This question was uploaded on 25/09/21 on social media accounts.
One right-angled triangle with sides 8, 6 as height and base and another isosceles right-angled triangle with hypotenuse 10 on the base (side 6) of the other triangle. Find fraction covered by the triangle formed by both triangles.
Step 1: Area of the blue triangle.
Step 2: Total area of both triangles.
Step 3: Fraction of blue triangle.
Solution:
Step 1:
\[tan(\theta)=\frac{8}{6}=\frac{h}{6-h}\]
\[\Rightarrow h=\frac{24}{7}\]
Area of blue triangle:
\[A = \frac{1}{2}\cdot6\cdot\frac{24}{7}=\frac{72}{7}\]
Step 2:
Area of the first triangle (sides: 6, 8, 10):
\[A_1 = \frac{1}{2}\cdot6\cdot8=24\]
\[A_1 = \frac{1}{2}\cdot6\cdot8=24\]
Area of the second triangle (hypotenuse: 10):
\[A_2 = \frac{\left(5\sqrt2\right)^2}{2}=25\]
Total area:
\[A_T = A_1+A_2-A =\frac{271}{7}\]
Step 3:
Fraction:
\[\frac{A}{A_T}=\frac{72}{271}\]
Puzzle related to Geometry, Triangle, Area.
