## RS Aggarwal Class 9 Ex 10A Chapter 10

## Exercise 10.1: Area

**Q.1: In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P.**

**Prove that: ar.( △ABP ) = ar. (quad. ABCD)**

**Sol:**

**Given:** ABCD is a quadrilateral in which through D, a line is drawn parallel to AC which meets BC produced in P.

**To prove:**

**Proof:**

Therefore,

Adding

Therefore,

**Q.2: In the adjoining figure, AD and BC are 2 diagonals of the parallelogram ABCD. The point at which the diagonals intersect is O. Prove that OA = OD.**

**Sol:**

**Given:** Two triangles, i.e.

**To prove:** OA = OD

**Construction:** Draw

**Proof:** We have:

So, 1/2 x BC x AP = 1/2 x BC x DQ [From (1)]

Therefore, AP = DQ . . . . . . . . . (2)

Now, in

And,

AP = DQ [From (2)]

Thus, by Angle-Angle-side criterion of congruence, we have:

Therefore,

The corresponding parts of the congruent triangles are equal.

Therefore, OA = OD [C.P.C.T]

**Q.3: In the adjoining figure, AD is one of the medians of a △ABC and P is appointed on AD.**

**Prove that:**

**(i) ar.(**

**(ii)ar.(**

**Sol:**

**Given:** A

**To prove:**

**(i)**

**(ii)**

**Proof:**

**(i)** In

So,

**(ii)** In

So,

But,

Subtracting

Therefore,

Therefore,

**Q.4: In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.**

**If BO = OD, prove that**

**ar.(**

**Sol:**

**Given:** A quadrilateral ABCD in which diagonals AC and BD intersect at O and BO = OD

**To Prove:**

**Proof:** Since OB = OD [given]

So, AO is the median of

Therefore,

As OC is the median of

Therefore,

Adding both sides of (i) and (ii), we get:

Therefore,

**Q.5: ABC is a triangle in which D is the midpoint of BC and E is the midpoint of AD.**

**Prove that:**

**ar.( △BED)= 14ar(△ABC)**

**Sol:**

**Given:** A

**To prove:**

**Proof:** Since,

i.e.

Now, as BE is the median of

Since,

Therefore,

=

=

=

**Q.6: The vertex A of △ABC is joined to a point D on the side BC. The midpoint AD is E.**

**Prove that:**

**ar( △BEC) = 12ar(△ABC)**

**Sol: **

**Given:** A

**To Prove:**

**Proof:** Since, BE is a median of

So,

i.e.

As, CE is the median of

So,

**Adding (i) and (ii), we get:**

=

**Q.7: D is the midpoint of side BC of △ABC and E is the midpoint of BD. if O is the midpoint of AE.**

**Prove that:**

**ar.(**

**Sol:**

**Given:** A

**To prove:**

**Proof:** Since O is the mid-point of AE.

So, BO is the medium of

Therefore,

Now, E is the mid-point of BD

So AE divides

Therefore,

As D is the mid-point of BC

So,

Therefore,

=

=

=

=

**Q.8: In the adjoining figure, ABCD parallelogram and O is any point on the diagonal AC.**

**Show that: ar.( △AOB )=ar.( △AOD)**

**Sol:**

**Given:** A parallelogram ABCD in which O is any point on the diagonal AC

**To Prove:**

**Construction:** Join BD which intersects AC at P.

**Proof:** As diagonals of a parallelogram bisect each other therefore, OP is the median of

Therefore,

Also, AP is the median of

Therefore,

Adding both sides, we get:

Therefore,

**Q.9: P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of parallelogram ABCD. Show that PQRS is parallelogram and also show that**

**ar.(parallelogram PQRS)= 12 * ar.(parallelogram ABCD).**

**Sol:**

**Given:** ABCD is a parallelogram and P,Q,R and S are the midpoints of AB,BC,CD,DA respectively.

**To prove: **PQRS is a parallelogram and parallelogram PQRS=

**Construction: **Join AC,BD and SQ.

**Proof: **As S and R are the midpoints of AD and CD, so in

Also, as P and Q are the midpoints of AB and BC. So, in

Therefore,

So,

Similarly, we can prove

Thus PQRS is a parallelogram as its opposite sides are parallel.

Since the diagonals of a parallelogram bisect each other.

So in

O is the midpoint of AC and S is the midpoint of AD.

Therefore,

Similarly in

Thus, ABQS is a parallelogram.

Now,

Similarly, we can prove that:

Adding (i) and (ii) we get,

Therefore,

**Q.10: The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F. Show that ar.(pentagon ABCDE) = ar.(**

**Sol:**

**Given: **ABCDE is a pentagon,

**To prove:** ar(pentagon ABCDE) = ar(

**Proof:**

ar(pentagon ABCDE) = ar(

Also, ar(

Now,

Therefore,

Similarly,

Therefore,

**From (iii) and (iv) we have:**

Adding

By substituting the values from (i) and (ii) we get:

ar(pentagon ABCDE) = ar(

**Q.11: Prove that a median divides a triangle into two triangles of equal area.**

**Sol:**

Let, AD is a median of

**To prove:**

**Construction:** Draw

**Proof:**

Since, D is the midpoint of BC, we have

Multiplying with

Therefore,

**Q.12: Show that a diagonal divides a parallelogram into two triangles of equal area.**

**Sol:**

Let, ABCD be a parallelogram and BD be its diagonal.

**To prove:**

Proof:

In

AB = CD (opposite sides of a parallelogram)

AD = CD (opposite sides of a parallelogram)

BD = DB (common side)

Therefore,

**Q.13: The base BC of △ABC is divided at D such that BD = 12DC . Prove that**

**ar( △ABD) = 13×ar(△ABC)**

**Sol:**

**Given:** D is a point on BC of

**To prove:**

**Construction:** Draw

**Proof:**

In

BC = BD + DC

Now we have:

Therefore,

**Q.14: In the adjoining figure, the point D divides the side BC of **

**Prove that:**

**ar( △ABD):ar(△ADC) = m:n.**

**Sol:**

**Given:** D is a point on BC of

**To prove:**

**Construction :** Draw

**Proof:**

**Dividing (i) and (ii) we have:**

Therefore,