Question : In a triangle DEF, DP is the bisector of $\angle D$, meeting EF at P. If DE = 14 cm, DF = 21 cm and EF = 9 cm, find EP.
Option 1: 3.6 cm
Option 2: 5.4 cm
Option 3: 6.3 cm
Option 4: 2.7 cm
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 3.6 cm
Solution : To find EP, we can use the angle bisector theorem, which states that the angle bisector of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle. Given: DE = 14 cm, DF = 21 cm and EF = 9 cm DP is the bisector of $\angle D$ Now, $\frac{EP}{PF} = \frac{DE}{DF}$ Substituting the given values, we get: $\frac{EP}{9 - EP} = \frac{14}{21}$ ⇒ 21 EP = 126 – 14 EP ⇒ 35 EP = 126 ⇒ EP = $\frac{126}{35}$ ⇒ EP = 3.6 Hence, the correct answer is 3.6 cm.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : $\Delta ABC$ and $\Delta DEF$ are similar. Also $\angle A=\angle D$ and $\angle B=\angle E$. If $4AB=DE$ and $BC=\operatorname{12 cm}$, then $EF$ is equal to:
Question : $\Delta ABC$ and $\Delta DEF$ are similar. Also $\angle A=\angle D$ and $\angle B=\angle E$. If $4AB=DE$ and $BC=\operatorname{12 cm}$, then $EF$ is equal to
Question : It is given that $\triangle \mathrm{ABC} \cong \triangle \mathrm{FDE}$ and $\mathrm{AB}=5 \mathrm{~cm}, \angle \mathrm{B}=40^{\circ}$ and $\angle \mathrm{A}=80^{\circ}$. Then which of the following is true?
Question : In $\triangle$ABC, $\angle$A = $\angle$B = 60°, AC = $\sqrt{13}$ cm, the lines AD and BD intersect at D with $\angle$D = 90°. If DB = 2 cm, then the length of AD is:
Question : Two medians DM and EN of $\triangle$DEF intersect each other at O at right angles. If EF = 20 cm and EN = 12 cm, then what is the length of DM?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile