Question

In ΔABC, ∠ABC = 90°, AB = 20 cm, AC = 25 cm, DE is perpendicular to AC such that ∠DEA = 90° and DE = 3 cm as shown in the given figure.

1. Prove that ΔABC ~ ΔAED.
2. Find the lengths of BC, AD, and AE.
3. If BCED represents a plot of land on a map whose actual area on the ground is 576 m², then find the scale factor of the map.

Answer 1

(a) In triangles ΔABC and ΔAED:

∠A is common to both triangles.

∠ABC = ∠AED = 90°.

Therefore, ΔABC is similar to ΔAED by the A similarity criterion. Hence proved.

(b) Applying the Pythagoras theorem in ΔABC:

(AC)² = (AB)² + (BC)² 

25² = 20² + (BC)²

625 = 400 + (BC)²

(BC)² = 225

BC = 15 cm

Since ΔABC ∼ ΔAED, the corresponding sides are proportional:

`(DE)/(BC) = (AD)/(AC) = (AE)/(AB)`

Using `(DE)/(BC) = (AD)/(AC)`

`3/15 = (AD)/25`

AD = `(3 xx 25)/15`

AD = 5 cm

Now Using `(AD)/(AC) = (AE)/(AB)`

`5/25 = (AE)/20`

AE = `(5 xx 20)/25` = 4 cm

Hence, BC = 15 cm, AD = 5 cm, AE = 4cm

(c) Given that the actual area is 576 m² = 57,600 cm².

Area of ΔABC: 

`1/2 xx 15 xx 20 = 150 cm^2`

Area of ΔAED:

`1/2 ​× 4 × 3 = 6 cm^2`

Area of BEDC:

150 − 6 = 144 cm²

The scale factor is given by:

`k^2 = "Area on map"/"Actual area" = 144/57600 = 1/400`

Therefore,

k = `1/20`

Hence, the scale of the map is 1 : 200.