Triangles | Exercise 6.3

Question 15

A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower.

Solution

Understand the Question

At any given moment, the sun casts shadows at the same angle of elevation for all objects in the vicinity.
A vertical object and its shadow on horizontal ground form a right-angled triangle.
Because both the pole and the tower form right-angled triangles with the same sun angle, the two triangles are similar by the AA similarity criterion.
Since corresponding sides of similar triangles are in proportion: $\dfrac{\text{Height of tower}}{\text{Height of pole}} = \dfrac{\text{Shadow of tower}}{\text{Shadow of pole}}$

Step 1 · Establish Similarity of the Two Triangles

Let $\text{CD} = 6\text{ m}$ be the vertical pole and $\text{DF} = 4\text{ m}$ be its shadow. Let $\text{AB} = h$ be the vertical tower and $\text{BE} = 28\text{ m}$ be its shadow. Diagram 1

In $\triangle \text{ABE}$ and $\triangle \text{CDF}$ : $\angle \text{ABE} = \angle \text{CDF} = 90^\circ \quad (\text{Both are vertical to the ground})$ $\angle \text{AEB} = \angle \text{CFD} \quad (\text{Angle of elevation of the Sun at the same time})$

By AA similarity criterion: $\triangle \text{ABE} \sim \triangle \text{CDF}$

Step 2 · Calculate the Height of the Tower

Since corresponding sides of similar triangles are proportional: $\dfrac{\text{AB}}{\text{CD}} = \dfrac{\text{BE}}{\text{DF}}$

Substitute the given values: $\dfrac{h}{6} = \dfrac{28}{4}$

\begin{aligned} h &= \dfrac{28}{4} \times 6 \\[0.6em] &= 7 \times 6 \\[0.6em] &= 42\text{ m} \end{aligned}

Answer

42\text{ m}

Common Mistakes

Mismatched Ratios: Mixing up corresponding sides, such as writing $\dfrac{\text{Height}}{\text{Shadow}} = \dfrac{\text{Shadow}}{\text{Height}}$ . Ensure corresponding terms align: $\dfrac{\text{Height}_1}{\text{Height}_2} = \dfrac{\text{Shadow}_1}{\text{Shadow}_2}$ .
Missing Justification for Similarity: Assuming triangles are similar without mentioning that shadows cast at the same time share the same angle of elevation of the sun.

More questions in Exercise 6.3

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 :

In Fig. 6.35, $\Delta \text{ODC} \sim \Delta \text{OBA}$ , $\angle \text{BOC} = 125^\circ$ and $\angle \text{CDO} = 70^\circ$ . Find $\angle \text{DOC}$ , $\angle \text{DCO}$ and $\angle \text{OAB}$ .

Diagonals $\text{AC}$ and $\text{BD}$ of a trapezium $\text{ABCD}$ with $\text{AB} \parallel \text{DC}$ intersect each other at the point $\text{O}$ . Using a similarity criterion for two triangles, show that $\dfrac{\text{OA}}{\text{OC}} = \dfrac{\text{OB}}{\text{OD}}$ .

In Fig. 6.36, $\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}}$ and $\angle 1 = \angle 2$ . Show that $\Delta \text{PQS} \sim \Delta \text{TQR}$ .

S and T are points on sides $PR$ and $QR$ of $\Delta \text{PQR}$ such that $\angle \text{P} = \angle \text{RTS}$ . Show that $\Delta \text{RPQ} \sim \Delta \text{RTS}$ .

In Fig. 6.37, if $\Delta \text{ABE} \cong \Delta \text{ACD}$ , show that $\Delta \text{ADE} \sim \Delta \text{ABC}$ .

In Fig. 6.38, altitudes $AD$ and $CE$ of $\Delta \text{ABC}$ intersect each other at the point $P$ . Show that:

(i) $\Delta \text{AEP} \sim \Delta \text{CDP}$

(ii) $\Delta \text{ABD} \sim \Delta \text{CBE}$

(iii) $\Delta \text{AEP} \sim \Delta \text{ADB}$

(iv) $\Delta \text{PDC} \sim \Delta \text{BEC}$

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\Delta \text{ABE} \sim \Delta \text{CFB}$ .

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) $\Delta \text{ABC} \sim \Delta \text{AMP}$

(ii) $\dfrac{\text{CA}}{\text{PA}} = \dfrac{\text{BC}}{\text{MP}}$

Q10

CD and GH are respectively the bisectors of $\angle \text{ACB}$ and $\angle \text{EGF}$ such that D and H lie on sides AB and FE of $\Delta \text{ABC}$ and $\Delta \text{EFG}$ respectively. If $\Delta \text{ABC} \sim \Delta \text{FEG}$ , show that:

(i) $\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}$

(ii) $\Delta \text{DCB} \sim \Delta \text{HGE}$

(iii) $\Delta \text{DCA} \sim \Delta \text{HGF}$

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 $\Delta ABD \sim \Delta ECF$ .

Q12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\Delta PQR$ (see Fig. 6.41). Show that $\Delta ABC \sim \Delta PQR$ .

Q13

D is a point on the side $BC$ of a triangle $ABC$ such that $\angle \text{ADC} = \angle \text{BAC}$ . Show that $\text{CA}^2 = \text{CB} \cdot \text{CD}$ .

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 $\Delta ABC \sim \Delta PQR$ .

Q15

A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower.

Q16

If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$ , respectively where $\Delta ABC \sim \Delta PQR$ , prove that

$\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{AD}}{\text{PM}}$