A RIGHT ANGLED TRIANGLE ABC IS GIVEN,WHERE AB IS 5CM,AC IS 12CM,BC IS 13CM,AD IS AN ALTITUDE FROM A TO BC,FIND THE AREA OF THE TRIANGLE AND FIND AD ALSO.
area = (1/2)base x height
= (1/2)(5)(12) = 30
We could also consider BC as the base, then AD becomes the height
so (1/2)(13)(AD) = 30
AD = 2(30)/13
= 60/13
30cm and60\13
your ans will may be wrong because height and base could not accept on those questions database
Here, area=30
Ad=4.6
Ad=4.6
To find the area of a right-angled triangle, we can use the formula:
Area = (base * height) / 2
In this case, AB is the base and AD is the height. To find AD, we can use the Pythagorean Theorem, which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Now, let's calculate AD and find the area of the triangle.
Step 1: Calculate AD
Using the Pythagorean theorem, we can say that:
AC^2 = AD^2 + CD^2
where CD represents the segment of BC, known as the base.
In a right-angled triangle, when the altitude is dropped from the right angle to the hypotenuse, it divides the hypotenuse into two segments, which we can call p and q. In our case, AD is the altitude, so we can write CD as BC - AD.
Plugging in the given values, we have:
12^2 = AD^2 + (13 - AD)^2
Simplifying the equation:
144 = AD^2 + (169 - 26AD + AD^2)
144 = 2AD^2 - 26AD + 169
2AD^2 - 26AD + 25 = 0
This equation is quadratic. We can solve it using factoring, completing the square, or the quadratic formula.
Solving the equation using factoring:
(2AD - 1)(AD - 25) = 0
Since AD cannot be negative, we have:
AD = 1/2 cm or AD = 25 cm
But in our case, AD represents the altitude of the triangle, so we take AD = 5 cm as the correct answer.
Step 2: Calculate the area of the triangle
Using the formula for the area of a triangle, we have:
Area = (AB * AD) / 2
Area = (5 * 5) / 2
Area = 25 / 2
Area = 12.5 cm^2
Therefore, the area of the given right-angled triangle ABC is 12.5 square centimeters, and the altitude AD is 5 centimeters.