3. An isosceles triangle has sides length xcm xcm and 9cm.
B its perimeter is less than 24 and x is a whole number.
I find the lowest value of x.
II find the highest value of x
x+x+9 < 24
2x < 15
x < 7.5
so, x <= 7
The sides of length x must each be > 9/2, so x >= 5
5 <= x <= 7
what is the answer?
Answers
maths
2×x+9<24=2x+9÷2<24÷2=2x+9<12
X+x+9<24
2x<15
x<7 1/2
Lowest value 0
Highest 7
X+x+9<24
2x<15
5
To find the lowest value of x, we need to determine the maximum possible sum of the two equal sides of the isosceles triangle. Since the perimeter of the triangle is less than 24, the sum of the two equal sides must be less than half of the perimeter.
Let's assume the length of the two equal sides is x cm. Then the inequality representing the sum of the sides is:
2x + 9 < 24
Simplifying the inequality, we get:
2x < 24 - 9
2x < 15
Now, divide both sides of the inequality by 2:
x < 15/2
Since x must be a whole number, the highest whole number that is less than 15/2 is 7. Therefore, the lowest value of x in this case is 7.
To find the highest value of x, we need to determine the minimum possible sum of the two equal sides of the triangle. Since the perimeter of the triangle is less than 24, the sum of the two equal sides must be greater than half of the perimeter.
Again, assuming the length of the two equal sides is x cm, the inequality representing the sum of the sides is:
2x + 9 > 24
Simplifying the inequality, we get:
2x > 24 - 9
2x > 15
Now, divide both sides of the inequality by 2:
x > 15/2
The lowest whole number greater than 15/2 is 8. Therefore, the highest value of x in this case is 8.