In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (ACB) = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)

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Solution

(i) ΔACB and ΔACF lie on the same base AC and are between

The same parallels AC and BF.

∴ Area (ΔACB) = Area (ΔACF)

(ii) It can be observed that

Area (ΔACB) = Area (ΔACF)

⇒ Area (ΔACB) + Area (ACDE) = Area (ACF) + Area (ACDE)

⇒ Area (ABCDE) = Area (AEDF)

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(ii) ar(AEDF) = ar(ABCDE)

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a r e a (A E D F) = a r e a (A B C D E)

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In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar (ACB) = ar (ACF) (ii) ar (AEDF) = ar (ABCDE)

(i) ΔACB and ΔACF lie on the same base AC and are between The same parallels AC and BF. ∴ Area (ΔACB) = Area (ΔACF) (ii) It can be observed that Area (ΔACB) = Area (ΔACF) ⇒ Area (ΔACB) + Area (ACDE) = Area (ACF) + Area (ACDE) ⇒ Area (ABCDE) = Area (AEDF)

In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (ACB) = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)

(i) ΔACB and ΔACF lie on the same base AC and are between

The same parallels AC and BF.

∴ Area (ΔACB) = Area (ΔACF)

(ii) It can be observed that

Area (ΔACB) = Area (ΔACF)

⇒ Area (ΔACB) + Area (ACDE) = Area (ACF) + Area (ACDE)

⇒ Area (ABCDE) = Area (AEDF)