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प्रश्न
In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).
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उत्तर
Draw a line through A parallel to BC
Given that, BD= DE = EC
We observe that the triangles ABD and AEC are on the equal bases and between the same
parallels C and BC. Therefore, Their areas are equal.
Hence, ar ( ABD) = ar (ΔADE) = ar ( ΔACDE)
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संबंधित प्रश्न
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, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that
ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
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.
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Prove that ar (LZY) = ar (MZYX)
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
In the following figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ).
