PUMPA - SMART LEARNING

எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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In real life, not every object is in the shape of triangles. Some objects are in the shape of a quadrilateral. We use Heron's formula not only for finding the area of triangles but also we can use it for finding the area of quadrilaterals. For that, we need to divide the quadrilateral into two triangular parts and then use the formula of the area of the triangle.
 
Let's see one real-life problem based on the shape of quadrilateral.
Example:
Sakthi have one notebook, he tears one paper from the notebook. The torn paper is a kind of quadrilateral ABCD, has \angle D = 90^{\circ}, AB = 9 \ cm, BC = 8 \ cm, CD = 5 \ cm and AD = 12 \ cm. What is the area of the torn paper?
(Use: \sqrt{35} = 5.91)
 
Solution
 
Given: \angle D = 90^{\circ}, AB = 9 \ cm, BC = 8 \ cm, CD = 5 \ cm and AD = 12 \ cm
 
Construction: Join diagonal AC.
 
paper.png
 
Diagonal AC divides a quadrilateral into two triangular parts ACD and ABC.
 
Now, we need to find the length of AC.
 
Since \angle D = 90^{\circ}, \Delta ACD is a right angled triangle.
 
Using Pythagoras theorem:
 
AC^{2} = AD^{2} + DC^{2}
 
AC^{2} = 12^{2} + 5^{2}
 
AC^{2} = 144 + 25
 
AC^{2} = 169
 
Taking square root on both sides, we get:
 
AC = 13 \ cm
 
Let AD = a = 12 \ cm, DC = b = 5 \ cm and AC = c = 13 \ cm.
 
s=a+b+c2
 
s=12+5+132=302=15 cm
 
Area of \Delta ADC = ssasbsc
 
= 1515121551513
 
= 15×3×10×2
 
= 900
 
= \sqrt{30^2}
 
= 30 \ cm^2
 
Now, we need to find the area of \Delta ADC.
 
Let AB = a = 9 \ cm, BC = b = 8 \ cm and AC = c = 13 \ cm
 
s=9+8+132=302=15 cm
 
Area of \Delta ABC = ssasbsc
 
= 151591581513
 
= 15×6×7×2
 
= 32×22×5×7
 
= 3 \times 2 \sqrt{35}
 
= 6 \times 5.91       ( Since \sqrt{35} = 5.91)
 
= 35.46 \ cm^2
 
Area of ABCD = Area of \Delta ADC + Area of \Delta ABC
 
= 30 + 35.46
 
= 65.46 \ cm^2
 
Therefore, the area of the torn paper is 65.46 \ cm^2.