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In the given figure, AC || BH and AD || EF. Show area (ΔAHF) = area (pentagon ABCDE). [Hint: Join BC and DE.] - Mathematics

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Question

In the given figure, AC || BH and AD || EF. Show area (ΔAHF) = area (pentagon ABCDE). [Hint: Join BC and DE.]


[Hint: Join BC and DE.]

Sum
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Solution

Join BC and DE.


From the given figure,

ar(ΔAHF) = ar(ΔACD) + ar(ΔAHC) + ar(ΔADF)

Also, ar(pentagon ABCDE) = ar(ΔACD) + ar(ΔABC) + ar(ΔAED)

It is given that, AC || BH and AD || EF

Since the area of triangles on the same base and between the same parallels are equal.

So, we can say

ar(ΔABC) = ar(ΔAHC) and ar(ΔAED) = ar(ΔADF)

Now we can write,  ar(ABCDE) = ar(ΔACD) + ar(ΔAHC) = ar(ΔADF)

Both ar(ΔAHF) and ar(pentagon ABCDE) are the same.

Hence, area area of ΔAHF = area of pentagon ABCDE

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Chapter 13: Theorems on Area - EXERCISE 13 [Page 162]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 13 Theorems on Area
EXERCISE 13 | Q 9. | Page 162
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