Question : In a $\triangle ABC$, D and E are two points on AB and AC respectively such that DE || BC, DE bisects the $\triangle ABC$ in two equal areas. Then the ratio DB : AB is:
Option 1: $1:\sqrt2$
Option 2: $1:2$
Option 3: $\left ( \sqrt2-1 \right ):\sqrt2$
Option 4: $\sqrt2:1$
Correct Answer: $\left ( \sqrt2-1 \right ):\sqrt2$
Solution : Given, area of ($\triangle ADE)$ = area of $BDEC$ ⇒ $\triangle 2ADE = \triangle ABC$ Since, $DE \parallel BC$ ∴ $\triangle ADE$ ~ $\triangle ABC$ Now, $\frac{area (\triangle ADE)}{area (\triangle ABC)} = \frac{1}{2} = \frac{AD^2}{AB^2}$ ⇒ $\frac{AB}{AD} = \sqrt2$....................(1) ⇒ $\frac{AB}{AD}-1 = \sqrt2-1$ ⇒ $\frac{AB-AD}{AD}= \sqrt2-1$ ⇒ $\frac{BD}{AD} = \sqrt2-1$......................(2) $\therefore \frac{BD}{AB} = \frac{BD}{AD}× \frac{AD}{AB}= \frac{\sqrt2-1}{\sqrt2}$ Hence, the correct answer is $(\sqrt2-1):\sqrt2$.
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Question : For a triangle ABC, D and E are two points on AB and AC such that $\mathrm{AD}=\frac{1}{6} \mathrm{AB}$, $\mathrm{AE}=\frac{1}{6} \mathrm{AC}$. If BC = 22 cm, then DE is _______. (Consider up to two decimals)
Option 1: 1.33 cm
Option 2: 1.67 cm
Option 3: 3.67 cm
Option 4: 3.33 cm
Question : $\triangle$ABC is an equilateral triangle in which D, E, and F are the points on sides BC, AC, and AB, respectively, such that AD $\perp$ BC, BE $\perp$ AC and CF $\perp$ AB. Which of the following is true?
Option 1: 4AC$^2$ = 5BE$^2$
Option 2: 3AC$^2$ = 4BE$^2$
Option 3: 2AB$^2$ = 3AD$^2$
Option 4: 7AB$^2$ = 9AD$^2$
Question : ABC is an equilateral triangle points D, E and F are taken in sides AB, BC and CA respectively so that, AD = BE = CF. Then DE, EF, and FD enclose a triangle which is:
Option 1: equilateral
Option 2: isosceles
Option 3: right angled
Option 4: none
Question : $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are two triangles such that $\triangle \mathrm{ABC} \cong \triangle \mathrm{FDE}$. If AB = 5 cm, $\angle$B = 40° and $\angle$A = 80°, then which of the following options is true?
Option 1: DF = 5 cm, $\angle$E = 60°
Option 2: DE = 5 cm, $\angle$F = 60°
Option 3: DE = 5 cm, $\angle$D = 60°
Option 4: DE = 5 cm, $\angle$E = 60°
Question : In $\triangle{ABC}, \angle A=90^{\circ}, {AB}=16\ cm $ and $AC =12 \ cm$. $D$ is the midpoint of $AC$ and $DE \perp CB$ at $E$. What is the area (in ${cm}^2$ ) of $\triangle{CDE}$?
Option 1: 8.64
Option 2: 7.68
Option 3: 5.76
Option 4: 6.25
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