# RBSE Solutions for Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Ex 10.2

RBSE Solutions for Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Ex 10.2 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 10 Area of Triangles and Quadrilaterals Ex 10.2.

 Board RBSE Class Class 9 Subject Maths Chapter Chapter 10 Chapter Name Area of Triangles and Quadrilaterals Exercise Ex 10.2 Number of Questions Solved 6 Category RBSE Solutions

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 10 Area of Triangles and Quadrilaterals Ex 10.2

Question 1.
In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Solution.
Area of parallelogram ABCD = AB x AE
⇒ 16 cm x 8 cm = 128 cm²
Also area of parallelogram ABCD, when (RBSESolutions.com) base is AD and corresponding altitude is CF = AD x CF
i.e. AD x 10 cm = 128 cm²
⇒ AD = $$\frac { 128 }{ 10 }$$ cm = 12.8 cm

Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD. Show that
ar (EFGH) = $$\frac { 1 }{ 2 }$$ ar (||gm ABCD).
Solution.
Given: ABCD is a parallelogram in which E, F, G and H are respectively the (RBSESolutions.com) midpoints of sides AB, BC, CD and DA.

To prove:
ar (EFGH) = $$\frac { 1 }{ 2 }$$ ar (||gm ABCD).
Construction: Join AC and HF.
Proof: In ∆ABC,
E and F are mid-points of sides AB and BC respectively.
Therefore
EF || AC and EF = $$\frac { 1 }{ 2 }$$AC …(i)
(by mid-point theorem)
Similarly, ∆ADC, H and G are (RBSESolutions.com) mid-points of sides AD and DC respectively
∴HG || AC and HG = $$\frac { 1 }{ 2 }$$AC …(ii)
From (i) and (ii), we get
EF = HG and EF || HG
∴EFGH is a parallelogram.
HA = FB and HA || FB
⇒ ABFH is a parallelogram.
Now in ∆HEF and parallelogram HABF are on the same base HF and between the same parallel HF and AB.
∴Area of ∆HEF = $$\frac { 1 }{ 2 }$$ ar (||gm HABF) …(iii)
Similarly,
ar (∆HGF) = $$\frac { 1 }{ 2 }$$ ar (||gm HFCD) …(iv)
Adding (iii) and (iv), we get
ar (∆HEF) + ar (∆HGF)
= $$\frac { 1 }{ 2 }$$ ar (||gm HABF) + $$\frac { 1 }{ 2 }$$ ar (||gm HFCD)
⇒ ar (||gm EFGH) = $$\frac { 1 }{ 2 }$$ ar (||gm ABCD).

Question 3.
P and Q are any two points lying on (RBSESolutions.com) the sides DC and AD respectively of a parallelogram ABCD. Show that:
ar (∆APB) = ar (∆BQC).
Solution.
Since ∆APB and parallelogram ABCD lie on the same base AB and between the same parallels AB and DC
ar (∆APB) = $$\frac { 1 }{ 2 }$$ ar (||gm ABCD) ….(i)

Also, ∆BQC and parallelogram ABCD lie on the same base BC and between the same parallels BC and AD.
ar (∆BQC) = $$\frac { 1 }{ 2 }$$ ar (||gm ABCD) …(ii)
From (i) and (ii), we have
ar (∆APB) = ar (∆BQC)

Question 4.
In figure, P is a point in the interior (RBSESolutions.com) of a parallelogram ABCD. Show that
(i) ar (∆APB) + ar (∆PCD) = $$\frac { 1 }{ 2 }$$ ar (||gm ABCD)
(ii) ar (∆APD) + ar (∆PBC) = ar (∆APB) + ar (∆PCD).
Solution.
Through P, draw a line EF parallel to AB.
(i) Since ∆APB and parallelogram ABFE are on the same base AB and between the same parallels AB and EF. Therefore
ar. (∆APB) = $$\frac { 1 }{ 2 }$$ ar (||gm ABFE) …(i)
Similarly,
ar (∆PCD) = $$\frac { 1 }{ 2 }$$ ar (||gm EFCD) …(ii)

On adding (i) and (ii), we get ar (∆APB) + ar (∆PCD)
= $$\frac { 1 }{ 2 }$$ ar (||gm ABFE) + $$\frac { 1 }{ 2 }$$ ar (||gm EFCD)
⇒ ar (∆APB) + ar (∆PCD) = $$\frac { 1 }{ 2 }$$ ar (||gm ABCD) …(3)
(ii) ∵ ar (∆APB) + ar (∆PBC) + ar (∆PCD) + ar (∆APD) = ar (||gm ABCD)
⇒ {ar (∆APB) + ar (∆PCD)} + ar (∆PBC) + ar (∆APD) = ar (||gm ABCD)
⇒ $$\frac { 1 }{ 2 }$$ ar (||gm ABCD) + ar (∆PBC) + ar (∆APD) = ar (||gm ABCD)
⇒ ar (∆PBC) + ar (∆APD) = ar (||gm ABCD) $$\frac { 1 }{ 2 }$$  – ar(||gm ABCD)
⇒ ar (∆PBC) + ar (∆APD) = $$\frac { 1 }{ 2 }$$ ar (||gm ABCD) …(iv)
From (iii) and (iv), we get
⇒ ar (∆APB) + ar (∆PCD)
= ar (∆PBC) + ar (∆APD).

Question 5.
In figure, PQRS and ABRS are (RBSESolutions.com) parallelograms and X is any point on side BR.
Show that:

(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = $$\frac { 1 }{ 2 }$$ ar (PQRS)
Solution.
(i) As we know that parallelograms on (RBSESolutions.com) the same base and between the same parallels are equal in area.
Here parallelograms PQRS and ABRS are on the same base SR and between the same parallels PB and SR. Therefore,
ar (PQRS) = ar (ABRS)
(ii) Again ∆AXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR.
∴ ar (∆AXS) = $$\frac { 1 }{ 2 }$$ ar (∆BRS)
But ar (ABRS) = ar (PQRS)
(proved earlier)
⇒ ar (AXS) = $$\frac { 1 }{ 2 }$$ ar (PQRS)

Question 6.
A farmer was having a field in the form of (RBSESolutions.com) a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Solution.
The field is divided into three parts. The parts are ∆APS, ∆APQ and ∆AQR.

Since ∆APQ and parallelogram, PQRS are (RBSESolutions.com) on the same base PQ and between the same parallels PQ and RS.
∴ ar (∆APQ) = $$\frac { 1 }{ 2 }$$ ar (||gm PQRS)
⇒ 2 ar (∆APQ) = ar (||gm PQRS)
But ar (|| gm PQRS)
= ar (∆APQ) + ar (∆APS) + ar (∆AQR) ⇒ 2ar (∆APQ)
= ar (∆APQ) + ar (∆APS) + ar (∆AQR) ⇒ ar (∆APQ) = ar (∆APS) + ar (∆AQR)
To sow wheat and pulses in equal portions of the field separately, farmer sow wheat in ∆APQ and pulses in other two triangular region or pulses in ∆APQ and wheat in other two triangular regions.

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