Question : $\triangle$LMN is right angled at M. If $\angle$N = 45$^\circ$, what is the length of MN (in cm), if NL = 9$\sqrt2$cm?
Option 1: $9\sqrt2$
Option 2: $\frac{9}{\sqrt2}$
Option 3: $18$
Option 4: $9$
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Correct Answer: $9$
Solution : $\triangle$LMN is right angled at M. $\angle$M = 90$^\circ$ $\angle$N = 45$^\circ$ $\angle$L = 180$^\circ$ – (45$^\circ$ + 90$^\circ$) = 45$^\circ$ So, $\triangle$LMN is an isosceles right-angled triangle. ⇒ LM = MN Also, NL = 9$\sqrt2$ cm Using the Pythagoras theorem, LM$^2$ + MN$^2$ = LN$^2$ ⇒ 2MN$^2$ = (9$\sqrt2$)$^2$ ⇒ MN = 9 cm Hence, the correct answer is $9$.
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Question : $\triangle PQR$ is right-angled at $Q$. The length of $PQ$ is 5 cm and $\angle P R Q=30^{\circ}$. Determine the length of the side $QR$.
Option 1: $5 \sqrt{3}~cm$
Option 2: $3 \sqrt{3}~cm$
Option 3: $\frac{1}{\sqrt{3}}~cm$
Option 4: $\frac{5}{\sqrt{3}}~cm$
Question : Suppose $\triangle A B C$ be a right-angled triangle where $\angle A=90^{\circ}$ and $A D \perp B C$. If area $(\triangle A B C)$ $=80 \mathrm{~cm}^2$ and $BC=16$ cm, then the length of $AD$ is:
Option 1: 10 cm
Option 2: 24 cm
Option 3: 18 cm
Option 4: 12 cm
Question : In $\triangle$LMN, LM = $5\sqrt{2}$ cm, LN = 13 cm and $\angle$LMN=135$^{\circ}$. What is the length (in cm) of MN?
Option 1: $7$
Option 2: $8$
Option 3: $8\sqrt{2}$
Option 4: $7\sqrt{2}$
Question : In a right-angled triangle $\Delta PQR, PR$ is the hypotenuse of length 20 cm, $\angle PRQ = 30^{\circ}$, the area of the triangle is:
Option 1: $50\sqrt{3}\text{ cm}^{2}$
Option 2: $100\sqrt{3}\text{ cm}^{2}$
Option 3: $25\sqrt{3}\text{ cm}^{2}$
Option 4: $\frac{100}{\sqrt{3}}\text{ cm}^{2}$
Question : Suppose $\triangle ABC$ be a right-angled triangle where $\angle A=90°$ and $AD\perp BC$. If the area of $\triangle ABC =40$ cm$^{2}$ and $\triangle ACD =10$ cm$^{2}$ and $\overline{AC}=9$ cm, then the length of $BC$ is:
Option 1: 12 cm
Option 2: 18 cm
Option 3: 4 cm
Option 4: 6 cm
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