Find the area of the Figur given in ncert book Exercise 12.2 Question Number 3 Page no. 206 Please illustrate the calculations.

Asked by Shivam Saxena | 14th Aug, 2010, 04:41: PM

Expert Answer:

There are 5 areas.Area I : It is an isosceles triangle with equal side 5cm and base 1cm.Using Heron's formula , s=5+5+12=112s(s-a)(s-b)(s-c)=112(112-5)(112-5)(112-1)                                    = 112121292=9916Therefore Area I = square root of 9916=3114=2.5(approx)Area II :It is a rectangle with sides 6.5cm and 1cm.Therefore its area = 6.5x1=132x1=132=6.5sq cmArea III : This is an isosceles trapezium with parallel sides 1cm and 2 cm. Therefore its area = 12xheight (sum of parallel sides)Find the height of the trapezium by drawing perpendicular on the base and using pythagoras theorem.In the right -angled triangle, hyotenuse = 1 cm and base = 0.5 cmtherefore (height)2 = 12-122= 34.  height =h = sq rt 32Area III =12xsq rt 32x(1+2)=1.3sq cm (aprox)Area IV and V :These are two right angled triangles with base 6 cm and height 1.5 cm Therefore the area of one triangle=12x6x32=92=4.5.The total Area IV and V = 2x 4.5=9sq cmThus the area of the complete figure is sum of all the areas=2.5+6.5+1.3+9 =19.3 sq cm (aprox)">

Answered by  | 14th Aug, 2010, 05:51: PM

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