Question : If $\mathrm{D}$ is the midpoint of $\mathrm{BC}$ in $\triangle \mathrm{ABC}$ and $\angle A=90^{\circ}$, then $AD=$ ______.
Option 1: $\frac{\mathrm{BC}}{4}$
Option 2: $2 \mathrm{BC}$
Option 3: $\frac{\mathrm{BC}}{2}$
Option 4: $\mathrm{BC}$
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Correct Answer: $\frac{\mathrm{BC}}{2}$
Solution :
Given: In $\triangle \mathrm{ABC}$, the midpoint of $\mathrm{BC}$ is $\mathrm{D}$ and $\angle A=90^{\circ}$. AD is the median to the BC. We know that the median to the hypotenuse of a right-angled triangle is half of the hypotenuse. $AD = \frac{1}{2} BC$ Hence, the correct answer is $\frac{\mathrm{BC}}{2}$.
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Question : In a $\triangle ABC$, if $\angle A=90^{\circ}, AC=5 \mathrm{~cm}, BC=9 \mathrm{~cm}$ and in $\triangle PQR, \angle P=90^{\circ}, PR=3 \mathrm{~cm}, QR=8$ $\mathrm{cm}$, then:
Question : In $\triangle \mathrm{ABC}$, $\angle \mathrm{ABC} = 90^{\circ}$, $\mathrm{BP}$ is drawn perpendicular to $\mathrm{AC}$. If $\angle \mathrm{BAP} = 50^{\circ},$ what is the value of $\angle \mathrm{PBC}?$
Question : If $\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}, \mathrm{BC}=6 \mathrm{~cm}$, and $\angle \mathrm{A}=75^{\circ}$, then which one of the following is true?
Question : If in $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}, \frac{A B}{D E}=\frac{B C}{F D}$, then they will be similar when:
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