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Byju's Answer

Standard IX

Mathematics

Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

In a triangle...

Question

In a triangle $A B C$ , the lengths of the sides $A B$ and $A C$ equal $17.5 c m$ and $9 c m$ respectively. Let $D$ be a point on the line segment $B C$ such that $A D$ is perpendicular to $B C$ . If $A D = 3 c m$ , then what is the radius (in $c m$ ) of the circle circumscribing the triangle $A B C$ ?

17.05

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27.85

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22.45

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26.25

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Solution

The correct option is B $26.25$
Circumradius, $R = \frac{a b c}{4 Δ} = \frac{(A B \times A C \times B C)}{4 Δ}$

Area of triangle, $Δ = \frac{1}{2} \times B C \times A D$

$R = \frac{(A B \times A C)}{2 \times A D}$

$⟹ R = \frac{(17.5 \times 9)}{(2 \times 3)} = 26.25 c m$

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In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD=3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?

The correct option is B 26.25Circumradius, R=abc4Δ=(AB×AC×BC)4ΔArea of triangle, Δ=12×BC×ADR=(AB×AC)2×AD⟹R=(17.5×9)(2×3)=26.25cm

The correct option is B $26.25$
Circumradius, $R = \frac{a b c}{4 Δ} = \frac{(A B \times A C \times B C)}{4 Δ}$

Area of triangle, $Δ = \frac{1}{2} \times B C \times A D$

$R = \frac{(A B \times A C)}{2 \times A D}$

$⟹ R = \frac{(17.5 \times 9)}{(2 \times 3)} = 26.25 c m$