# Solution - Triangles on the Same Base and Between the Same Parallels

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ConceptTriangles on the Same Base and Between the Same Parallels

#### Question

In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

#### Solution

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#### Similar questions VIEW ALL

In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).

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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)

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D and E are points on sides AB and AC respectively of ΔABC such that

ar (DBC) = ar (EBC). Prove that DE || BC.

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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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Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

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#### BooksVIEW ALL [1]

Solution for concept: Triangles on the Same Base and Between the Same Parallels. For the course 8th-10th CBSE
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