Question

Can anyone help me with this Q.3

Answer 1

Answer

Length of the sides of the triangle section I = 5cm, 1cm and 5cm
Perimeter of the triangle = 5 + 5 + 1 = 11cm
Semi perimeter = 11/2 cm = 5.5cm
Using heron's formula,
Area of section I  = √s (s-a) (s-b) (s-c)
                                       = √5.5(5.5 - 5) (5.5 - 5) (5.5 - 1) cm2
                                       = √5.5 × 0.5 × 0.5 × 4.5 cm2
                                       = 0.75√11 cm2 = 0.75 × 3.317cm2 = 2.488cm2 (approx)
Length of the sides of the rectangle of section I = 6.5cm and 1cm
Area of section II = 6.5 × 1 cm2 =  6.5 cm2
Section III is an isosceles trapezium which is divided into 3 equilateral of side 1cm each.
Area of the trapezium = 3 × √3/4 × 12 cm2 = 1.3 cm2 (approx)
Section IV and V are 2 congruent right angled triangles with base 6cm and height 1.5cm
Area of region IV and V = 2 × 1/2 × 6 × 1.5cm2 = 9cm2
Total area of the paper used = (2.488 + 6.5 + 1.3 + 9)cm2 = 19.3 cm2

Answer 2

first you have to find the area of the triangle above with the heroines formula that is a you will take the a side as 5,5 and one moving on you will find the area of the second part which is a rectangle using the formula length × breadth in which you will take the length as 6.5 cm and the breadth as 1 cm then you will find the area of the third portion with the help of the formula to find the area of trapezieum
then find the area of the 5th part using the formula height X half X base.because the IV portion is equal to V portion then both of their area will be same and then you have to add up the area of all the portions and then you will get the answer
HOPE IT HELPED YOU....