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1) P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar(BQC).
2) ABCD is a parallelogram, AE perpendicular to DC and CF perpendicular AD.If AB = 16 cm, AE= 8cm and CF = 10cm, find AD. diagram
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Answer:
If a triangle and a parallelogram are on the same base and between the same parallel lines, the area of the triangle will be half of the area of the parallelogram.
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC)
It can be observed that ΔBQC and parallelogram ABCD lie on the same base BC and these are between the same set of parallel lines AD and BC.
∴ Area (ΔBQC) = 1/2 Area (ABCD).....(1)
Similarly, ΔAPB and parallelogram ABCD lie on the same base AB and between the same set of parallel lines AB and DC.
∴ Area (ΔAPB) = 1/2 Area (ABCD).....(2)
From Equations (1) and (2), we see that,
Area (ΔBQC) = Area (ΔAPB)
Step-by-step explanation:
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[tex]\large\underline{\sf{Solution-1}}[/tex]
Given that, P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
Now, Triangle APB and parallelogram ABCD are on the same base AB and between same parallels AB and CD.
We know, Area of triangle which is on the same base of parallelogram and between same parallels is half of the parallelogram.
[tex]\implies\sf\:Area( \triangle \: APB) = \dfrac{1}{2} \: Area(ABCD) - - - (1) \\ [/tex]
Further, Triangle BQC and parallelogram ABCD are on the same base AB and between same parallels AD and BC.
[tex]\implies\sf\:Area( \triangle \: BQC) = \dfrac{1}{2} \: Area(ABCD) - - - (2) \\ [/tex]
So, from equation (1) and (2), we get
[tex]\implies\bf\:Area(\triangle \:APB) = Area( \triangle \: BQC) \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex]\large\underline{\sf{Solution-2}}[/tex]
Given that, ABCD is a parallelogram, AE perpendicular to DC and CF perpendicular AD.
Further given that, AB = 16 cm, AE= 8cm and CF = 10cm.
Now,
[tex]\sf\: Area( { \parallel}^{gm} \: ABCD) = AD \times CF - - - (1)\\ [/tex]
Also,
[tex]\sf\: Area( { \parallel}^{gm} \: ABCD) = AB \times AE - - - (2)\\ [/tex]
From equation (1) and (2), we get
[tex]\sf\: AD \times CF = AB \times AE \\ [/tex]
[tex]\sf\: AD \times 10 = 16 \times 8 \\ [/tex]
[tex]\sf\: AD \times 10 = 128 \\ [/tex]
[tex]\implies\bf\:AD = 12.8 \: cm \\ [/tex]