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प्रश्न
The area of the parallelogram ABCD is 90 cm2 (see figure). Find
- ar (ΔABEF)
- ar (ΔABD)
- ar (ΔBEF)
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उत्तर
Given, area of parallelogram, ABCD = 90 cm2.
i. We know that, parallelograms on the same base and between the same parallel are equal in areas.
Here, parallelograms ABCD and ABEF are on same base AB and between the same parallels AB and CF.
So, ar (ΔBEF) = ar (ABCD) = 90 cm2
ii. We know that, if a triangle and a parallelogram are on the same base and between the same parallels, then area of triangle is equal to half of the area of the parallelogram.
Here, ΔABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
So, ar (ΔABD) = `1/2` ar (ABCD)
= `1/2 xx 90` ...[∴ ar (ABCD) = 90 cm2]
= 45 cm2
iii. Here, ABEF and parallelogram ABEF are on the same base EF and between the same parallels AB and EF.
ar (ΔBEF) = `1/2` ar (ABEF)
= `1/2 xx 90` ...[∴ ar (ABEF) = 90 cm2, from part (i)]
= 45 cm2
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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).
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).
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
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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