How much paper of each shade is needed to make a kite given in Fig. 12.4, in which ABCD is a square with diagonal 44 cm.
Solution:
Given, ABCD is a square with diagonal 44 cm
We have to find the quantity of paper of each shade needed to make a kite as shown in
the given figure.
We know that all the sides of a square are equal.
AB = BC = CD = AD
In triangle ACD,
Given, AC = 44 cm
By using Pythagorean theorem,
AC² = AD² + DC²
Since AD = DC
(44)² = AD² + AD²
2AD² = 1936
AD² = 1936/2
AD² = 968
Taking square root,
AD = √22 × 44
= √2 × 11 × 4 × 11
= (2 × 11)√2
AD = 22√2 cm
Area of square = side × side
Area of square ABCD = AB × CD
= 22√2 × 22√2
= 968 cm²
From the figure,
We observe that the square is divided into four equal parts.
2 yellow parts, 1 green part and 1 red part.
Area of green region = 968/4 = 242 cm²
Area of red region = 968/4 = 242 cm²
Area of 2 yellow regions = 968/2 = 484 cm²
Area of green part = area of triangle PCQ
In triangle PCQ,
PC = QC = 20 cm
PQ = 14 cm
This implies PCQ is an isosceles triangle
Area of isosceles triangle = a/4 √4b² - a²
Here, a = 14 cm and b = 20 cm
= 14/4 √4(20)² - (14)²
= 7/2 √4(400) - 196
= 7/2 √1600 - 196
= 7/2 √1404
= 7/2 (37.46)
= 131.04 cm²
Area of green part = 131.04 cm²
Therefore, area of green part = 242 + 131.04
= 373.04 cm²
Therefore, the paper required for each shade to make the kite is
Yellow shade = 484 cm²
Red shade = 242 cm²
Green shade = 373.04 cm²
✦ Try This: Two parallel sides of a trapezium are 60 cm and 77 cm and the other sides are 25 cm and 26 cm. Find the area of the trapezium.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 12
NCERT Exemplar Class 9 Maths Exercise 12.4 Problem 1
How much paper of each shade is needed to make a kite given in Fig. 12.4, in which ABCD is a square with diagonal 44 cm.
Summary:
The quantity of paper of each shade is needed to make a kite given in Fig. 12.4, in which ABCD is a square with diagonal 44 cm is Yellow shade = 484 cm², Red shade = 242 cm², Green shade = 373.04 cm²
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