Question

Maggie puts together two isosceles triangles so that they share a base, creating a kite. Each leg of the upper triangle measures 41 inches and each leg of the lower one measures 50 inches. If the length of the base of both triangles measures 80 inches, what is the length of the kite’s shorter diagonal?

Answer 1

Answer:

The answer is 39 on edge. :)

Step-by-step explanation:

Answer 2

Answer:

Length of the kite’s shorter diagonal = 39.03 inches

Step-by-step explanation:

Refer the given figure, we need to find DB,

Consider ΔABC,

Using cosine rule

     $cos B=\frac{a^2+c^2-b^2}{2ac}$

     a = c = 41 inches

     b = 80 inches

     We need to find ∠B

Substituting

      $cos B=\frac{41^2+41^2-80^2}{2\times 41\times 41}=-0.903\\\\B=154.56^0$

We can see that BD divides ∠B equally,

  So, $\angle ABD=\frac{\angle B}{2}=\frac{154.56}{2}=77.28^0$

Now consider ΔABD,

Using sine rule

       $\frac{AB}{sinD}=\frac{AD}{sinB}=\frac{DB}{sinA}$

       AB = 41 inches, AD = 50 inches, ∠B = 77.28°

       Substituting

       $\frac{41}{sinD}=\frac{50}{sin77.28}=\frac{DB}{sinA}\\\\sinD=0.7999\\\\D=53.12^0\\\\A=180-53.12-77.28=49.6^0\\\\\frac{50}{sin77.28}=\frac{DB}{sin49.6}\\\\DB=39.03inch$

Length of the kite’s shorter diagonal = 39.03 inches