Question : In $\triangle X Y Z$, L and M are the middle points of the sides XY and XZ, respectively. N is a point on the segment LM such that LN : NM = 1 : 2. If LN = 5 cm, then YZ is equal to:
Option 1: 30 cm
Option 2: 24 cm
Option 3: 28 cm
Option 4: 26 cm
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Correct Answer: 30 cm
Solution : In ΔXYZ, L and M are the middle points of the sides XY and XZ, respectively. LN : NM = 1 : 2 and LN = 5 cm In $\triangle$XYZ and in $\triangle$XLM $\angle$X = $\angle$X [common] $\angle$XYZ = $\angle$XLM $\angle$XZY = $\angle$XML $\therefore \triangle$XYZ ~ $\triangle$XLM LN : NM = 1 : 2 LN = 5 cm $\frac{\text{LN}}{\text{NM}} = \frac{1}{2}$ $⇒\frac{5}{\text{NM}} = \frac{1}{2}$ $⇒\text{NM} = 10$ cm LM = LN + NM = 5 + 10 = 15 cm Since, $\triangle$XYZ ~ $\triangle$XLM $\frac{\text{LM}}{\text{YZ}}=\frac{\text{XL}}{\text{XY}}$ $⇒\frac{15}{\text{YZ}}=\frac{\text{XL}}{\text{2XL}}$ [∵ L is midpoint of XY] $⇒\frac{15}{\text{YZ}}=\frac{1}{2}$ $\therefore\text{YZ} = 30$ cm Hence, the correct answer is 30 cm.
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Question : In $\triangle X Y Z, P$ is a point on side YZ and XY = XZ. If $\angle X P Y=90°$ and $Y P=9\ \text{cm}$, then what is the length of $YZ$?
Option 1: 17 cm
Option 2: 18 cm
Option 3: 12 cm
Option 4: 15 cm
Question : If $xy+yz+zx=0$, then $(\frac{1}{x^2–yz}+\frac{1}{y^2–zx}+\frac{1}{z^2–xy})$$(x,y,z \neq 0)$ is equal to:
Option 1: $3$
Option 2: $1$
Option 3: $x+y+z$
Option 4: $0$
Question : If $x : y$ is the ratio of two whole numbers and $z$ is their HCF, then the LCM of those two numbers is:
Option 1: $yz$
Option 2: $\frac{xz}{y}$
Option 3: $\frac{xy}{z}$
Option 4: $xyz$
Question : In $\triangle A B C,$ P and Q are the middle points of the sides AB and AC, respectively. R is the point on the segment PQ such that PR : RQ = 1 : 3. If PR = 6 cm, then BC is:
Option 1: 46 cm
Option 2: 48 cm
Option 3: 44 cm
Option 4: 50 cm
Question : In $\triangle X Y Z, \angle YXZ=90°$, P is a point on side YZ such that XP is perpendicular to YZ. If XP = YP = 10 cm then what will be the value of PZ?
Option 1: 8 cm
Option 2: 9 cm
Option 4: 10 cm
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