Triangles | Exercise 6.3

Question 15

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.

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Solution

We will use the concept of similar triangles to solve this problem.

Step 1 — Draw the diagram

Let's imagine the pole and the tower. They both stand straight up from the ground. The sun casts shadows for both. This forms two right-angled triangles. Let the pole be CD. Its height is 6 m. Its shadow is DF. Its length is 4 m. Let the tower be AB. Its height is hh. Its shadow is BE. Its length is 28 m.

Diagram 1

Step 2 — Identify similar triangles

We have two triangles formed. One is CDF\triangle CDF for the pole. The other is ABE\triangle ABE for the tower. The pole and tower are vertical. So, CDF=90\angle CDF = 90^\circ. Also, ABE=90\angle ABE = 90^\circ. The sun's angle is the same. So, DFC=BEA\angle DFC = \angle BEA. These are angles of elevation of the sun. By AA similarity criterion, the triangles are similar. So, CDFABE\triangle CDF \sim \triangle ABE.

Step 3 — Set up the proportion

Corresponding sides of similar triangles are proportional. We can write a ratio of their heights. We can also write a ratio of their shadows.

Height of towerHeight of pole=Shadow of towerShadow of pole\frac{\text{Height of tower}}{\text{Height of pole}} = \frac{\text{Shadow of tower}}{\text{Shadow of pole}}

Let hh be the height of the tower. We substitute the given values.

ABCD=BEDF\frac{AB}{CD} = \frac{BE}{DF}

h6=284\frac{h}{6} = \frac{28}{4}

Step 4 — Calculate the height

Now we solve for hh. We multiply both sides by 6.

h=284×6h = \frac{28}{4} \times 6

h=7×6h = 7 \times 6

h=42 m\boxed{h = 42 \text{ m}}

Answer

The height of the tower is 42 m.

More questions in Exercise 6.3

Q1
  1. 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 :
Q2
  1. In Fig. 6.35, ΔODCΔOBA\Delta \text{ODC} \sim \Delta \text{OBA}, BOC=125\angle \text{BOC} = 125^\circ and CDO=70\angle \text{CDO} = 70^\circ. Find DOC\angle \text{DOC}, DCO\angle \text{DCO} and OAB\angle \text{OAB}.
Q3
  1. Diagonals AC and BD of a trapezium ABCD with ABDC\text{AB} \parallel \text{DC} intersect each other at the point O. Using a similarity criterion for two triangles, show that OAOC=OBOD\frac{\text{OA}}{\text{OC}} = \frac{\text{OB}}{\text{OD}}.
Q4
  1. In Fig. 6.36, QRQS=QTPR\frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}} and 1=2\angle 1 = \angle 2. Show that ΔPQSΔTQR\Delta \text{PQS} \sim \Delta \text{TQR}.
Q5
  1. S and T are points on sides PR and QR of ΔPQR\Delta \text{PQR} such that P=RTS\angle \text{P} = \angle \text{RTS}. Show that ΔRPQΔRTS\Delta \text{RPQ} \sim \Delta \text{RTS}.
Q6
  1. In Fig. 6.37, if ΔABEΔACD\Delta \text{ABE} \cong \Delta \text{ACD}, show that ΔADEΔABC\Delta \text{ADE} \sim \Delta \text{ABC}.
Q7
  1. In Fig. 6.38, altitudes AD and CE of ΔABC\Delta \text{ABC} intersect each other at the point P. Show that:

(i) ΔAEPΔCDP\Delta \text{AEP} \sim \Delta \text{CDP} (ii) ΔABDΔCBE\Delta \text{ABD} \sim \Delta \text{CBE} (iii) ΔAEPΔADB\Delta \text{AEP} \sim \Delta \text{ADB} (iv) ΔPDCΔBEC\Delta \text{PDC} \sim \Delta \text{BEC}

Q8
  1. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABEΔCFB\Delta \text{ABE} \sim \Delta \text{CFB}.
Q9
  1. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) ΔABCΔAMP\Delta \text{ABC} \sim \Delta \text{AMP} (ii) CAPA=BCMP\frac{\text{CA}}{\text{PA}} = \frac{\text{BC}}{\text{MP}}

Q10
  1. CD and GH are respectively the bisectors of ACB\angle \text{ACB} and EGF\angle \text{EGF} such that D and H lie on sides AB and FE of ΔABC\Delta \text{ABC} and ΔEFG\Delta \text{EFG} respectively. If ΔABCΔFEG\Delta \text{ABC} \sim \Delta \text{FEG}, show that:

(i) CDGH=ACFG\frac{\text{CD}}{\text{GH}} = \frac{\text{AC}}{\text{FG}} (ii) ΔDCBΔHGE\Delta \text{DCB} \sim \Delta \text{HGE} (iii) ΔDCAΔHGF\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 Δ\Delta ABD \sim Δ\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 Δ\Delta PQR (see Fig. 6.41). Show that Δ\Delta ABC \sim Δ\Delta PQR.

Q13

D is a point on the side BC of a triangle ABC such that \angle ADC = \angle BAC. Show that CA2=CB.CD\text{CA}^2 = \text{CB.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 Δ\Delta ABC \sim Δ\Delta PQR.

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.

Q16

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

ABPQ=ADPM\frac{\text{AB}}{\text{PQ}} = \frac{\text{AD}}{\text{PM}}

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