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प्रश्न
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).
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उत्तर
AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas.
∴ Area (ΔABD) = Area (ΔACD)
⇒ Area (ΔABD) = 1/2Area (ΔABC)... (1)
In ΔABD, E is the mid-point of AD. Therefore, BE is the median.
∴ Area (ΔBED) = Area (ΔABE)
⇒ Area (ΔBED) = 1/2Area (ΔABD)
⇒ Area (ΔBED) = 1/2*1/2Area (ΔABC) [From equation (1)]
⇒ Area (ΔBED) = 1/4Area (ΔABC)
संबंधित प्रश्न
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
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.
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 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)
In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)
If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)
