Suppose ABC is a triangle and D, E are points | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $\angle A = 	heta$.The area of $	riangle ABC$ is given by $	ext{ar}(	riangle ABC) = \frac{1}{2} \cdot AB \cdot AC \cdot \sin 	heta$.The a

Vedclass · Accepted Answer

$\left(2, \frac{5}{2}\right]$

Vedclass · Answer

(B) Let $\angle A = 	heta$.The area of $	riangle ABC$ is given by $	ext{ar}(	riangle ABC) = \frac{1}{2} \cdot AB \cdot AC \cdot \sin 	heta$.The area of $	riangle ADE$ is given by $	ext{ar}(	riangle ADE) = \frac{1}{2} \cdot AD \cdot AE \cdot \sin 	heta$.Therefore,the ratio of the areas is:$\frac{	ext{ar}(	riangle ABC)}{	ext{ar}(	riangle ADE)} = \frac{\frac{1}{2} \cdot AB \cdot AC \cdot \sin 	heta}{\frac{1}{2} \cdot AD \cdot AE \cdot \sin 	heta} = \frac{AB}{AD} 	imes \frac{AC}{AE}$.Given $AD : AB = 3 : 5$,we have $\frac{AB}{AD} = \frac{5}{3}$.Given $AE : AC = 2 : 3$,we have $\frac{AC}{AE} = \frac{3}{2}$.Thus,the ratio is $\frac{5}{3} 	imes \frac{3}{2} = \frac{5}{2} = 2.5$.The value $2.5$ lies in the interval $\left(2, \frac{5}{2}ight]$.Hence,the correct option is $B$.

Suppose $ABC$ is a triangle and $D, E$ are points on the sides $AB$ and $AC$ respectively. If $AD : AB = 3 : 5$ and $AE : AC = 2 : 3$,then the ratio of the areas of the triangles $ABC$ and $ADE$ lies in the interval.

Similar Questions

If $\tan \theta = \frac{a}{b},$ then $\frac{\sin \theta}{\cos^8 \theta} + \frac{\cos \theta}{\sin^8 \theta} = $

$\sec 2 \theta - \tan 2 \theta =$

The value of $\tan 67 \frac{1}{2}^{\circ} + \cot 67 \frac{1}{2}^{\circ}$ is

Find the value of $\sin \frac{31 \pi}{3}$.

Find the value of $\operatorname{cosec} 750^{\circ} - 2 \cot 765^{\circ}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module