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प्रश्न
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
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उत्तर
It is given that
Area (ΔAOD) = Area (ΔBOC)
Area (ΔAOD) + Area (ΔAOB) = Area (ΔBOC) + Area (ΔAOB)
Area (ΔADB) = Area (ΔACB)
We know that triangles on the same base having areas equal to each other lie between the same parallels.
Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels.
i.e., AB || CD
Therefore, ABCD is a trapezium.
संबंधित प्रश्न
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)
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)
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]
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.
In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :
(1) ar (Δ PBQ) = ar (Δ ARC)
(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)
(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
In the following figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).
