Problems on Isosceles Triangles with Detailed Solutions

Problems on isosceles triangles are presented along with their detailed solutions.

Isosceles Triangle Formulas

Problems with Solutions

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Solutions to the Above Questions

1. 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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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

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3. 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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4. 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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5. 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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6. 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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7. 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 = π r ² = 100π / 34

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8. 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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9. 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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10. 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(α) and A₂ = (1 / 2) a₂ ² sin(α) , corresponding angles are equal in similar triangles.
    
    Hence the ratio: A₁ / A₂ = (1 / 2) a₁ ²sin(α) / (1 / 2)a₂ ²sin(α) = 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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