हिंदी

◻PQRS is a rectangle. If A, B and C are the mid-points of sides PQ, PS and QR, respectively. Prove that AB + AC = 1/2 (PR + SQ). - Mathematics

Advertisements
Advertisements

प्रश्न

◻PQRS is a rectangle. If A, B and C are the mid-points of sides PQ, PS and QR, respectively. Prove that AB + AC = `1/2` (PR + SQ).

प्रमेय
Advertisements

उत्तर

Given:

PQRS is a rectangle.

A is midpoint of PQ, B is midpoint of PS, C is midpoint of QR.

To Prove: AB + AC = `1/2` (PR + SQ).

Proof [Step-wise]:

1. Place the rectangle in a coordinate plane for convenience:

Let P = (0, 0)

Q = (2a, 0)

S = (0, 2b)

R = (2a, 2b)

2. Find coordinates of midpoints:

A, midpoint of PQ:

`A = ((0 + 2a)/2, (0 + 0)/2)`

= (a, 0)

B, midpoint of PS:

`B = ((0 + 0)/2, (0 + 2b)/2)`

= (0, b)

C, midpoint of QR:

`C = ((2a + 2a)/2, (0 + 2b)/2)`

= (2a, b)

3. Compute lengths:

AB = Distance between A(a, 0) and B(0, b):

`AB = sqrt((a - 0)^2 + (0 - b)^2)`

= `sqrt(a^2 + b^2)`

AC = Distance between A(a, 0) and C(2a, b):

`AC = sqrt((2a - a)^2 + (b - 0)^2)`

= `sqrt(a^2 + b^2)`

Thus, AB = AC = `sqrt(a^2 + b^2)`.

So, AB + AC = `2sqrt(a^2 + b^2)`.

4. Diagonals:

PR: Distance between P(0, 0) and R(2a, 2b)

= `sqrt((2a)^2 + (2b)^2)`

= `2sqrt(a^2 + b^2)`

SQ: Distance between S(0, 2b) and Q(2a, 0)

= `sqrt((2a)^2 + (-2b)^2)`

= `2sqrt(a^2 + b^2)`

Thus, `PR + SQ = 4 sqrt(a^2 + b^2)`

And `1/2 (PR + SQ) = 2sqrt(a^2 + b^2)`

5. Compare: 

AB + AC

= `2sqrt(a^2 + b^2)`

= `1/2 (PR + SQ)`

Therefore, AB + AC = `1/2` (PR + SQ).

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Mid-point Theorem - Exercise 9A [पृष्ठ १९५]

APPEARS IN

नूतन Mathematics [English] Class 9 ICSE
अध्याय 9 Mid-point Theorem
Exercise 9A | Q 4. | पृष्ठ १९५
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×