Problems on Isosceles Triangles with Detailed Solutions
Problems on isosceles triangles are presented along with their detailed solutions.
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

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}

Solution
Use formula of area of isosceles triangle
A = (1/2) a^{ 2} sin(α)
to find α as follows
sin(α) = 2 A / a^{ 2} = 2 * 6 / 5^{ 2} = 12 / 25
α = arcsin(12 / 25)
Using right triangle CC'B (see figure at top), we can write
sin(α/2) = (b/2) / a = b / 2a
b = 2 a sin (α/2)
b = 2 a sin( (1/2) arcsin(12 / 25) ) = 10 sin( (1/2) arcsin(12 / 25) ) ≅ 2.48

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

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

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°

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°

Solution
Radius of inscribed circle to an isosceles triangle of base b = 10 and lateral side a = 12 is given by
r = \( \dfrac{b}{2} \sqrt{\dfrac{2ab}{2a+b}} = 5 \sqrt{\dfrac{4}{34}} \)
Area A of circle of radius r is given by: A = π r^{ 2} = 100π / 34

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 = \( \dfrac{a^2}{\sqrt{4a^2b^2}} \) and r = \( \dfrac{b}{2} \sqrt{\dfrac{2ab}{2a+b}} \)
Substitute a by 2 b (given) in both formulas and simplify
R = \( \dfrac{(2b)^2}{\sqrt{4(2b)^2b^2}} = \dfrac{4b^2}{\sqrt{15b^2}} =\dfrac{4b}{\sqrt15} \)
r = \( \dfrac{b}{2} \sqrt{\dfrac{4bb}{4b+b}} = (b/2) \sqrt{\dfrac{3}{5}} \)
Simplify ratio: R / r = 8/3

Solution
Radius R of the circumscibed circle an isosceles triangle of base b and lateral side a are given by
R = \( \dfrac{a^2}{\sqrt{4a^2b^2}} \)
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.

Solution
Let A_{1} and A_{2} be the areas of the triangle with lateral sides a1 = 2 and a2 = 10 respectively.
Formulas for A_{1} and A_{2}
A_{1} = (1 / 2) a_{1}^{ 2}sin(α) and A_{2} = (1 / 2) a_{2}^{ 2} sin(α) , corresponding angles are equal in similar triangles.
Hence the ratio: A_{1} / A_{2} = (1 / 2) a_{1}^{ 2}sin(α) / (1 / 2)a_{2}^{ 2}sin(α) = a_{1}^{ 2} / a_{2}^{ 2}
We now need to find A_{2}
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 A_{2} = (1 / 2) b h = (1 / 2) 12×8 = 48 unit^{ 2}
a_{1}^{ 2} / a_{2}^{ 2} = A_{1} / A_{2}
A_{1} = A_{2} × a_{1}^{ 2} / a_{2}^{ 2} = 48 × 2^{ 2}/ 10^{ 2} = 1.92 unit^{ 2}

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