In a right triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. The chord joining A with the point of intersection D of the hypotenuse and the semicircle, then the length AC equals to -

Question

In a right triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. The chord joining A with the point of intersection D of the hypotenuse and the semicircle, then the length AC equals to - (A) ( AB ⋅ AD /√ AB 2+ AD 2) (B) ( AB ⋅ AD / AB + AD ) (C) √ AB ⋅ AD  (D) ( AB ⋅ AD /√ AB 2- AD 2)

Tardigrade Admin · Accepted Answer

Triangles BAC and BDA are similar ∴ ( AC / AD )=( BC / AB ) A C=(B C ⋅ A D/A B) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ed62600ebee49888213e6d9df9cafde7-.png" /> AB 2= BD . BC =(A B ⋅ A D/B D)=(A B ⋅ A D/√A B2-A D2)

Q. In a right triangle $A BC$ , right angled at $A$ , on the leg $A C$ as diameter, a semicircle is described. The chord joining $A$ with the point of intersection $D$ of the hypotenuse and the semicircle, then the length $A C$ equals to -

$\frac{A B \cdot A D}{A B ^{2} + A D ^{2}}$

$\frac{A B \cdot A D}{A B + A D}$

$A B \cdot A D$

$\frac{A B \cdot A D}{A B ^{2} - A D ^{2}}$

Triangles $B A C$ and $B D A$ are similar
$∴ \frac{A C}{A D} = \frac{BC}{A B}$
$A C = \frac{BC \cdot A D}{A B}$

${A B^{2} = B D . BC}$
$= \frac{A B \cdot A D}{B D} = \frac{A B \cdot A D}{A B ^{2} - A D ^{2}}$

Q. In a right triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. The chord joining A with the point of intersection D of the hypotenuse and the semicircle, then the length AC equals to -

AB2+AD2​AB⋅AD​

AB+ADAB⋅AD​

AB⋅AD​

AB2−AD2​AB⋅AD​