हिंदी

Show that ΔABC with vertices A(–2, 0), B(0, 2) and C(2, 0) is similar to ΔDEF with vertices D(–4, 0), E(0, 4) and F(4, 0).

Advertisements
Advertisements

प्रश्न

Show that ΔABC with vertices A(–2, 0), B(0, 2) and C(2, 0) is similar to ΔDEF with vertices D(–4, 0), E(0, 4) and F(4, 0).

योग
Advertisements

उत्तर

Given: A(–2, 0), B(0, 2), C(2, 0) and D(–4, 0), E(0, 4), F(4, 0).

Step-wise calculation:

1. Notice DEF is obtained by multiplying each coordinate of ABC by 2:

2 × A = 2(–2, 0)

= (–4, 0)

= D

2 × B = 2(0, 2)

= (0, 4)

= E

2 × C = 2(2, 0)

= (4, 0)

= F 

So, DEF is a dilation of ABC about the origin with scale factor 2.

Hence, corresponding angles are equal.

2. Compute side lengths of ΔABC:

`AB = sqrt(0 - (-2)^2 + (2 - 0)^2` 

= `sqrt(4 + 4)`

= `sqrt(8)`

= `2sqrt(2)`

`BC = sqrt((2 - 0)^2 + (0 - 2)^2)` 

= `sqrt(4 + 4)`

= `sqrt(8)`

= `2sqrt(2)`

`CA = sqrt((2 - (-2))^2 + (0 - 0)^2)`

 = `sqrt(16)`

= 4

3. Compute side lengths of ΔDEF:

`DE = sqrt((0 - (-4))^2 + (4 - 0)^2)`

= `sqrt(16 + 16)`

= `sqrt(32)`

= `4sqrt(2)`

`EF = sqrt((4 - 0)^2 + (0 - 4)^2)`

= `sqrt(16 + 16)`

= `4sqrt(2)`

`FD = sqrt((4 - (-4))^2 + (0 - 0)^2)`

= `sqrt(64)`

= 8

4. Compare ratios (or observe factor 2):

`(AB)/(DE) = (2sqrt(2))/(4sqrt(2))`

= `1/2`

`(BC)/(EF) = (2sqrt(2))/(4sqrt(2))`

= `1/2`

`(CA)/(FD) = 4/8`

= `1/2`

All corresponding side ratios are equal (common ratio `1/2`), so sides are proportional. This verifies similarity by the SSS similarity criterion.

ΔABC ~ ΔDEF with correspondence A ↔ D, B ↔ E, C ↔ F. The similarity is a dilation about the origin with scale factor 2 (equivalently the side-length scale factor from ABC to DEF is 2).

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Coordinate Geometry - EXERCISE 6A [पृष्ठ ३१३]

APPEARS IN

आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 6 Coordinate Geometry
EXERCISE 6A | Q 33. | पृष्ठ ३१३
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×