o

 

Now, let us join the points A & C.  

So, we get two triangles namely △ABC (a right angled triangle) and △ACD.  

Applying Pythagoras theorem in △ABC, we get  

AC=  

AB  

2

+BC  

2

 

​

=  

6  

2

+8  

2

 

​

=  

36+64

​

=  

100

​

=10cm  

So, the area of △ABC=  

2

1

​

×base×height

=  

2

1

​

×AB×BC  

=  

2

1

​

×6×8=24  

Now, in △ACD, we have  

AC=10cm,CD=12cm,AD=14cm.  

According to Heron's formula the area of triangle (A)=  

[s(s−a)(s−b)(s−c)]

​

 

where, 2s=(a+b+c).  

Here, a=10cm,b=12cm,c=14cm  

s=  

2

(10+12+14)

​

=  

2

36

​

=18  

Area of △ACD=  

[18×(18−10)(18−12)(18−14)]

​

 

=  

(18×8×6×4)

​

 

=  

(2×3×3×2×2×2×2×3×2×2)

​

 

=  

[(2×2×2×2×2×2×3×3)×2×3]

​

 

=2×2×2×3×  

6

​

 

=24  

6

​

 

So, total area of quadrilateral ABCD =△ABC+△ACD

=24+24  

6

​

 

=24(  

6

​

+1)

Answer 2

Given:

Length of one side = 12m

Length of second side = 9m

Length of third side = 10m

Length of fourth side = 7m

Angle between the first two sides = 90°

To find:

Area of the field.

Solution:

Let the field be in the shape of a quadrilateral ABCD where

AB = 12cm, BC = 9m, CD = 10m and DA = 7m

The angle between AB and BC,

∠ABC = 90°

On drawing a diagonal from point A to point C, we get two triangles,

ΔABC and ΔADC

ΔABC is a right-angled triangle. By Pythagoras theorem,

$AC^{2} =AB^{2} +BC^{2}$

$AC^{2} =12^{2}+ 9^{2}$

$AC^{2} =144+81$

$AC^{2} =225$

$AC=15m$

Area of ΔABC = $\frac{1}{2}$ × base × height

$ar(ABC)=\frac{1}{2}*9*12=54m^{2}$

In the second triangle, ΔADC all the 3 sides are given. Using Heron's formula, we can find the area of the triangle.

Perimeter of ΔADC = $7+10+15=32$

Semiperimeter, $s=\frac{Perimeter}{2} =\frac{32}{2}=16$

Area of ΔADC = $\sqrt{s(s-a)(s-b)(s-c)}$

This is Heron's formula.

where $a,b,c$ are the three sides of the triangle

$ar(ADC)=\sqrt{16(16-15)(16-10)(16-7)}$

$ar(ADC)=\sqrt{16(1*6*9)}$

$ar(ADC)=\sqrt{16*54}=\sqrt{864}$

$ar(ADC)=29.4m^{2}$

Area of the field = Area of ΔABC + Area of ΔADC

Area of the field = $54+29.4=83.4m^{2}$

Hence, the area of the field is 83.4m²

Thus, option (b) is correct.

The area of the field is 83.4m²

Option (b) is the correct answer.