Can you solve this question . if you can solve this question then you are perfect in Heron's formula chapter .
Answers
Step-by-step explanation:
It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm
Perimeter = 5+5+1 = 11 cm
So, semi perimeter = 11/2 cm = 5.5 cm
Using Heron’s formula,
Area = √[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)
For the quadrilateral II section:
This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.
∴ Area = 6.5×1 cm2=6.5 cm2
For the quadrilateral III section:
It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.
Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle
The perpendicular height of the parallelogram will be

= 0.86 cm
And, the area of the equilateral triangle will be (√3/4×a2) = 0.43
∴ Area of the trapezoid = 0.86+0.43 = 1.3 cm2 (approximately).
Answer:
Step-by-step explanation:
= 5+5+1 = 11 cm
So, semi perimeter = 11/2 cm = 5.5 cm
Using Heron’s formula,
Area = √[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)
For the quadrilateral II section:
This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.
∴ Area = 6.5×1 cm2=6.5 cm2
For the quadrilateral III section:
It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.
Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle
The perpendicular height of the parallelogram will be

= 0.86 cm
And, the area of the equilateral triangle will be (√3/4×a2) = 0.43
∴ Area of the trapezoid = 0.86+0.43 = 1.3 cm2 (approximately).