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Q15:
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution :
Given:
1. A vertical pole $AB$ of length $6\text{ m}$ casts a shadow $BC$ of length $4\text{ m}$ on the ground.
2. A tower $DE$ of unknown height $h$ casts a shadow $EF$ of length $28\text{ m}$ on the ground at the same time.
To Find:
The height of the tower ($DE = h$).
Step 1: Establishing Geometric Similarity
Let $\triangle ABC$ be the triangle formed by the pole and its shadow, and $\triangle DEF$ be the triangle formed by the tower and its shadow.
In $\triangle ABC$ and $\triangle DEF$:
1. $\angle B = \angle E = 90^\circ$ (Since both the pole and the tower are vertical to the ground).
2. $\angle C = \angle F$ (Since the sun's rays fall at the same angle at the same time for both objects).
Therefore, by the AA (Angle-Angle) Similarity Criterion, $\triangle ABC \sim \triangle DEF$.
Step 2: Applying the Property of Similar Triangles
Since the triangles are similar, the ratios of their corresponding sides are equal:
$\frac{AB}{DE} = \frac{BC}{EF}$
Step 3: Substituting Known Values
Given $AB = 6\text{ m}$, $BC = 4\text{ m}$, and $EF = 28\text{ m}$. Let $DE = h$.
$\frac{6}{h} = \frac{4}{28}$
Step 4: Solving for $h$
Simplify the fraction on the right side:
$\frac{4}{28} = \frac{1}{7}$
Now, the equation is:
$\frac{6}{h} = \frac{1}{7}$
By cross-multiplication:
$h \times 1 = 6 \times 7$
$h = 42$
Final Answer: The height of the tower is 42 m.
More Questions from Class 10 Mathematics Triangles EXERCISE 6.3
- Q1(i): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (i)
- Q1(ii): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (ii)
- Q1(iii): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (iii)
- Q1(iv): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (iv)
- Q1(v): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (v)
- Q1(vi): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (vi)
- Q2: In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^{\circ}$ and $\angle CDO = 70^{\circ}$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.
- Q3: Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.
- Q4: In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.
- Q5: $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.
- Q6: In Fig. 6.37, if $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$.
- Q7(i): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (i) $\triangle AEP \sim \triangle CDP$
- Q7(ii): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (ii) $\triangle ABD \sim \triangle CBE$
- Q7(iii): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iii) $\triangle AEP \sim \triangle ADB$
- Q7(iv): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iv) $\triangle PDC \sim \triangle BEC$
- Q8: $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\triangle ABE \sim \triangle CFB$.
- Q9(i): In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (i) $\triangle ABC \sim \triangle AMP$
- Q9(ii): In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (ii) $\frac{CA}{PA} = \frac{BC}{MP}$
- Q10(i): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (i) $\frac{CD}{GH} = \frac{AC}{FG}$
- Q10(ii): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (ii) $\triangle DCB \sim \triangle HGE$
- Q10(iii): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (iii) $\triangle DCA \sim \triangle HGF$
- Q11: In Fig. 6.40, $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, prove that $\triangle ABD \sim \triangle ECF$.
- Q12: Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \triangle PQR$.
- Q13: $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.
- Q14: Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC \sim \triangle PQR$.
- Q16: If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectively where $\triangle ABC \sim \triangle PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$.
CBSE Solutions for Class 10 Mathematics Triangles
Chapters in CBSE - Class 10 Mathematics
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