ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.13). Show that
(i) ∆ APB ≅ ∆ CQD
(ii) AP = CQ
(i) In ∆APB and ∆CQD,
∠APB = CQD (Each 90°)
AB = CD (Opposite sides of parallelogram ABCD) ∠ABP
= ∠CDQ (Alternate interior angles for AB || CD)
∠∆APB ∠∆CQD (By AAS congruency)
(ii) By using the above result
∆APB ∠∆CQD, we obtain
AP = CQ (By CPCT)
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.12). Show that:
(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?