Question : BL and CM are medians of $\triangle$ABC right-angled at A and BC = 5 cm. If BL=$\frac{3\sqrt5}{2}$ cm, then the length of CM is
Option 1: $2\sqrt5$ cm
Option 2: $5\sqrt2$ cm
Option 3: $10\sqrt2$ cm
Option 4: $4\sqrt5$ cm
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Correct Answer: $2\sqrt5$ cm
Solution : Given: BC = 5 cm. and BL=$\frac{3\sqrt5}{2}$ cm According to the question, $4(BL^2+CM^2)=5BC^2$ ⇒ $4[(\frac{3\sqrt5}{2})^2+CM^2]=5×5^2$ ⇒ $4×\frac{45}{4}+4CM^2=125$ ⇒ $4CM^2=125-45$ ⇒ $4CM^2=80$ ⇒ $CM^2=\frac{80}{4}$ ⇒ $CM=\sqrt20=2\sqrt5$ Hence, the correct answer is $2\sqrt5$ cm.
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Question : $\angle A$ of $\triangle ABC$ is a right angle. $AD$ is perpendicular on $BC$. If $BC= 14$ cm and $BD= 5$ cm, then measure of $AD$ is:
Option 2: $\sqrt5$ cm
Option 3: $3\sqrt5$ cm
Option 4: $3.5\sqrt5$ cm
Question : If $\triangle ABC$~$\triangle PQR$, the ratio of perimeter of $\triangle ABC$ to perimeter of $\triangle PQR$ is 36 : 23 and QR = 3.8 cm, then the length of BC is:
Option 1: $4 \frac{103}{121} cm$
Option 2: $3 \frac{109}{121} cm$
Option 3: $5 \frac{109}{115} cm$
Option 4: $3 \frac{107}{115} cm$
Question : $\triangle{ABC}$ is a right angled triangle. $\angle \mathrm{C}=90°$, AB = 25 cm and BC = 20 cm. What is the value of $\mathrm{sec}\; A$?
Option 1: $\frac{5}{3}$
Option 2: $\frac{4}{5}$
Option 3: $\frac{4}{3}$
Option 4: $\frac{5}{4}$
Question : $\triangle \mathrm{RMS}$ is right-angled at M. The length of the base, RM = 4 cm and the length of perpendicular MS = 3 cm. Find the value of $\sec R$.
Option 1: $\frac{3}{4}$
Option 2: $\frac{5}{4}$
Option 3: $\frac{4}{5}$
Option 4: $\frac{2}{5}$
Question : In $\triangle$ABC, $\angle$A = $90^{\circ}$, BP and CQ are two medians. Then the value of $\frac{BP^2 + CQ^2}{BC^2}$ is:
Option 1: $\frac{4}{5}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{3}{5}$
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