A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?
Solution:
We divide the kite into three different triangles, and by using Heron’s formula, we can calculate the area of triangle.
Heron's formula for the area of a triangle = √s(s - a)(s - b)(s - c)
Where a, b, and c are the sides of the triangle, and
s = Semi-perimeter = Half the Perimeter of the triangle = (a + b + c)/2
We know that the diagonals of a square are perpendicular bisectors of each other.
Given diagonal BD = AC = 32 cm, then OA = 1/2 AC = 16 cm.
So square ABCD is divided into two isosceles triangles ABD and CBD of base 32 cm and height 16 cm.
Area of ∆ABD = 1/2 × base × height
= (32 × 16)/2
= 256 cm2
Since the diagonal divides the square into two equal triangles. Therefore, Area of ∆ABD = Area of ∆CBD = 256 cm2
Now, for ∆CEF
Semi Perimeter(s) = (a + b + c)/2
s = (6 + 6 + 8)/2
s = 20/2
s = 10 cm
By using Heron’s formula,
Area of ∆CEF = √s(s - a)(s - b)(s - c)
= √10(10 - 6)(10 - 6)(10 - 8)
= √10 × 4 × 4 × 2
= 8√5
= 8 × 2.24
= 17.92 cm2
Thus, the area of the paper used to make region I = 256 cm2, region II = 256 cm2, and region III = 17.92 cm2.
☛ Check: NCERT Solutions Class 9 Maths Chapter 12
Video Solution:
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?
Class 9 Maths NCERT Solutions Chapter 12 Exercise 12.2 Question 7
Summary:
It is given that there is a kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades. We have found that the area of the paper of shade I = 256 cm2, shade II = 256 cm2, and shade III = 17.92 cm2 has been used in it.
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