Question

What is the radius of the circumscribed circle of △ABC?
A. 1.3
B. 5.8
C. 4.0
D. 5.6

Answer 1

Answer: The radius of the circumscribed circle of △ABC. IS 5.8

the correct answer is  (B) 5.8.

Step-by-step explanation:

For a triangle △ABC, let s = 12 (a+b+ c). Then the radius R of its circumscribed circle is R=abc4√s(s−a)(s−b)(s−c). In addition to a circumscribed circle, every triangle has an inscribed circle, i.e. a circle to which the sides of the triangle are tangent.

Let there is a circle having center O touches the sides AB and AC of the triangle at point E and F respectively.

Let the length of the line segment AE is x.

Now in △ABC,

CF=CD=6 (tangents on the circle from point C)

BE=BD=6 (tangents on the circle from point B)

AE=AF=x (tangents on the circle from point A)

$Now AB=AE+EB⟹AB=x+8=cBC=BD+DC⟹BC=8+6=14=aCA=CF+FA⟹CA=6+x=b$

$NowSemi-perimeter, s= 2(AB+BC+CA)s= 2(x+8+14+6+x)​s= 2(2x+28)​⟹s=x+14$

$Area of the △ABC= s(s−a)(s−b)(s−c)​= (14+x)((14+x)−14)((14+x)−(6+x))((14+x)−(8+x))​= (14+x)(x)(8)(6)= (14+x)(x)(2×4)(2×3)Area of the △ABC=4 3x(14+x)$

$​Area of △OBC= 21 ×OD×BC= 21 ×4×14=28Area of △OBC= 21 ×OF×AC= 21 ×4×(6+x)=12+2xArea of ×OAB= 21​ ×OE×AB= 21 ×4×(8+x)=16+2x$

$Now, Area of the △ABC=Area of △OBC+Area of △OBC+Area of △OAB4 3x(14+x) =28+12+2x+16+2x4 3x(14+x)​ =56+4x=4(14+x)3x(14+x)​ =14+x$

$On squaring both sides, we get3x(14+x)=(14+x) 2 3x=14+x ------------- (14+x=0⟹x=−14 is not possible)3x−x=142x=14x= 214​x=7$

Hence

$AB=x+8AB=7+8AB=15AC=6+xAC=6+7AC=13$

So, the value of AB is 15 cm and that of AC is 13 cm. Option B.

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