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# ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig. 9.6). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is

a. a : b

b. (3a + b) : (a + 3b)

c. (a + 3b) : (3a + b)

d. (2a + b) : (3a + b)

**Solution:**

It is given that

AB = a cm

DC = b cm

AB || DC

E and F are the mid-points of AD and BC

Consider h as the distance between AB, CD and EF

Now join BD which intersects EF at M

In ∆ ABD,

E is the midpoint of AD and EM || AB

M is the midpoint of BD

Using the midpoint theorem

EM = 1/2 AB … (1)

In ∆ CBD

MF = 1/2 CD … (2)

From equations (1) and (2)

EM + MF = 1/2 AB + 1/2 CD

EF = 1/2 (AB + CD)

EF = 1/2 (a + b)

Here

Area of trapezium ABFE = 1/2 [sum of parallel sides] × [distance between parallel sides]

Substituting the values

= 1/2 [a + 1/2 (a + h)] × h

= 1/4 [3a + b] h

Similarly

Area of trapezium EFCD = 1/2 [b + 1/2 (a + h)] × h = 1/4 [3b + a] h

Required ratio = Area of trapezium ABFE/ Area of trapezium EFCD

By substituting the values

= 1/4 [3a + b] h/ 1/4 [3b + a] h

= [3a + b]/ [3b + a]

= [3a + b]: [3b + a]

Therefore, the ratio of ar (ABFE) and ar (EFCD) is (3a + b) : (a + 3b).

**✦ Try This: **The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 9 cm and 6 cm is :

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

**NCERT Exemplar Class 9 Maths Exercise 9.1 Problem 10**

## ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig. 9.6). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is a. a : b, b. (3a + b) : (a + 3b), c. (a + 3b) : (3a + b), d. (2a + b) : (3a + b)

**Summary:**

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig. 9.6). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is (3a + b) : (a + 3b)

**☛ Related Questions:**

- If P is any point on the median AD of a ∆ ABC, then ar (ABP) ≠ ar (ACP). Is the given statement true . . . .
- If in Fig. 9.7, PQRS and EFRS are two parallelograms, then ar (MFR) = 1/2 ar (PQRS). Is the given st . . . .
- ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm² , then ar (ABC) = 24 cm² . . . .

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