Question

In the quadrilateral ABCD, we have AB = 3 cm, BC = 4 cm, AC = 5 cm, ∠A = 120°, ∠C = 70°. If we draw the circle with AC as diameter, which of the four vertices of ABCD would be inside the circle? Which of them would be outside this circle? Is any vertex on the circle? What about the circle with BD as diameter.

Answer 1

Given: AB = 3 cm, BC = 4 cm, AC = 5 cm, ∠A = 120° and ∠C = 70°

Consider the following:

AB² = 3² = 9

BC² = 4² = 16

AC² = 5² = 25

⇒ AC² = AB² + BC²

Therefore, by the converse of Pythagoras theorem, we get that ∠B = 90°.

Using angle sum property in quadrilateral ABCD:

∠A + ∠B + ∠C + ∠D = 360°

⇒ 120° + 90° + 70° + ∠D = 360°

⇒ 280° + ∠D = 360°

⇒ ∠D = 360° − 280° = 80°

We know that angle in a semi circle is a right angle.

If we draw a circle with AC as the diametre then vertex B will lie on the circle.

As the measure of ∠D is less than 90°, vertex D will lie outside the circle.

Thus, vertices A, B and C will lie on the circle.

Now, let us draw a circle with BD as the diametre.

As ∠A = 120° and ∠C = 70°, vertex A will lie inside the circle and vertex C will lie outside the circle.

Only vertices B and D will lie on the circle.