Suppose A B C is triangle and D, E are points on the sides AB and… - Vidyadip

MCQ

Suppose $A B C$ is 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.

A
$(1,2]$
✓
$\left(2, \frac{5}{2}\right]$
C
$\left(\frac{5}{2}, 3\right]$
D
$\left(3, \frac{7}{2}\right]$

✓

Answer

Correct option: B.

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

b
(b)

$\frac{\operatorname{ar} \triangle ABC }{\operatorname{ar} \triangle ADE }=\frac{\frac{1}{2} \cdot AB \cdot AC \sin \theta}{\frac{1}{2} \cdot AD \cdot AE \sin \theta}$

$=\frac{ AB }{ AD } \times \frac{ AC }{ AE }$

$=\frac{5}{3} \times \frac{3}{2}=\frac{5}{2}$

correct option is $\frac{5}{2}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The number of values of $'r'$ satisfying $^{69}C_{3r-1} - ^{69}C_{r^2}=^{69}C_{r^2-1} - ^{69}C_{3r}$ is :-

If $A =$ [$x:x$ is a multiple of $3$] and $B =$ [$x:x$ is a multiple of $5$], then $A -B$ is ($\bar A$ means complement of $A$)

If $f(x) = \frac{{{x^2} - 1}}{{{x^2} + 1}}$, for every real numbers. then the minimum value of $f$

$\mathop {\lim }\limits_{\theta \to {0^ + }} \,{\left( {\sin \theta } \right)^{\left( {\sin \theta - {{\sin }^2}\theta } \right)}}$ is

Lines are drawn parallel to the line $4x -3y + 2 = 0$ at a distance $\frac {3}{5}$ from the origin. Then which one of the following points lies on any of these lines?

Let $z_1, z_2 \in C$ such that $| z_1 + z_2 |= \sqrt 3$ and $|z_1| = |z_2| = 1,$ then the value of $|z_1 - z_2|$ is

Solution of differential equation $x\,dy - y\,dx = 0$ represents

If the co-ordinates of two points $A$ and $B$ are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively and $P$ is any point on the conic, $9 x^{2}+16 y^{2}=144,$ then $PA + PB$ is equal to

The area (in sq. units) of the region $\left\{ {x \in R:x \ge } \right.0,\,y \ge 0,\,y \ge x - 2\,and\,y \le \sqrt x \} \,,\,$ is

Let $[\alpha]$ denote the greatest integer $\leq \alpha$. Then $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots .+[\sqrt{120}]$ is equal to.