Lesson Solved problems on area of right-angled triangles

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<H2>Solved problems on area of right-angled triangles</H2>

<H3>Problem 1</H3>Find the area of the right-angled triangle, &nbsp;if its legs are of &nbsp;5 cm&nbsp; and &nbsp;8 cm&nbsp; long.

Solution

The area of a right angled triangle is half the product of the measures of its legs. &nbsp;So, &nbsp;in our case the area is &nbsp;{{{5*8/2}}} = 20 {{{cm^2}}}.

Answer. &nbsp;20 {{{cm^2}}}.

<H3>Problem 2</H3>Find the length of the altitude of a right-angled triangle drawn to the hypotenuse, &nbsp;if the legs have the measures of &nbsp;{{{a}}}&nbsp; and &nbsp;{{{b}}}&nbsp; units.

Solution

Since the measures of the legs of the right-angled triangle are &nbsp;{{{a}}}&nbsp; and &nbsp;{{{b}}}, &nbsp;its hypotenuse has the measure &nbsp;{{{c}}} = {{{sqrt(a^2 + b^2)}}}&nbsp; in accordance with the &nbsp;Pythagorean Theorem&nbsp; (see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Pythagorean-theorem/The-Pythagorean-Theorem.lesson>The Pythagorean Theorem</A>&nbsp; under the topic &nbsp;Pythagorean Theorem&nbsp; of the section &nbsp;Geometry&nbsp; in this site). &nbsp;Now, &nbsp;we can write the equation

{{{a*b/2}}} = {{{c*h/2}}}

for the area of a right-angled triangle &nbsp;(lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Formulas-for-area-of-a-triangle.lesson>Formulas for area of a triangle</A>&nbsp; under the topic &nbsp;Area and Surface Area&nbsp; of the section &nbsp;Geometry&nbsp; in this site). &nbsp;In this equation &nbsp;{{{h}}}&nbsp; is the length of the altitude of the right-angled triangle drawn to the hypotenuse. &nbsp;From the equation

{{{h}}} = {{{a*b/c}}} = {{{a*b/sqrt(a^2 + b^2)}}}.

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

There are two other solutions of the &nbsp;Problem 2&nbsp; in this site.

The solution in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/Problems-on-similarity-for-right-angled-triangles.lesson>Problems on similarity for right-angled triangles</A>&nbsp; is based on the triangles similarity.

The solution in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Altitude-drawn-to-the-hypotenuse-in-a-right-triangle.lesson>Altitude drawn to the hypotenuse of a right triangle</A>&nbsp; uses the "first principles". &nbsp;It is based on the &nbsp;Pythagorean Theorem.

<H3>Problem 3</H3>In a right-angled triangle the altitude drawn to the hypotenuse divides it in segments of &nbsp;{{{p}}}&nbsp; and &nbsp;{{{q}}}&nbsp; units long.

Prove that the measure of the altitude &nbsp;{{{h}}}&nbsp; is the &nbsp;Geometric mean&nbsp; of the measures of these segments: &nbsp;{{{h}}} = {{{sqrt(p*q)}}}.

<TABLE> 
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 <TD>
Solution

Let &nbsp;ABC&nbsp; be a right-angled triangle with the right angle &nbsp;C &nbsp;(Figure 3). &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

Let &nbsp;CD&nbsp; be the altitude in the triangle &nbsp;ABC&nbsp; drawn from the right angle

vertex &nbsp;C&nbsp; to the hypotenuse &nbsp;AB, &nbsp;which divides the hypotenuse in the

segments &nbsp;AD&nbsp; and &nbsp;BD&nbsp; of the length &nbsp;{{{p}}}&nbsp; and &nbsp;{{{q}}}&nbsp; respectively.

We need to prove that the measure &nbsp;{{{h}}}&nbsp; of the altitude &nbsp;CD&nbsp; is the

Geometric mean&nbsp; of the measures &nbsp;{{{p}}}&nbsp; and &nbsp;{{{q}}}&nbsp;: &nbsp;{{{h}}} = {{{sqrt(p*q)}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)

</TD>
 <TD>

{{{drawing( 350, 150, -1.0, 6.0, -0.5, 2.5,

line(  0,  0,  5,  0), 
            line(  0,  0,  3.43,  2.08),
            line(  5,  0,  3.43,  2.08),

locate ( -0.1,  -0.05,    A),
            locate (  4.9,  -0.05,    B),
            locate (  3.33, 2.45, C),

line(  3.42,  2.08,  3.42,  0),
            circle ( 3.43, 0.0, 0.05, 0.05),
            locate ( 3.33, -0.05, D),

locate (3.47, 1.27, h),
            locate (1.45, 1.27, a),
            locate (4.3,  1.27, b),
            locate (1.7, -0.05, p),
            locate (4.1, -0.05, q)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 3. &nbsp;To the Problem 3

</TD>
 </TR>
</TABLE>

Let &nbsp;{{{a}}}&nbsp; and &nbsp;{{{b}}}&nbsp; are the measures of the legs &nbsp;AC&nbsp; and &nbsp;BC&nbsp; of the triangle &nbsp;ABC.

From the solution of the previous &nbsp;Problem 1&nbsp; {{{h}}} = {{{a*b/sqrt(a^2 + b^2)}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)

Hence, &nbsp;from the right-angled triangle &nbsp;ADC&nbsp; &nbsp;&nbsp;{{{p^2}}} = {{{a^2}}} - {{{a^2*b^2/(a^2 + b^2)}}} = {{{a^4/(a^2 + b^2)}}}&nbsp; &nbsp;&nbsp;in accordance with the Pythagorean Theorem.

From the right-angled triangle &nbsp;BDC&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{q^2}}} = {{{b^2}}} - {{{a^2*b^2/(a^2 + b^2)}}} = {{{b^4/(a^2 + b^2)}}}&nbsp; &nbsp;&nbsp;by the same reason.

It implies that &nbsp;&nbsp;&nbsp;{{{sqrt(p*q)}}} = {{{a*b/sqrt(a^2 + b^2)}}}. &nbsp;&nbsp;&nbsp;&nbsp;In turn, &nbsp;&nbsp;{{{a*b/sqrt(a^2 + b^2)}}} = {{{h}}}&nbsp; &nbsp;&nbsp;due to &nbsp;(2).

Thus it is proved that &nbsp;&nbsp;h = {{{sqrt(p*q)}}}. &nbsp;&nbsp;The solution is completed.

There are two other solutions of the &nbsp;Problem 3&nbsp; in this site.

<H3>Problem 4</H3>Find the area of a right-angled triangle, &nbsp;if the altitude drawn to the hypotenuse divides the hypotenuse in segments of &nbsp;18 cm&nbsp; and &nbsp;32 cm&nbsp; long.

Solution

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

Second, &nbsp;calculate the measure of the altitude drawn to the hypotenuse from the right-angle vertex.

Apply the result of the &nbsp;Problem 3&nbsp; above. &nbsp;The measure of the altitude drawn to the hypotenuse in a right-angled triangle is the &nbsp;Geometric mean&nbsp; of the measures of

the segments the altitude divides the hypotenuse.

So, &nbsp;{{{h}}} = {{{sqrt(18*32)}}} = {{{sqrt(9*2*2^5)}}} = {{{sqrt(3^2*2^6)}}} = {{{3*2^3}}} = 3*8 = 24 cm.

Now, &nbsp;the area of the triangle is &nbsp;{{{50*24/2}}} = 600 {{{cm^2}}}.

Answer. &nbsp;The area of the triangle is &nbsp;600 {{{cm^2}}}.

My other lessons on the topic &nbsp;Area&nbsp; in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/What-is-area.lesson>WHAT IS area?</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Formulas-for-area-of-a-triangle.lesson>Formulas for area of a triangle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/-Proof-of-the-Heron%27s-formula-for-the-area-of-a-triangle.lesson>Proof of the Heron's formula for the area of a triangle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/One-more-proof-of-the-Heron%27s-formula-for-the-area-of-a-triangle.lesson>One more proof of the Heron's formula for the area of a triangle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Proof-of-the-formula-for-the-area-of-a-triangle-via-the-radius-of-the-inscribed-circle.lesson>Proof of the formula for the area of a triangle via the radius of the inscribed circle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Proof-of-the-formula-for-the-radius-of-the-circumscribed-circle.lesson>Proof of the formula for the radius of the circumscribed circle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-parallelogram.lesson>Area of a parallelogram</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-trapezoid.lesson>Area of a trapezoid</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-quadrilateral.lesson>Area of a quadrilateral</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-sircumscribed-quadrilateral.lesson>Area of a quadrilateral circumscribed about a circle</A> &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-quadrilateral-inscribed-in-a-circle.lesson>Area of a quadrilateral inscribed in a circle</A>

under the topic &nbsp;Area and surface area&nbsp; of the section &nbsp;Geometry, &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-triangles.lesson>Solved problems on area of triangles</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-regular-triangles.lesson>Solved problems on area of regular triangles</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-the-radius-of-inscribed-circles-and-semicircles.lesson>Solved problems on the radius of inscribed circles and semicircles</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-the-radius-of-a-circumscribed-circle.lesson>Solved problems on the radius of a circumscribed circle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-Math-circle-level-problem-on-area-of-a-triangle.lesson>A Math circle level problem on area of a triangle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-parallelograms.lesson>Solved problems on area of parallelograms</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-rhombis-rectangles-and-squares.lesson>Solved problems on area of rhombis, rectangles and squares</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-trapezoids.lesson>Solved problems on area of trapezoids</A> &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-quadrilaterals.lesson>Solved problems on area of quadrilaterals</A>

under the topic &nbsp;Geometry&nbsp; of the section &nbsp;Word problems.

For navigation over the lessons on &nbsp;Area of Triangles&nbsp; use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Surface-area/REVIEW-OF-LESSONS-ON-AREA-OF-TRIANGLES.lesson>OVERVIEW of lessons on area of triangles</A>.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.