MCQ
In the given figure, $\text{DE}\parallel\text{BC} , AB = 15\ cm, BD = 6\ cm, AC = 25\ cm,$ then $AE$ is equal to :
- A$20\ cm.$
- B$18\ cm.$
- C$10\ cm.$
- ✓$15\ cm$.
✓
Answer
Correct option: D.
$15\ cm$.
Since $\text{DE} \parallel\text{BC},$ then using Thales theorem,
$\Rightarrow\frac{\text{AB}}{\text{DB}}=\frac{\text{AC}}{\text{EC}}$
$\Rightarrow\frac{15}{6}=\frac{25}{\text{EC}}$
$\Rightarrow\text{EC}=10 \text{ cm}$
Now $, AE = AC -EC $
$= 25-10 = 15\ cm$
$\Rightarrow\frac{\text{AB}}{\text{DB}}=\frac{\text{AC}}{\text{EC}}$
$\Rightarrow\frac{15}{6}=\frac{25}{\text{EC}}$
$\Rightarrow\text{EC}=10 \text{ cm}$
Now $, AE = AC -EC $
$= 25-10 = 15\ cm$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Two different coins are tossed simultaneously. The probability of getting at least one head is:
From a point on the ground, which is $15 m$ away from the foot of a vertical tower, the angle of elevation of the top of the tower, is found to be $60^{\circ}$. The height of the tower (in metres) is
→The sum of areas of a major sector and the corresponding minor sector of a circle is equal to
The angles of elevation of the top of a tower from two points on the ground at distances $8m$ and $18m$ from the base of the tower and in the same straight line with it are complementary. The height of the tower is :
→If the points $(k, 2k), (3k, 3k)$ and $(3, 1)$ are collinear, then $k$ :
→In the given figure, PQ and PR are tangents drawn from P to a circle with centre O. If $\angle\text{OPQ}=35^\circ$, then:
→If the sum of $n$ terms of two A.P.'s are in the ratio $(2 n+3):(3 n+2)$, then the ratio of then $m^{\text {th }}$ terms is
→The system of linear equations $a_1 x+b_1 y_1+c_1=0$ and $a_2 x+b_2 y+c_2=0$ has no solution if :
→If the diagonals of a quadrilateral divide each other proportionally, then it is a
If the polynomial $f(x)=a x^3+b x-c$ is divisible by the polynomial $g(x)=x^2+b x+c$, then $a b=$
→