Consider a circle of diameter '17'. Let B, C, E and D (taken in order) be points on the circle such that
BD and CE when extended intersect at A. If AD and AE have lengths 10 and 6 respectively and
DBC is right angle, then 10 BD is___
plz give quick solution
Answers
Answer:
Hii
Step-by-step explanation:
(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=AB2+CA2⇒BC=62+62−−−−−−√ ⇒BC=62–√ cm
Thus, d(B, C) = 62–√ 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.
I hope it helps you
Given : Consider a circle of diameter '17'. Let B, C, E and D (taken in order) be points on the circle such that BD and CE when extended intersect at A. AD and AE have lengths 10 and 6 . DBC is right angle
To find : length of BD
Solution:
∠DEA = ∠DBC = 90° ( property of cyclic Quadrilateral)
=> DE² = AD² - AE²
=> DE² = 10² - 6²
=> DE = 8
∠DBC = 90°
=> DC is Diameter
=> DC = 17 cm
∠DEC = 90° ( as sum of opposite angle of cyclic quadrilateral = 180°)
=> CE² = DC² - DE²
=> CE² = 17² - 8²
=> CE = 15
Comparing Δ DEA & Δ CBA
∠A = ∠A ( common)
∠DBC = ∠DEA = 90°
=> Δ DEA ≈ Δ CBA
=> EA/AB = AD/AC
=> 6/(AD + BD) = 10/(AE + EC)
=> 6/(10 + BD) = 10/(6 + 15)
=> 6/(10 + BD) = 10/21
=> 126 = 100 + 10BD
=> 10BD = 26
=> BD = 2.6
Length of BD = 2.6
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