RBSE Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2

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

Answer:
We have Area of a ||gm = base × height 
Area of ||gm ABCD = AB × AE
= (16 × 8) cm² = 128 cm²
Area of ||gm ABCD = AD × CF
= (AD × 10) cm²
From (1) and (2), we get 
128 = AD × 10
So, 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 (ABCD).
Answer:

∆HGF and ||gm HDCF stand on the same base HF and lie between the same parallels HF and DC.
∴ ar (HGF) = \(\frac{1}{2}\) ar (HDCF) ........ (1)
Similarly, ∆HEF and || gmABFH stand on the same base HF and lie between the same parallels HF and AB.
ar (HEF) = \(\frac{1}{2}\) ar (ABFH)
∴ Adding (1) and (2), we get
ar (HGF) + ar (∆HEF) = \(\frac{1}{2}\) [ar (HDCF) + ar (ABFH)
or ar (EFGH) = \(\frac{1}{2}\) ar (ABCD).    

Question 3.
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC). 
Answer:

∆ APB and ||gm ABCD stand on the same base AB and lie between the same parallels AB and DC.
∴ ar (APB) = \(\frac{1}{2}\) ar (ABCD) ........... (1)
Similarly, ∆BQC and ||gm ABCD stand on the same base BC and lie between the same parallels BC and AD.
∴ ar (BQC) = \(\frac{1}{2}\) ar (ABCD)
From (1) and (2), we have :
ar (APB) = ar (BQG).

Question 4.
In the figure, P is a point in the ulterior of a parallelogram ABCD. Show that:

(i) ar (APB) + ar (PCD) = \(\frac{1}{2}\) ar (ABCD)
(ii) ar (APD) + ar (PBO = ar (APB) + ar (PCD)
[Hint: Through P, draw a line parallel to AS.]
Answer:
Draw EPF parallel to AB or DC and GPH parallel to AD or BC.    
Now, AGHD is a ||gm        [∵ GBT || DA and AG || DB]    
Similarly, HCBG, EFCD and ABFE are parallelograms.    

(i)  ∆APB and || gm ABFE stand on the same base AB and lie between the same parallels AB and EF. 
∴ ar (APB) = \(\frac{1}{2}\) ar (ABFE)  ........... (1)
Similarly, ar (PCD)  =  \(\frac{1}{2}\) ar (EFCD) ............ (2)
Adding (1) and (2), we get
ar (APB) + ar(PCD)  = \(\frac{1}{2}\) [ar (ABFE) + ar (EFCD)]
=  \(\frac{1}{2}\) ar (ABCD) ......... (3)

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

(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = \(\frac{1}{2}\)ar (PQRS)
Answer:
(i) ||gm PQRS and ||gm ABRS stand on the same base RS and lie between the same parallels SR and PAQB.
∴ ar (PQRS) = ar (ABRS) ............. (1)