Problems on Isosceles Triangles with Detailed Solutions

Question: Find the area of an isosceles triangle with base \(b = 8 \, \text{cm}\) and lateral side \(a = 5 \, \text{cm}\).

Solution:

• Using Pythagorean theorem: \(h = \sqrt{a^2 - (b/2)^2} = \sqrt{5^2 - 4^2} = 3 \, \text{cm}\)
• Area: \(A = \dfrac{1}{2} b h = \dfrac{1}{2} \times 8 \times 3 = 12 \, \text{cm}^2\)

Question: Find the base of an isosceles triangle with lateral side \(a = 5 \, \text{cm}\) and area \(A = 6 \, \text{cm}^2\).

Solution:

• Using area formula: \(A = \dfrac{1}{2} a^2 \sin\alpha \implies \sin\alpha = \dfrac{2A}{a^2} = \dfrac{12}{25}\)
• \(\alpha = \arcsin(12/25)\)
• From geometry: \(\sin(\alpha/2) = \dfrac{b/2}{a} = \dfrac{b}{2a}\)
• \(b = 2a \sin(\alpha/2) = 10 \sin\left(\dfrac{1}{2} \arcsin\dfrac{12}{25}\right) \approx 2.48 \, \text{cm}\)

Question: Find the lateral side of an isosceles triangle with area \(A = 20 \, \text{units}^2\) and base \(b = 10 \, \text{units}\).

Solution:

• From area: \(A = \dfrac{1}{2} b h \implies 20 = \dfrac{1}{2} \times 10 \times h \implies h = 4\)
• Lateral side: \(a = \sqrt{(b/2)^2 + h^2} = \sqrt{5^2 + 4^2} = \sqrt{41}\)

Question: Find the lateral side of an isosceles triangle where the height \(h\) is 4 cm shorter than the base \(b\), and the area is \(30 \, \text{cm}^2\).

Solution:

• Given \(h = b - 4\) and \(A = \dfrac{1}{2} b h = 30\)
• \(\dfrac{1}{2} b (b - 4) = 30 \implies b^2 - 4b - 60 = 0\)
• Positive solution: \(b = 10 \, \text{cm}, \, h = 6 \, \text{cm}\)
• Lateral side: \(a = \sqrt{(b/2)^2 + h^2} = \sqrt{25 + 36} = \sqrt{61} \, \text{cm}\)

Question: \(\triangle ABC\) and \(\triangle BCD\) are isosceles triangles. Find \(\angle BDE\).

Solution:

• In \(\triangle ABC\): \(\angle CAB = \angle CBA = \dfrac{180^\circ - 66^\circ}{2} = 57^\circ\)
• In \(\triangle BCD\): \(\angle CBD = \angle CDB = 45^\circ\)
• Angles on straight line: \(\angle DBE = 180^\circ - 57^\circ - 45^\circ = 78^\circ\)
• In \(\triangle BDE\): \(\angle BDE = 90^\circ - 78^\circ = 12^\circ\)

Question: \(\triangle ABC\) and \(\triangle CDE\) are isosceles triangles. Find \(\angle CED\).

Solution:

• \(\angle ABC = 180^\circ - 116^\circ = 64^\circ\)
• In \(\triangle ABC\): \(\angle BCA = 180^\circ - 2 \times 64^\circ = 52^\circ\)
• \(\angle DCE = \angle BCA = 52^\circ\) (vertical angles)
• In \(\triangle CDE\): \(\angle CED = 180^\circ - 2 \times 52^\circ = 76^\circ\)

Question: Find the area of the circle inscribed in an isosceles triangle with base \(b = 10\) and lateral side \(a = 12\).

Solution:

• Inradius: \(r = \dfrac{b}{2} \sqrt{\dfrac{2a-b}{2a+b}} = 5 \sqrt{\dfrac{4}{34}} = 5\sqrt{\dfrac{2}{17}}\)
• Circle area: \(A = \pi r^2 = \pi \left(5\sqrt{\dfrac{2}{17}}\right)^2 = \dfrac{50\pi}{17}\)

Question: Find the ratio \(R/r\) for an isosceles triangle with \(a = 2b\).

Solution:

• Circumradius: \(R = \dfrac{a^2}{\sqrt{4a^2 - b^2}} = \dfrac{(2b)^2}{\sqrt{4(2b)^2 - b^2}} = \dfrac{4b}{\sqrt{15}}\)
• Inradius: \(r = \dfrac{b}{2} \sqrt{\dfrac{2a-b}{2a+b}} = \dfrac{b}{2} \sqrt{\dfrac{3}{5}}\)
• Ratio: \(\dfrac{R}{r} = \dfrac{4b/\sqrt{15}}{(b/2)\sqrt{3/5}} = \dfrac{8}{\sqrt{9}} = \dfrac{8}{3}\)

Question: Find the lateral side and base of an isosceles triangle with height \(h = 16 \, \text{cm}\) and circumradius \(R = 9 \, \text{cm}\).

Solution:

• From Pythagorean theorem: \(4a^2 - b^2 = 4h^2\)
• Circumradius formula: \(R = \dfrac{a^2}{\sqrt{4a^2 - b^2}} = \dfrac{a^2}{2h}\)
• \(9 = \dfrac{a^2}{32} \implies a^2 = 288 \implies a = 12\sqrt{2} \, \text{cm}\)
• \(b = \sqrt{4a^2 - 4h^2} = \sqrt{4 \times 288 - 4 \times 256} = \sqrt{128} = 8\sqrt{2} \, \text{cm}\)

Question: Find the area of an isosceles triangle with lateral side \(2\) that is similar to another isosceles triangle with lateral side \(10\) and base \(12\).

Solution:

• For larger triangle: \(h = \sqrt{10^2 - 6^2} = 8\)
• Area: \(A_2 = \dfrac{1}{2} \times 12 \times 8 = 48\)
• Ratio of similarity: \(k = \dfrac{2}{10} = \dfrac{1}{5}\)
• Area ratio: \(k^2 = \dfrac{1}{25}\)
• Smaller triangle area: \(A_1 = 48 \times \dfrac{1}{25} = 1.92 \, \text{units}^2\)