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Isosceles Triangle Formulas
An Isosceles triangle has two equal sides with the angles opposite to them equal. The relationship between the lateral side \( a \), the based \( b \) of the isosceles triangle, its area A, height h, inscribed and circumscribed radii r and R respectively are give by:
Problem 1
What is the area of an isosceles triangle with base b of 8 cm and lateral a side 5 cm?
Problem 2
What is the base of an isosceles triangle with lateral side a = 5 cm and area 6 cm 2?
Problem 3
What is the lateral side of an isosceles triangle with area 20 unit 2 and base 10 units?
Problem 4
What is the lateral side of an isosceles triangle such that its height h ( perpendicular to its base b) is 4 cm shorter than its base b and its area is 30 cm 2?
Problem 5
ABC and BCD are isosceles triangles. Find the size of angle BDE.
Problem 6
ABC and CDE are isosceles triangles. Find the size of angle CED.
Problem 7
Find the area of the circle inscribed to an isosceles triangle of base 10 units and lateral side 12 units.
Problem 8
Find the ratio of the radii of the circumscribed and inscribed circles to an isosceles triangle of base b units and lateral side a units such that a = 2 b.
Problem 9
Find the lateral side and base of an isosceles triangle whose height ( perpendicular to the base ) is 16 cm and the radius of its circumscribed circle is 9 cm.
Problem 10
What is the area of an isosceles triangle of lateral side 2 units that is similar to another isosceles triangle of lateral side 10 units and base 12 units?
Solutions to the Above Questions
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Solution
Apply Pythagora's theorem to the right triangle CC'B (see figure at top) to write
a 2 = (b/2) 2 + h 2
h = a 2 - (b/2) 2 = 5 2 - 4 2 which gives h = 3
Area A = (1/2) b h = (1/2) 8×3 = 12 unit 2
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Solution
Use formula of area of isosceles triangle
A = (1/2) a 2 sin(?)
to find ? as follows
sin(?) = 2 A / a 2 = 2 * 6 / 5 2 = 12 / 25
? = arcsin(12 / 25)
Using right triangle CC'B (see figure at top), we can write
sin(?/2) = (b/2) / a = b / 2a
b = 2 a sin (?/2)
b = 2 a sin( (1/2) arcsin(12 / 25) ) = 10 sin( (1/2) arcsin(12 / 25) ) ? 2.48
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Solution
Use formula of area of isosceles triangle to write
A = (1/2) b h = 20
Given b = 10, find h
h = 40 / 10 = 4
Pythagora's theorem used in the right triangle CC'B (see figure at top) to write
a 2 = (b/2) 2 + h 2 = √ ( 5 2 + 4 2) = √41
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Solution
Use formula of area of isosceles triangle to write
A = (1/2) b h = 30
h = b - 4 (given)
Substitute h by b - 4 in the formula for A
A = (1/2) b (b - 4) = 30
Gives the equation: b 2 - 4 b - 60 = 0
Solutions to the equation: b = 10 and b = - 6
b is a length and therefore is positive b = 10 , h = b - 4 = 6
a = √ (h 2 + (b/2) 2) = √ (36 + 25) = √61
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Solution
ABC is an isosceles triangle and therefore
?CAB = ?ABC
Also: ?CAB + ?ABC = 180 - 66
2 ?ABC = 180 - 66
gives: ?ABC = 57°
BCD is a right isosceles triangle; hence
?CBD = ?CDB = (180 - 90)/2 = 45°
Note that ?ABC, ?CBD and ?DBE make a straight angle. Hence
?ABC + ?CBD + ?DBE = 180°
gives ?DBE = 180 - ?ABC - ?CBD = 180 - 57 - 45 = 78°
DBE is a right triangle ; hence
?BDE = 90 - 78 = 12°
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Solution
Straight angle at B hence
?ABC = 180 - 116 = 64°
ABC is an isosceles triangle and therefore
?CAB = ?ABC = 64°
In trangle ABC: ?BCA + ?CAB + ?ABC = 180°
Hence: ?BCA = 180 - 64 - 64 = 52°
?DCE = ? BCA = 52°
CDE isosceles: ?DCE = ?CDE = 52°
In triangle CDE we have: ?DCE + ?CDE + ?CED = 180°
?CED = 180 - 52 - 52 = 76°
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Solution
Radius of inscribed circle to an isosceles triangle of base b = 10 and lateral side a = 12 is given by
r = \( \dfrac{b}{2} \sqrt{\dfrac{2a-b}{2a+b}} = 5 \sqrt{\dfrac{4}{34}} \)
Area A of circle of radius r is given by: A = ? r 2 = 100? / 34
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Solution
Radius R of the circumscibed circle and radius r of the inscribed circle to the same isosceles triangle of base b and lateral side a are given by
R = \( \dfrac{a^2}{\sqrt{4a^2-b^2}} \) and r = \( \dfrac{b}{2} \sqrt{\dfrac{2a-b}{2a+b}} \)
Substitute a by 2 b (given) in both formulas and simplify
R = \( \dfrac{(2b)^2}{\sqrt{4(2b)^2-b^2}} = \dfrac{4b^2}{\sqrt{15b^2}} =\dfrac{4b}{\sqrt15} \)
r = \( \dfrac{b}{2} \sqrt{\dfrac{4b-b}{4b+b}} = (b/2) \sqrt{\dfrac{3}{5}} \)
Simplify ratio: R / r = 8/3
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Solution
Radius R of the circumscibed circle an isosceles triangle of base b and lateral side a are given by
R = \( \dfrac{a^2}{\sqrt{4a^2-b^2}} \)
Relationship between h, b and a (Pythagora in triangle CC'B see figure above)
a 2 = (b/2) 2 + h 2
Gives: a 2 = b 2 / 4 + h 2 or 4 h 2 = 4 a 2 - b 2
Substitute R by 9 (given) and 4 a 2 - b 2 by 4 h 2 in the formula for R above to get the equation
9 = \( \dfrac{a^2}{\sqrt{4h^2}} = a^2 / (2h) \)
a 2 = 9 × 2 × 16
gives a = 12 √2 cm.
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Solution
Let A1 and A2 be the areas of the triangle with lateral sides a1 = 2 and a2 = 10 respectively.
Formulas for A1 and A2
A1 = (1 / 2) a1 2sin(?) and A2 = (1 / 2) a2 2 sin(?) , corresponding angles are equal in similar triangles.
Hence the ratio: A1 / A2 = (1 / 2) a1 2sin(?) / (1 / 2)a2 2sin(?) = a1 2 / a2 2
We now need to find A2
Use Pythagora's theorem in the right triangle CC'B (see figure at top) to write
a 2 = (b/2) 2 + h 2
h = a 2 - (b/2) 2 = 10 2 - 6 2 which gives h = 8
Area A2 = (1 / 2) b h = (1 / 2) 12×8 = 48 unit 2
a1 2 / a2 2 = A1 / A2
A1 = A2 × a1 2 / a2 2 = 48 × 2 2/ 10 2 = 1.92 unit 2
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