Lesson Solved problems on area of right-angled triangles

This Lesson (Solved problems on area of right-angled triangles) was created by by ikleyn(52750)

: View Source, Show
About ikleyn:

Solved problems on area of right-angled triangles

Problem 1

Find the area of the right-angled triangle, if its legs are of 5 cm and 8 cm long.

Solution

The area of a right angled triangle is half the product of the measures of its legs. So, in our case the area is 5%2A8%2F2

= 20

.
Answer. 20 cm%5E2

Problem 2

Find the length of the altitude of a right-angled triangle drawn to the hypotenuse, if the legs have the measures of

and

units.

Solution

Since the measures of the legs of the right-angled triangle are

and

, its hypotenuse has the measure

in accordance with the Pythagorean Theorem (see the lesson The Pythagorean Theorem under the topic Pythagorean Theorem of the section Geometry in this site). Now, we can write the equation

a%2Ab%2F2

for the area of a right-angled triangle (lesson Formulas for area of a triangle under the topic Area and Surface Area of the section Geometry in this site). In this equation

is the length of the altitude of the right-angled triangle drawn to the hypotenuse. From the equation

.

This is the required expression for the altitude of a right-angled triangle via its legs. The solution is completed.

There are two other solutions of the Problem 2 in this site.
The solution in the lesson Problems on similarity for right-angled triangles is based on the triangles similarity.
The solution in the lesson Altitude drawn to the hypotenuse of a right triangle uses the "first principles". It is based on the Pythagorean Theorem.

Problem 3

In a right-angled triangle the altitude drawn to the hypotenuse divides it in segments of

and

units long.
Prove that the measure of the altitude

is the Geometric mean of the measures of these segments:

Solution

Let ABC be a right-angled triangle with the right angle C (Figure 3).
Let CD be the altitude in the triangle ABC drawn from the right angle
vertex C to the hypotenuse AB, which divides the hypotenuse in the
segments AD and BD of the length

and

respectively.

We need to prove that the measure

of the altitude CD is the
Geometric mean of the measures

and

. (1)

Figure 3. To the Problem 3

Let

and

are the measures of the legs AC and BC of the triangle ABC.
From the solution of the previous Problem 1

. (2)
Hence, from the right-angled triangle ADC p%5E2

in accordance with the Pythagorean Theorem.
From the right-angled triangle BDC q%5E2

by the same reason.
It implies that sqrt%28p%2Aq%29

. In turn, a%2Ab%2Fsqrt%28a%5E2+%2B+b%5E2%29

due to (2).
Thus it is proved that h = sqrt%28p%2Aq%29

. The solution is completed.

There are two other solutions of the Problem 3 in this site.
The solution in the lesson Problems on similarity for right-angled triangles is based on the triangles similarity.
The solution in the lesson Altitude drawn to the hypotenuse of a right triangle uses the "first principles". It is based on the Pythagorean Theorem.

Problem 4

Find the area of a right-angled triangle, if the altitude drawn to the hypotenuse divides the hypotenuse in segments of 18 cm and 32 cm long.

Solution

First, calculate the full length of the hypotenuse of the triangle. It is 18 cm + 32 cm = 50 cm.

Second, calculate the measure of the altitude drawn to the hypotenuse from the right-angle vertex.
Apply the result of the Problem 3 above. The measure of the altitude drawn to the hypotenuse in a right-angled triangle is the Geometric mean of the measures of
the segments the altitude divides the hypotenuse.

So,

= 3*8 = 24 cm.

Now, the area of the triangle is 50%2A24%2F2

= 600

.

Answer. The area of the triangle is 600 cm%5E2

This lesson has been accessed 3408 times.