Question
In parallelogram ABCD shown below, the vertical distance between the lines AD and BC is 5 cm and length of BC is 4 cm. P and Q are the midpoints of AB and AD respectively. Area of triangle AQP is ____.
Answer: Option B
:
B
Given:
ABCD is a parallelogram.
Base BC = 4 cm
Height of a parallelogram = 5 cm
Area of parallelogram ABCD = Base xHeight
=5cm×4cm
=20cm2
A diagonal divides a parallelogram in two congruent triangles which have equal areas.
∴AreaΔABD=AreaΔBDC
∴AreaΔABD =12Area of ABCD
=10cm2
Let R be the midpoint of the diagonal. Join P and Q to R.
P and Q are midpoints of AB and AD.
Therefore PQ is parallel to BD (Midpoint theorem)
Similarly, QR is parallel to AB and PR is parallel to AD
Therefore, APQR, DQPR and BPQR are all parallelograms
∴AreaΔAQP=AreaΔRQP
=AreaΔBPR=AreaΔQPR...(i)
AreaΔABD=AreaΔAQP+AreaΔRQP+AreaΔBPR+AreaΔQPR
AreaΔABD=4×AreaΔAQP=14×AreaΔABD
∴AreaΔAQP=14×10
=2.5cm2
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:
B
Given:
ABCD is a parallelogram.
Base BC = 4 cm
Height of a parallelogram = 5 cm
Area of parallelogram ABCD = Base xHeight
=5cm×4cm
=20cm2
A diagonal divides a parallelogram in two congruent triangles which have equal areas.
∴AreaΔABD=AreaΔBDC
∴AreaΔABD =12Area of ABCD
=10cm2
Let R be the midpoint of the diagonal. Join P and Q to R.
P and Q are midpoints of AB and AD.
Therefore PQ is parallel to BD (Midpoint theorem)
Similarly, QR is parallel to AB and PR is parallel to AD
Therefore, APQR, DQPR and BPQR are all parallelograms
∴AreaΔAQP=AreaΔRQP
=AreaΔBPR=AreaΔQPR...(i)
AreaΔABD=AreaΔAQP+AreaΔRQP+AreaΔBPR+AreaΔQPR
AreaΔABD=4×AreaΔAQP=14×AreaΔABD
∴AreaΔAQP=14×10
=2.5cm2
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