Question

please answer Monday is my exam

Answer 1

Q.6

In ΔADC, S and R are the midpoints of AD and DC respectively. 

Recall that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and half of it.

 Hence SR || AC and SR = (1/2) AC → (1)

 Similarly, in ΔABC, P and Q are midpoints of AB and BC respectively.

 ⇒ PQ || AC and PQ = (1/2) AC → (2)

[By midpoint theorem] 

From equations (1) and (2),

we getPQ || SR and PQ = SR → (3) 

Clearly, one pair of opposite sides of quadrilateral PQRS is equal and parallel. 

Hence PQRS is a parallelogram Hence the diagonals of parallelogram PQRS bisect each other. Thus PR and QS bisect each other.




7.



Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. 

To Prove: 

(i) D is the mid - point of AC 

(ii) MD | AC 

(iii) CM = MA = 1 / 2 AB. 

Proof : (i) Since M is the mid point of hyp. AB and MD | | BC  .

⇒ D is the mid - point of AC .

(ii) Since ∠BCA = 90° 

and MD  | | BC  [given] 

⇒  ∠MDA = ∠BCA 

= 90° [corresp ∠s]

⇒ MD | AC 

(iii) Now, in △ADM and △CDM 

MD = MD [common]

∠MDA = ∠MDC [each = 90°]

AD = CD [∵ D the mid - point of AC]

⇒ △ADM ≅ △CDM  [by SAS congruence axiom]

⇒ AM = CM 

Also, M is the mid - point of AB [given]

⇒ CM = MA = 1 / 2 = AB.