he i shall be giving some impt question on maths that i think can come in the exam.
Q.1 Find the area of an Airplane made by Radha such that it can be divided in 5 portions. A triangle with sides 5cm, 5cm and 1cm. A rectangle with side 6.5cm and 1cm. 2 right angled triangles with base as 1.5cm and 6cm. and the last is a trapezium with sides 1cm 1cm 1cm and 2cm where the parallel sides are 1 and 2cm.
Soln 1.
Region 1.
s = 5 + 5 + 1 / 2 (by herons formula)
s = 11/2
Area = under root of [11/2 * (11/2 - 5)(11/2 - 5)(11/2 - 1)]
Area = under root of [11/2 * 1/2 * 1/2 * 9/2]
Area = 3/4 * root of 11
Area = 2.49cm2
Region 3 n 3
Again easy…….
Area of right angled triangle = 1/2 * base * height
Area of right angled triangle = 1/2 * 6 * 1.5
Area of right angled triangle = 4.5cm2
that means, area of region 4 and 5 will be same. so Area of region 5 will also be 4.5cm2
Region 5
Attention needed !!!
a trapezium can be divided into 1 triangle and 1 parallelogram.
Triangle
Area of equilateral triangle = root 3 / 4 * side * side
Area of equilateral triangle = root 3 / 4 * 1 * 1
Area of equilateral triangle = root 3 / 4
now
root 3 / 4 = 1/2 * b * h …(bcoz area of same triangle by any method will be same)
root 3 / 4 = 1/2 * 1 * h
root 3 / 4 * 2 = h
h = root 3 / 2
now we have got our altitude…. now we can find out the area of 5
Area of trapezium = 1/2 * sum of parallel sides * altitude
Area of trapezium = 1/2 * (2 + 1) * root 3 / 2
Area of trapezium = 1/2 * 3 * 1.73/2 (put 1.73 in place of root 3)
Area of trapezium = 1.3cm2 (after solving)




















