Question

Question 3
Radha made a picture of an aero plane with colored papers as shown in the given figure. Find the total area of the paper used.

Answer 1

For triangle I
This triangle is an isosceles triangle.
Perimeter = 2s = (5 +5 + 1) cm=11cm
$s=\frac{11cm}{2}=5.5cm$
Area of triangle
= $\sqrt{s(s-a)(s-b)(s-c)}=\lfloor \sqrt{5.5(5.5-5)(5.5-5)(5.5-1)}\rfloor c{m}^2=\lfloor \sqrt{\left(5.5\right)\left(0.5\right)\left(0.5\right)\left(4.5\right)}\rfloor c{m}^2=2.488c{m}^2\left(approximately\right)$

For quadrilateral II
This quadrilateral is a rectangle.
$Area=l\times b=(6.5\times 1)c{m}^2=6.5c{m}^2$

For quadrilateral III
This quadrilateral is a trapezium.
(Divide the trapezium into a triangle and parallelogram)
Perpendicular height of parallelogram=$\left(\sqrt{1^2-(0.5{)}^2}\right)cm$
$=\sqrt{0.75}cm=0.866cm$
Area of trapezium = Area of parallelogram +Area of equilateral triangle

$=\left(0.866\right)1+\frac{\sqrt{3}}{4}(1{)}^2=0.866+0.433=1.299c{m}^2$



Area of triangle(IV) =Area of triangle in (V)
$=(\frac{1}{2}\times 1.5\times 6)c{m}^2=4.5c{m}^2$
Total area of the paper used = $2.488+6.5+1.299+4.5\times 2$
$=19.287c{m}^2$