Home
Class 12
MATHS
"Another Is related. I A triangle (v) AP...

"Another Is related. I A triangle (v) APCQ a Samatara Pumana. . 10. ABCD is a parallelogram and AP and CQ Diagonals from vertices A and C are perpendicular to BD respectively. (See figure 8.21). Show that (i) AAPBEACQD Shot on realme X = CQ DO Ankit Keiji Figure 8.21 EF and lets see.

Promotional Banner

Similar Questions

Explore conceptually related problems

ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD (see figure) show that : /_\APB ~= /_\CQD

ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its diagonal BD (See fig.) Show that AP = CQ.

ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its diagonal BD (See fig.) Show that AP = CQ.

ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure)..Show that : i) DeltaAPB = DeltaCQD .

ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its diagonal BD (See fig.) Show that Delta APB ~= Delta CQD .

ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that (i) DeltaAPB ~= DeltaCQD (ii) AP = CQ

ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that (i) DeltaAPB ~= DeltaCQD (ii) AP = CQ

ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that (i) DeltaAPB ~= DeltaCQD (ii) AP = CQ

ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that (i) DeltaAPB ~= DeltaCQD (ii) AP = CQ

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . Show that (i) DeltaA P B~=\ DeltaC Q D (ii) A P\ =\ C Q