find the area..
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Answers
Step-by-step explanation:
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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
Ncert solutions class 9 chapter 12-15
= 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).
For triangle IV and V:
These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm
Area triangles IV and V = 2×(½×6×1.5) cm2 = 9 cm2
So, the total area of the paper used = (2.488+6.5+1.3+9) cm2 = 19.3 cm2
Answer:
StFor the triangle I section:
Ncert solutions class 9 chapter 12-14
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
Ncert solutions class 9 chapter 12-15
= 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).
For triangle IV and V:
These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm
Area triangles IV and V = 2×(½×6×1.5) cm2 = 9 cm2
So, the total area of the paper used = (2.488+6.5+1.3+9) cm2 = 19.3 cm2
4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:
Given,
It is given that the parallelogram and triangle have equal areas.
The sides of the triangle are given as 26 cm, 28 cm and 30 cm.
So, the perimeter = 26+28+30 = 84 cm
And its semi perimeter = 84/2 cm = 42 cm
Now, by using Heron’s formula, area of the triangle =
Ncert solutions class 9 chapter 12-16
= √[42(42-26)(46-28)(46-30)] cm2
= √[46×16×14×16] cm2
= 336 cm2
Now, let the height of parallelogram be h.
As the area of parallelogram = area of the triangle,
28 cm× h = 336 cm2
∴ h = 336/28 cm
So, the height of the parallelogram is 12 cm.
5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Solution:
Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.
Ncert solutions class 9 chapter 12-17
Consider the triangle BCD,
Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m
Using Heron’s formula,
Area of the ΔBCD =
Ncert solutions class 9 chapter 12-18
= 432 m2
∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m2 = 864 m2
Thus, the area of the grass field that each cow will be getting = (864/18) m2 = 48 m2ep-by-step explanation: