1. Find the cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of ₹ 3 per m².
a = 51 m, b = 37 m, c = 20 m
semiperimeter: p = (51+37+20):2 = 54 m
Area of triangle:
Rate: ₹ 3 per m².
Cost: ₹ 3•306 = ₹ 918
2. Find the area of the isosceles triangle whose perimeter is 11 cm and the base is 5 cm.
a = 5 cm
a+2b = 11 cm ⇒ 2b = 6 cm ⇒ b = 3 cm
p = 11:2 = 5.5
3. Find the area of the equilateral triangle whose each side is 8 cm.
a = b = c = 8 cm
p = (8•3):2 = 12 cm
4. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.
a = 2x
b = 3x
2x + 2•3x = 32 cm ⇒ 8x = 32 cm ⇒ x = 4 cm ⇒ a = 8 cm, b = 12 cm
p = 32:2 = 16 cm
The figure consists of a rectangle and a circle.
first lets find the sides of the triangle
sides of the triangle are 4+4=8 and 3+1=4
therefore area of the rectangle is 8*4=32
now area of semicircle =πr^2/2 = π2^2/2=2π=6.3
we have to add the area of rectangle with the area of semicircle
this is because it the given figure no part of area of circle and rectangle are in common.
therefore total area of given figure = 32+6.3=38.3
Answer: 11
Step-by-step explanation:
It is 11 because count from -6...
-6 = 0
-5 = 1
-4 = 2
-3 = 3
-2 = 4
-1 = 5
0 = 6
1 = 7
2 = 8
3 = 9
4 = 10
5 = 11
Hope this helpsand sorry if not
A+7=0
A+7-7=0-7
A=-7
B+3=7
B+3-3=7-3
B=4
Postulate1:A line contains atleast 2 points.Postulate1a:A plane contains at least 3 points not all on one line.Postulate1b:Space contains atleast four points not all on one plane
