A semicircle with diameter BE is drawn in triangle ABC, which is right-angled atB. If AB = 4 cm and BE = 6 cm, find DC : EC. About the author Madelyn
Answer: (1) It is given that line AB is tangent to the circle at A. ∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact) Thus, the measure of ∠CAB is 90º. (2) Distance of point C from AB = 6 cm (Radius of the circle) (3) ∆ABC is a right triangle. CA = 6 cm and AB = 6 cm Using Pythagoras theorem, we have BC2=FROM2+THAT2⇒BC= √ 62+62 ⇒BC=6 √ 2 cm Thus, d(B, C) = 6 √ 2 cm (4) In right ∆ABC, AB = CA = 6 cm ∴ ∠ACB = ∠ABC (Equal sides have equal angles opposite to them) Also, ∠ACB + ∠ABC = 90º (Using angle sum property of triangle) ∴ 2∠ABC = 90º ⇒ ∠ABC = 90° 2 = 45º Thus, the measure of ∠ABC is 45º Reply
Answer:
(1) It is given that line AB is tangent to the circle at A.
∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)
Thus, the measure of ∠CAB is 90º.
(2) Distance of point C from AB = 6 cm (Radius of the circle)
(3) ∆ABC is a right triangle.
CA = 6 cm and AB = 6 cm
Using Pythagoras theorem, we have
BC2=FROM2+THAT2⇒BC=
√
62+62
⇒BC=6
√
2
cm
Thus, d(B, C) = 6
√
2
cm
(4) In right ∆ABC, AB = CA = 6 cm
∴ ∠ACB = ∠ABC (Equal sides have equal angles opposite to them)
Also, ∠ACB + ∠ABC = 90º (Using angle sum property of triangle)
∴ 2∠ABC = 90º
⇒ ∠ABC =
90°
2
= 45º
Thus, the measure of ∠ABC is 45º