MCQ
A vertical stick $1.8m$ long casts a shadow $45\ cm$ long on the ground. At the same time, what is the lenght of the shadow of a pole $6m$ high?
- A$2.4m$
- B$1.35m$
- ✓$1.5m$
- D$13.5m$
✓
Answer
Correct option: C.
$1.5m$
Let $AN$ be the vertical stick and $AW$ be its shadow.
Let $OB$ be the pole and $OW$ be its shadows.
$AW = 45\ cm = 0.45m$
$AN = 1.8m$
$OB = 6m$
Ratio of actual lengths $=$ ratio of their shadows
$\Rightarrow\frac{\text{OB}}{\text{AN}}=\frac{\text{OW}}{\text{AW}}$
$\Rightarrow\frac{6}{1.8}=\frac{\text{OW}}{0.45}$
$\Rightarrow\text{OW}=\frac{6\times0.45}{1.8}$
$\Rightarrow\text{OW}=1.5\text{m}$
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
In the formula $\overline{\text{X}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$ for finding the mean of grouped data di’s are deviations from a of:
→In a cyclic $\square ABCD$, twice the measure of $\angle A$ is thrice the measure of $\angle C$. Find the measure of $\angle C$ ?
→The remainder when the square of any prime number greater than $3$ is divided by $6$, is:
→The median class for the following frequency distribution is
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 |
| Frequency | 5 | 7 | 13 | 18 |
$\Rightarrow\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $\text{ar}(\triangle\text{ABC})=36\text{cm}^2$ and $\text{ar}(\triangle\text{DEF})=49\text{cm}^2.$ Then, the ratio of their corresponding sides is:
→$\sqrt2$ is:
→One card is drawn at random from a well$-$shuffled deck of $52$ cards. What is the probability of getting a $6?$
→The slant height of a bucket is $45\ cm$ and the radii of its top and bottom are $28\ cm$ and $7\ cm$, respectively. The curved surface area of the bucket is:
→
In $\triangle RST , \angle S =90^{\circ}, RT =12 m$, $ST =8 m$ then $RS =$ ______
The diameters of the top and the bottom portions of a bucket are $42\ cm$ and $28\ cm$ respectively. If the height of the bucket is $24\ cm$, then the cost of painting its outer surface at the rate of $50\ paise/ cm^2$ is:
→