In Fig. 9.22, l, m, n, are straight lines such that l || m and n intersects l at P and m at Q. ABCD is a quadrilateral such that its vertex A is on l. The vertices C and D are on m and AD || n. Show that ar (ABCQ) = ar (ABCDP).

Solution:

Given, l, m, n are straight lines such that L || M

N intersects l at P and m at Q.

ABCD is a quadrilateral with its vertex A on l.

The vertices C and D are on m.

Given, AD || n

We have to show that ar(ABCQ) = ar(ABCDP)

We know that the area of triangles on the same base and between the same parallel lines are equal.

Triangles ADF and AQD are on the same base AD and between the same parallel lines AD and n.

So, ar(ADF) = ar(DFB) ------- (1)

Adding ar (ABCD) on both sides in (1),

ar (APD) + ar (ABCD) = ar (AQD) + ar (ABCD)

From the figure,

ar(APD) + ar(ABCD) = ar(ABCDP)

So, ar(ABCDP) = ar(AQD) + ar(ABCD)

From the figure,

ar(AQD) + ar(ABCD) = ar(ABCQ)

Therefore, ar(ABCDP) = ar(ABCQ)

✦ Try This: ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that area(ADx) = area(ACy)

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9

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NCERT Exemplar Class 9 Maths Exercise 9.4 Sample Problem 2

In Fig. 9.22, l, m, n, are straight lines such that l || m and n intersects l at P and m at Q. ABCD is a quadrilateral such that its vertex A is on l. The vertices C and D are on m and AD || n. Show that ar (ABCQ) = ar (ABCDP)

Summary:

In Fig. 9.22, l, m, n, are straight lines such that l || m and n intersects l at P and m at Q. ABCD is a quadrilateral such that its vertex A is on l. The vertices C and D are on m and AD || n. It is shown that ar (ABCQ) = ar (ABCDP)

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