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प्रश्न
In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).
विकल्प
True
False
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उत्तर
This statement is False.
Explanation:
In the given figure, join PG.
Since, G is the mid-point of CD.
Thus, PG is a median of ΔDPC and it divides the triangle into parts of equal areas.
Then, ar (ΔDPG) = ar (ΔGPC) = `1/2` ar (ΔDPC) ...(i)
Also, we know that, if a parallelogram and a triangle lie on the same base and between the same parallels, then area of triangle is equal to half of the area of parallelogram.
Here, parallelogram EFGD and ΔDPG lie on the same base DG and between the same parallels DG and EF.
So, ar (ΔDPG) = `1/2` ar (EFGD) ...(ii)
From equations (i) and (ii),
`1/2` ar (ΔDPG) = `1/2` ar (EFGD)
⇒ ar (ΔDPC) = ar (EFGD)
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संबंधित प्रश्न
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).
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)
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-
(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar (CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
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The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
