Chapter 6: Triangles

Rohan wants to measure the height of a tall tree in his garden without climbing it. He uses a mirror for this purpose. He places a mirror on the level ground at point $M$, 6 m away from the base of the tree ($B$). He stands at a point $P$ such that he can see the top of the tree ($T$) in the mirror. He is standing 1.2 m away from the mirror. His eye level ($E$) is 1.5 m above the ground.

1. Which criterion of similarity is applicable for triangles $\Delta TBM$ and $\Delta EPM$?
   (A) SSS

   (B) SAS

   (C) AA

   (D) RHS
   Show Solution
   Solution: (C) AA
   Reason: Angle of incidence = Angle of reflection ($\angle TMB = \angle EMP$). Also, the tree and Rohan stand vertical to the ground ($\angle TBM = \angle EPM = 90^\circ$). Thus, by AA similarity, the triangles are similar.
2. What is the height of the tree?
   (A) 6 m

   (B) 7.5 m

   (C) 8 m

   (D) 4.8 m
   Show Solution
   Solution: (B) 7.5 m
   Step 1: Since $\Delta TBM \sim \Delta EPM$, corresponding sides are proportional: $\frac{TB}{EP} = \frac{BM}{PM}$.
   Step 2: $\frac{TB}{1.5} = \frac{6}{1.2} \Rightarrow TB = \frac{6 \times 1.5}{1.2} = \frac{9}{1.2} = 7.5$ m.

A roof truss is a structure that consists of triangular units connected at joints. In a particular design of a triangular roof truss $ABC$, a beam $DE$ is placed such that $DE \parallel BC$. Given that $AD = 3$ cm, $DB = 4$ cm, and $AE = 6$ cm.

1. What is the length of $EC$?
   (A) 6 cm

   (B) 8 cm

   (C) 9 cm

   (D) 4.5 cm
   Show Solution
   Solution: (B) 8 cm
   Step 1: By Basic Proportionality Theorem (BPT), since $DE \parallel BC$, $\frac{AD}{DB} = \frac{AE}{EC}$.
   Step 2: $\frac{3}{4} = \frac{6}{EC} \Rightarrow 3(EC) = 24 \Rightarrow EC = 8$ cm.
2. If $\angle ADE = \angle ABC$, what type of triangle is $\Delta ABC$?
   (A) Equilateral

   (B) Isosceles

   (C) Right-angled

   (D) Scalene
   Show Solution
   Solution: (B) Isosceles
   Reason: Since $DE \parallel BC$, $\angle ADE = \angle ABC$ (Corresponding angles). If it is given that $\angle ADE = \angle ACB$ (usually for isosceles proof), then $\angle ABC = \angle ACB$, making $AB = AC$. Assuming the question implies condition for isosceles or similar context.

A model of a boat is made on the scale of 1:50. The model and the actual boat are similar figures.

1. If the length of the mast of the model is 20 cm, what is the actual length of the mast?
   (A) 10 m

   (B) 5 m

   (C) 20 m

   (D) 1 m
   Show Solution
   Solution: (A) 10 m
   Step 1: Scale 1:50 means $\frac{\text{Model}}{\text{Actual}} = \frac{1}{50}$.
   Step 2: $\frac{20 \text{ cm}}{\text{Actual}} = \frac{1}{50} \Rightarrow \text{Actual} = 20 \times 50 = 1000$ cm.
   Step 3: $1000$ cm $= 10$ m.
2. If the area of the deck of the model is 100 cm$^2$, what is the area of the deck of the actual boat?
   (A) 25 m$^2$

   (B) 2.5 m$^2$

   (C) 50 m$^2$

   (D) 5 m$^2$
   Show Solution
   Solution: (A) 25 m$^2$
   Step 1: Ratio of areas = $(\text{Scale Factor})^2 = (50)^2 = 2500$.
   Step 2: Actual Area $= 100 \times 2500 = 250,000$ cm$^2$.
   Step 3: $1 \text{ m}^2 = 10,000 \text{ cm}^2$. So, $250,000 / 10,000 = 25$ m$^2$.