A kite in the shape of a square with a diagonal 32cm and an isosceles triangle of base 8cm and side 6cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?
Answers
REQUIRED PROPERTIES OF SQUARE IN THIS QUESTION::
- Diagonals of square are equal
- Diagonals of square are perpendicular bisector of each other
So, we can take triangle CMB to consider the area of square.
In triangle CMB:-
MB = CM = 32 / 2 = 16cm
Area of CMB = 1/2 * base * Height = 1/2 * 16 * 16 = 118 sq cm
Area of triangle * 4 = Area of square
= 118 * 4
= 472 sq cm.
So in the kite area of paper for shaded region 1 and 2 = 236 sq cm for each.
If we look at the isosceles triangle at the bottom then lets draw a perpendicular at the base from A.
This line divides the base into two equal halves. Lets take the point on the base be O.
Lets mark the two points of the base of isosceles triangle as E and F.
In triangle AOF;
Angle O = 90°
So, if we apply pythagoreas theorem then;
AO² =OF² + AF²
AO² = 4² + 6²
AO = √52cm²
So the area of isosceles triangle = Area of paper used to shade =
2√52 cm².
HAPPY LEARNING MATE
Answer:
The perimeter of a triangle is equal to the sum of its three sides it is denoted by 2S.
2s=(a+b+c)
s=(a+b+c)/2
Here ,s is called semi perimeter of a triangle.
The formula given by Heron about the area of a triangle is known as Heron's formula.
According to this formula area of a triangle= √s (s-a) (s-b) (s-c)
Where a, b and c are three sides of a triangle and s is a semi perimeter.
This formula can be used for any triangle to calculate its area and it is very useful when it is not possible to find the height of the triangle easily .
Heron's formula is generally used for calculating area of scalene triangle.
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Solution:
Let the kite is made with square ABCD
& an isosceles ∆DEF.
Given, sides of a ∆DEF are DE=DF= 6cm & EF= 8cm
& Diagonal of a square ABCD= 32cm
We know that,
As the diagonals of a square bisect each other at right angle.
OA=OB=OC=OD=32/2=16cm
AO perpendicular BC & DO perpendicular BC
Area of region I = Area of ∆ABC= ½×BC×OA
[Area of right Triangle=1/2× base height]
Area of region I= ½×32×16=256cm²
Similarly area of region II = 256cm²
For the III section,
Now, in ∆DEF
let the sides a=6cm,b= 6cm & c=8cm
Semi perimeter of triangle,s = (6 + 6 + 8)/2 cm = 10cm
Using heron’s formula,
Area of the III triangular piece = √s (s-a) (s-b) (s-c)
= √10(10 – 6) (10 – 6) (10 – 8)
= √10 × 4 × 4 × 2
=√2×5×4×4×2
=√2×2×4×4×5
=2×4√5=8√5
= 8×2.24=17.92cm²
[√5= 2.24...]
Hence, area of paper of I colour used in making kite= 256cm²
Area of paper of II colour used in making kite= 256cm²
And area of paper of III colour used in making kite= 17.92cm²
Hope this will help you....
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