(D) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.Substituting the given values,$AC^2 = 5^2 + 12^2 =

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.Substituting the given values,$AC^2 = 5^2 + 12^2 = 25 + 144 = 169$.Thus,$AC = \sqrt{169} = 13$.In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Therefore,$BD = \frac{1}{2} \times AC = \frac{1}{2} \times 13 = 6.5$.

Class 10 Mathematics Triangles Similarity of Triangles and Basic Proportionality Theorem

Medium

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is a median. If $AB = 5$ and $BC = 12$,find $BD$.

A
$11$
B
$3.6$
C
$2$
D
$6.5$

Explore More

Browse All

Question Bank

All Subjects

Class 10 Questions

Class 10

Mathematics Questions

Chapter

Triangles

Subtopic

Similarity of Triangles and Basic Proportionality Theorem

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $a = 9$ and $c = 12$,then $b = \dots$

Easy

View Solution

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. If $AM = 7$ and $CM = 9$,then $BC = \ldots$.

Medium

View Solution

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $D$ is a point on $\overline{BC}$. Prove that $AD^{2} + BC^{2} = AC^{2} + BD^{2}$.

Medium

View Solution

In rectangle $ABCD$,$AB = 2.4$ and $BC = 3.2$. Then,$AC = \ldots$

Easy

View Solution

In $\Delta XYZ$,the bisector of $\angle Y$ intersects $\overline{XZ}$ at $M$. Then,which of the following holds true?

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is a median. If $AB = 5$ and $BC = 12$,find $BD$.

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $a = 9$ and $c = 12$,then $b = \dots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. If $AM = 7$ and $CM = 9$,then $BC = \ldots$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $D$ is a point on $\overline{BC}$. Prove that $AD^{2} + BC^{2} = AC^{2} + BD^{2}$.

In rectangle $ABCD$,$AB = 2.4$ and $BC = 3.2$. Then,$AC = \ldots$

In $\Delta XYZ$,the bisector of $\angle Y$ intersects $\overline{XZ}$ at $M$. Then,which of the following holds true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module