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Sagot :
Answer:
The height is 12 cm
Step-by-step explanation:
Hi there!
We are given a triangle with the 3 sides marked as 15, 20, and 25 and we want to find the height of it (marked as x in the problem).
This problem can seem a bit difficult, but let's see if the triangle is a right triangle first off.
One way to figure out if it is a right triangle is to apply the converse of the Pythagorean theorem.
Let's label the sides, where a is the shortest side, b is the second shortest side, and c is the longest side:
a=15
b=20
c=25
Now square a and b, then add the result together. If it's the same as c squared, then the triangle is a right triangle
15²+20²=25²
225+400=625
625=625
So we can safely say that the triangle is a right triangle
This makes the problem way easier, as there are 2 ways to find the area of a right triangle:
The first way is to multiply the legs (the sides that make up the right angle) together, then divide the result by 2
The other way is to multiply the height and the hypotenuse (the side OPPOSITE to the right angle) together, and then divide the result by 2
First, we need to figure out which sides are the legs, and which side is the hypotenuse
By the triangle inequality theorem, the hypotenuse of a right triangle is the longest side, which means that the 25 cm side is the hypotenuse, and that leaves 15 cm and the 20 cm sides as the legs
So let's find the area of the triangle using the legs
A=[tex]\frac{15*20}{2}[/tex]=[tex]\frac{300}{2}[/tex]=150
So the area of the triangle is 150 cm²
However, as mentioned above, we can also find the area of the triangle by multiplying the hypotenuse by the base, then dividing the result by 2
Which means that the area is also:
A=[tex]\frac{25x}{2}[/tex] cm²
As these both equal the area of the triangle, we can set them equal to each other. This is possible via a property known as transitivity (if a=b and b=c, then a=c)
[tex]\frac{25x}{2}=150[/tex]
Multiply both sides by 2
25x=300
Divide both sides by 25
x=12 cm
So the height of the triangle is 12 cm
Hope this helps!
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