C-06-Q183mediumsingle_mcq

In $\triangle ABC$ , $\angle B = 1$ right angle and $D$ is the midpoint of hypotenuse $AC$ . Then $BD =$

a $\tfrac{1}{2} AC$
b $\tfrac{1}{3} AC$
c $AC$
d $2 AC$

ব্যাখ্যা

The midpoint of the hypotenuse of a right triangle is equidistant from all three vertices (it is the circumcentre, with the hypotenuse as diameter). Hence $BD = AD = DC = \tfrac{1}{2}AC$ .

এই অধ্যায়ের আরও প্রশ্ন

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Lines, Angles and Triangles — সব প্রশ্ন দেখুন (191টি) →

In $\triangle ABC$ , $\angle B = 1$ right angle and $D$ is the midpoint of hypotenuse $AC$ . Then $BD =$

\tfrac{1}{2} AC

\tfrac{1}{3} AC

AC

2 AC

ব্যাখ্যা

The midpoint of the hypotenuse of a right triangle is equidistant from all three vertices (it is the circumcentre, with the hypotenuse as diameter). Hence $BD = AD = DC = \tfrac{1}{2}AC$ .

In △ABC\triangle ABC△ABC, ∠B=1\angle B = 1∠B=1 right angle and DDD is the midpoint of hypotenuse ACACAC. Then BD=BD =BD=

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In △ABC\triangle ABC△ABC, ∠B=1\angle B = 1∠B=1 right angle and DDD is the midpoint of hypotenuse ACACAC. Then BD=BD =BD=

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In $\triangle ABC$ , $\angle B = 1$ right angle and $D$ is the midpoint of hypotenuse $AC$ . Then $BD =$

In $\triangle ABC$ , $\angle B = 1$ right angle and $D$ is the midpoint of hypotenuse $AC$ . Then $BD =$