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प्रश्न
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
पर्याय
True
False
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उत्तर
This statement is False.
Explanation:
Given in the question, ABCD is a parallelogram and X is the mid-point of AB.
So, area(ABCD) = area(AXCD) + area(ΔXBC) ...(i)
Now, diagonal AC of a parallelogram divides it into two triangles of equal area.
area(ABCD) = 2area(ΔABC) ...(ii)
Similarly, X is the mid-point of AB,
So, area(ΔCXB) = `1/2`area(ΔABC) ...(iii) [Median divides the triangle in two triangles of equal area]
2area(ΔABC) = `24 + 1/2` area(ΔABC) ...[By using equation (i), (ii) and (iii)]
Now, 2area(ΔABC) – `1/2`area(ΔABC) = 24
`3/2`area(ΔABC) = 24
Therefore, area(ΔABC) = `(2 xx 24)/3` = 16 cm2.
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संबंधित प्रश्न
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
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).
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
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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- ar (ΔABEF)
- ar (ΔABD)
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A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)
In the following figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)
