Solve right triangle problems involving area, perimeter, hypotenuse, sides, and their relationships.
A right triangle has one angle equal to \(90^\circ\).
Area: \(A = \dfrac{1}{2} a \times b\)
Perimeter: \(P = a + b + h\)
By the Pythagorean theorem: \(h^2 = a^2 + b^2\) ⇒ \(h = \sqrt{a^2 + b^2}\)
Perimeter in terms of sides only: \(P = a + b + \sqrt{a^2 + b^2}\)
Find the area and perimeter of a right triangle with sides \(0.4\) ft and \(0.3\) ft.
Find the area and perimeter of an isosceles right triangle with hypotenuse \(50\) cm.
The first side of a right triangle is \(200\) m longer than the second side. The hypotenuse is \(1000\) m. Find the lengths of both sides, the area, and the perimeter.
A right triangle has one side that is \(\dfrac{4}{3}\) of the other. The hypotenuse is \(30\) ft. Find the sides, area, and perimeter.
Triangle \(ABC\) is a right triangle with perimeter \(60\) units and area \(150\) units\(^2\). Find its two sides and hypotenuse.
Right triangles \(ABC\) and \(DEF\) have \(\angle ACB = \angle DFE\). The hypotenuse ratio is \(AC/DF = 3/2\). The area of \(\triangle DEF\) is \(100\) units\(^2\). What is the area of \(\triangle ABC\)?
Let \(a = 0.4\) ft, \(b = 0.3\) ft.
Area: \(A = \dfrac{1}{2} \times 0.4 \times 0.3 = 0.06\) ft\(^2\)
Hypotenuse: \(h = \sqrt{0.4^2 + 0.3^2} = \sqrt{0.16 + 0.09} = \sqrt{0.25} = 0.5\) ft
Perimeter: \(P = 0.4 + 0.3 + 0.5 = 1.2\) ft
For an isosceles right triangle, \(a = b\).
Pythagorean theorem: \(a^2 + a^2 = 50^2\) ⇒ \(2a^2 = 2500\) ⇒ \(a^2 = 1250\)
Area: \(A = \dfrac{1}{2} a^2 = \dfrac{1}{2} \times 1250 = 625\) cm\(^2\)
Perimeter: \(P = 2a + 50 = 2\sqrt{1250} + 50 = 50\sqrt{2} + 50 = 50(\sqrt{2} + 1) \approx 120.7\) cm
Let second side = \(x\) m ⇒ first side = \(x + 200\) m, hypotenuse = \(1000\) m.
By Pythagorean theorem: \(x^2 + (x + 200)^2 = 1000^2\)
Expand: \(x^2 + x^2 + 400x + 40000 = 1000000\)
Simplify: \(2x^2 + 400x - 960000 = 0\)
Divide by 2: \(x^2 + 200x - 480000 = 0\)
Solve quadratic: \(x = \dfrac{-200 \pm \sqrt{200^2 + 4 \times 480000}}{2} = \dfrac{-200 \pm \sqrt{1960000}}{2} = \dfrac{-200 \pm 1400}{2}\)
Positive solution: \(x = \dfrac{-200 + 1400}{2} = 600\) m
First side: \(600 + 200 = 800\) m
Area: \(A = \dfrac{1}{2} \times 600 \times 800 = 240000\) m\(^2\)
Perimeter: \(P = 600 + 800 + 1000 = 2400\) m
Let second side = \(x\) ft ⇒ first side = \(\dfrac{4}{3}x\) ft, hypotenuse = \(30\) ft.
Pythagorean theorem: \(x^2 + \left(\dfrac{4}{3}x\right)^2 = 30^2\)
Simplify: \(x^2 + \dfrac{16}{9}x^2 = 900\) ⇒ \(\dfrac{25}{9}x^2 = 900\)
Solve: \(x^2 = \dfrac{900 \times 9}{25} = 324\) ⇒ \(x = \sqrt{324} = 18\) ft
First side: \(\dfrac{4}{3} \times 18 = 24\) ft
Area: \(A = \dfrac{1}{2} \times 18 \times 24 = 216\) ft\(^2\)
Perimeter: \(P = 18 + 24 + 30 = 72\) ft
Given: \(a + b + h = 60\), \(\dfrac{1}{2}ab = 150\), and \(a^2 + b^2 = h^2\).
From area: \(ab = 300\)
From perimeter: \(a + b = 60 - h\)
Square both sides: \((a + b)^2 = (60 - h)^2\) ⇒ \(a^2 + 2ab + b^2 = 3600 - 120h + h^2\)
Substitute \(a^2 + b^2 = h^2\) and \(ab = 300\): \(h^2 + 600 = 3600 - 120h + h^2\)
Simplify: \(600 = 3600 - 120h\) ⇒ \(120h = 3000\) ⇒ \(h = 25\)
Then \(a + b = 60 - 25 = 35\) and \(ab = 300\)
Solve quadratic: \(t^2 - 35t + 300 = 0\) ⇒ \((t - 15)(t - 20) = 0\)
Sides: \(a = 15\), \(b = 20\) (or vice versa)
Hypotenuse: \(h = 25\) (verified by Pythagorean theorem: \(15^2 + 20^2 = 625 = 25^2\))
Triangles are similar (AA similarity: right angles and \(\angle ACB = \angle DFE\)).
Ratio of corresponding sides: \(\dfrac{AC}{DF} = \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{3}{2}\)
Area ratio = square of side ratio: \(\dfrac{A_{ABC}}{A_{DEF}} = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}\)
Thus: \(A_{ABC} = \dfrac{9}{4} \times 100 = 225\) units\(^2\)