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
Correct Answer: 8.64
Solution : Given: 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 $E$. The area of the triangle = $\frac{1}{2}\times \text{Base} \times \text{Height}$ Use the Pythagoras's theorem in $\triangle ABC$, $(BC)^2=(AB)^2+(AC)^2$ ⇒ $(BC)^2=(16)^2+(12)^2$ ⇒ $(BC)^2=256+144=400$ ⇒ $BC=20\ cm$ The area of the $\triangle ABC= \frac{1}{2}\times 12 \times 16=96 \ cm^2$ Also, $\triangle ABC \sim \triangle EDC$. ⇒ $\frac{\text{The area of $\triangle EDC$}}{\text{The area of $\triangle ABC$}}=(\frac{\text{DC}}{\text{BC}})^2$ ⇒ $\frac{\text{The area of $\triangle EDC$}}{96}=(\frac{6}{20})^2$ ⇒ $\frac{\text{The area of $\triangle EDC$}}{96}=\frac{36}{400}$ ⇒ $\text{The area of $\triangle EDC$}=\frac{36\times 96}{400}=8.64 \ cm^2$ Hence, the correct answer is 8.64.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
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?
Question : If AB = 5 cm, AC = 12 cm, and AB$\perp$ AC, then the radius of the circumcircle of $\triangle ABC$ is:
Question : In $\triangle$ ABC, $\angle$ BCA = $90^{\circ}$, AC = 24 cm and BC = 10 cm. What is the radius (in cm) of the circumcircle of $\triangle$ ABC?
Question : In a triangle ABC, if $\angle B=90^{\circ}, \angle C=45^{\circ}$ and AC = 4 cm, then the value of BC is:
Question : In $\triangle \mathrm{ABC}, \mathrm{BD} \perp \mathrm{AC}$ at D. E is a point on BC such that $\angle \mathrm{BEA}=x^{\circ}$. If $\angle \mathrm{EAC}=62^{\circ}$ and $\angle \mathrm{EBD}=60^{\circ}$, then the value of $x$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile