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प्रश्न
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
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उत्तर
(i) ΔACB and ΔACF lie on the same base AC and are between
The same parallels AC and BF.
∴ Area (ΔACB) = Area (ΔACF)
(ii) It can be observed that
Area (ΔACB) = Area (ΔACF)
⇒ Area (ΔACB) + Area (ACDE) = Area (ACF) + Area (ACDE)
⇒ Area (ABCDE) = Area (AEDF)
संबंधित प्रश्न
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)]
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.
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = 1/2 ar(ARC)
(ii) ar(RQC) = 3/8 ar(ABC)
(iii) ar(PBQ) = ar(ARC)
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = `1/4` ar (ABC).
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Figure), then prove that ar (BPQ) = `1/2` ar (∆ABC).
In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)
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)
