In 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. -Maths 9th

1 Answer

Answer :

Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the same base AD and between the same parallels AD and EQ. So, ar (ΔADQ) = ar (ΔADE) ...(i) Similarly, ΔACP and ΔACB lie on the same base AC and between the same parallels AC and BP. So, ar (ΔACP) = ar (ΔACB) …(ii) On adding Eqs. (i) and (ii), we get ar (ΔADQ) + ar (ΔACP) = ar (ΔADE) + ar (ΔACB) On adding ar (ΔACD) both sides, we get ar (ΔADQ) + ar (ΔACP) + ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

Related questions

Description : In 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. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

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Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

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Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

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Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

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Last Answer : Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ ... Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.

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Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

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Description : Triangle P and pentagon Q have markings on them as shown in the figure. If they are placed over each other, which of the following arrangements is/are possible? 

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Description : ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Last Answer : In ||gm ABCD , ar(△APC) = ar(△BCP) ---i) [∵ triangles on the same base and between the same parallels have equal area] Similarly, ar( △ADQ) = ar(△ADC) ---ii) Now, ar(△ADQ) - ar(△ADP) = ar(△ADC) - ar(△ADP) ... ) From (i) and (iii) , we have ar(△BCP) = ar(△DPQ) or ar( △BPC) = ar(△DPQ)

Description : ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

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Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

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Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

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