Question
and are respectively the mid-point of sides and of a triangle and is the mid-point of , show that
(i)
(ii)
(iii)
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$$P$$ and $$Q$$ are respectively the mid- points of sides $$AB$$ and $$BC$$ of $$\Delta ABC$$ and $$R$$ is the mid- point of $$AP$$.
Join $$AQ$$ and $$PC$$.
(i) We have,
$$ar (PQR) = 12 ar (APQ)$$
[Since QR is a median of $$\Delta APQ$$ and it divides the $$\Delta$$ into two other $$\Delta s$$ of equal area]
$$=\dfrac{1}{2}\times \dfrac{1}{2} ar (ABQ) $$
[Since $$QP$$ is a median of $$\Delta ABQ$$]
$$=\dfrac{1}{4} ar (ABQ) = \dfrac{1}{4}×\dfrac{1}{2} ar (ABC) $$
[Since AQ is a median of $$Δ ABC$$]
[Since AQ is a median of $$Δ ABC$$]
$$=\dfrac{1}{8} ar (ABC)$$ ... (1)
Again, $$ar(ARC) = \dfrac{1}{2} ar (APC) $$
[Since $$CR$$ is a median of $$\Delta APC$$]
$$=\dfrac{1}{2}\times\dfrac{1}{2} ar (ABC) $$
[Since $$CP$$ is a median of $$\Delta ABC$$]
$$=\dfrac{1}{4} ar (ABC)$$ ... (2)
From (1) and (2) , we get
$$ar (PQR) = \dfrac{1}{8} ar (ABC) = \dfrac{1}{2}\times\dfrac{1}{4} ar (ABC) \\
=\dfrac{1}{2} ar (ARC).$$
(ii) We have,
$$ar (RQC) = ar(RQA) + ar (AQC) – ar (ARC)$$ ... (3)
Now, $$ar (\Delta RQA) = \dfrac{1}{2} ar (PQA) $$
$$ar (PQR) = \dfrac{1}{8} ar (ABC) = \dfrac{1}{2}\times\dfrac{1}{4} ar (ABC) \\
=\dfrac{1}{2} ar (ARC).$$
(ii) We have,
$$ar (RQC) = ar(RQA) + ar (AQC) – ar (ARC)$$ ... (3)
Now, $$ar (\Delta RQA) = \dfrac{1}{2} ar (PQA) $$
[Since RQ is a median of $$\Delta PQA$$]
$$=\dfrac{1}{2}\times\dfrac{1}{2} ar (AQB) $$
[Since $$PQ$$ is a median of $$Δ AQB$$]
$$=\dfrac{1}{4} ar (AQB) $$
$$=\dfrac{1}{4}\Delta\dfrac{1}{2} ar(ABC) $$
[Since $$AQ$$ is a median of $$\Delta ABC$$]
$$=\dfrac{1}{8} ar (ABC)$$ ... (4)
$$ar (AQC) = 12 ar (ABC)$$ ... (5)
[Since $$AQ$$ is a median of $$\Delta ABC$$]
$$ar (Δ ARC) = \dfrac{1}{2} ar (APC)$$
[Since $$CR$$ is a median of $$\Delta APC$$]
$$=\dfrac{1}{2}\times\dfrac{1}{2} ar(ABC)$$
[Since $$CP$$ is a median of $$\Delta ABC$$]
$$=\dfrac{1}{4} ar (ABC)$$ ... (6)
From (3), (4) , (5) and (6) we have
$$ar (RQC) = \dfrac{1}{8} ar (ABC) + \dfrac{1}{2} ar (ABC) – \dfrac{1}{4} ar (ABC) \\
=\left(\dfrac{1}{8}+\dfrac{1}{2}−\dfrac{1}{4}\right) ar (ABC) \\
=\dfrac{3}{8}ar (ABC)$$
(iii) We have,
$$ar (PBQ) = 12 ar (ABQ) $$
$$ar (RQC) = \dfrac{1}{8} ar (ABC) + \dfrac{1}{2} ar (ABC) – \dfrac{1}{4} ar (ABC) \\
=\left(\dfrac{1}{8}+\dfrac{1}{2}−\dfrac{1}{4}\right) ar (ABC) \\
=\dfrac{3}{8}ar (ABC)$$
(iii) We have,
$$ar (PBQ) = 12 ar (ABQ) $$
[Since $$PQ$$ is a median of $$\Delta ABQ$$]
$$=\dfrac{1}{2}\times\dfrac{1}{2}ar(ABC) $$
[Since $$AQ$$ is a median of $$\Delta ABC$$]
$$=14 ar (ABC) $$
$$= ar(ARC) $$ [Using (6)]
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Question Text | and are respectively the mid-point of sides and of a triangle and is the mid-point of , show that (i) (ii) (iii) |
Answer Type | Text solution:1 |
Upvotes | 150 |