It is a right angle triangle with an area of: 0.5*7*9 = 31.5 square units
A triangle with three sides all congruent to each other is an equilateral triangle.A triangle with two of its sides congruent to each other is an isosceles triangle.If you have a triangle with sides {a1,b1,c1} and another triangle with sides {a2,b2,c2} and: a1 = a2; b1 = b2 & c1 = c2, then these two triangles are congruent.
To find the SA of any prism, find the area of each face. (*Note: Depending on the shape of the face, there might be different formulas. Area of Square=length x length Area of Triangle=base x height divided by 2 Area of Trapezoid=A1+A2 (divide the trapezoid into two triangles and find the area of each triangle, then add them) ) After you find the areas of all the faces of the prism, add all the areas to get the SA.
Area of circle = PI * R2In A1 enter radius of circle (e.g. 4).In B1 enter this formula: =PI()*A1^2Observe in B1 the number as the area of a circle with radius of 4.
Since the area is (length of base)*(height)/2 {call these dimensions B & H) A1 = B*H/2 With dimensions doubled, A2 = (2*B)*(2*H)/2 = 4*B*H/2 = 4*A1. By not simplifying to 2*B*H, it's easier to see that it is four times the original area. It is 4 times because the two length dimensions are multiplied, and 2 * 2 = 4.
The area of A0 is 1 metre2. A1 has half that area, A2 is a half of A1, and so on. Thus the area of A4 is 1/16 m2. If the paper has a mass of 80 grams per m2 then 1 sheet of A4 will have a mass of 5 grams.
Suppose the polygon's vertices are denoted A1A2...An. We can fix one endpoint of the diagonal at A1 and count the number of ways to choose from the remaining vertices. There are n-3 vertices to choose from, because we cannot count A1, nor count A2 or An (since they are directly next to A1). There are n vertices total, so we multiply to get n(n-3). However, we have to divide by 2 because we count each diagonal twice. The diagonal AiAj is the same as AjAi, and each is counted. Therefore, the number of diagonals in an n-gon is n(n-3)/2.
the periods in this method are to organize the spacing. public String getTriangleType(int a1, int a2, int 23) { ....if (a1+a2+a3!=180) return "Invalid Triangle"; ....if (a1==90 a2==90 a3==90) return "Right"; ....if (a1>90 a2>90 a3>90) return "Obtuse"; ....return "acute"; }
A triangle with three sides all congruent to each other is an equilateral triangle.A triangle with two of its sides congruent to each other is an isosceles triangle.If you have a triangle with sides {a1,b1,c1} and another triangle with sides {a2,b2,c2} and: a1 = a2; b1 = b2 & c1 = c2, then these two triangles are congruent.
1. sort desc, so a1 >= a2 >= a3 2. check if (a1*a1 == a2*a2 + a3*a3) then true
To find the SA of any prism, find the area of each face. (*Note: Depending on the shape of the face, there might be different formulas. Area of Square=length x length Area of Triangle=base x height divided by 2 Area of Trapezoid=A1+A2 (divide the trapezoid into two triangles and find the area of each triangle, then add them) ) After you find the areas of all the faces of the prism, add all the areas to get the SA.
Area of circle = PI * R2In A1 enter radius of circle (e.g. 4).In B1 enter this formula: =PI()*A1^2Observe in B1 the number as the area of a circle with radius of 4.
First, there is Alaska Electric, serving primarily Juneau and its surrounding area. There is also A1 electric, which serves the Anchorage area. AA electric also serves the same area as A1 electric.
#include<stdio.h> main() { int radius; float a1; float area(int r); printf("Enter the r value"); scanf("%d",&radius); a1=area(radius); printf("The answer=%d",a1); } float area(int r) { float a2; a2=3.14*r*r; return(a2); } getch();
Right formula is = V = [L X B X d] + h / 3 [ A1 + A2 + ROUTE UNDER A1 A2 ] where , l = length , b = breadth , d = bottom depth , h = D - d , A1 = bottom area , A2= top area (ii)is = V = [L X B X d] + h/6(A1 + A2 + 4 Am) where , h = D - d , A1 = bottom area , A2= top area, Am = Mean of Top and Bottom AreaRoute Under means Square route
Let A1 be the area of the first square and X the length of its side Let A2 be the area of the second square and Y the length of its side A1 = X2 , A2 = Y2 , A1/A2 = X2/Y2 , Square Root (A1/A2) = Square Root ( X2/Y2), X/Y = Square Root (A1/A2) = Square Root (2/1) = 1.414
Since the area is (length of base)*(height)/2 {call these dimensions B & H) A1 = B*H/2 With dimensions doubled, A2 = (2*B)*(2*H)/2 = 4*B*H/2 = 4*A1. By not simplifying to 2*B*H, it's easier to see that it is four times the original area. It is 4 times because the two length dimensions are multiplied, and 2 * 2 = 4.
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