EXERCISE 9.2
Q 1, Ex. 9.2, Page No 159 - Areas of Parallelograms & Triangles, Class 9th Maths
Page No 159:
Question 1:
In the given figure, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Answer:
In parallelogram ABCD, CD = AB = 16 cm[Opposite sides of a parallelogram are equal]
We know that
Area of a parallelogram = Base × Corresponding altitude
Area of parallelogram ABCD = CD × AE = AD × CF
16 cm × 8 cm = AD × 10 cm
$\mathrm{AD}=\frac{16 \times 8}{10} \mathrm{cm}$
=12.8cm
Thus, the length of AD is 12.8 cm.
Q 2, Ex. 9.2, Page No 159 - Areas of Parallelogram & Triangles - NCERT Class 9th Maths Solutions
Question 2:
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show thatar (EFGH) $=\frac{1}{2}$ar (ABCD)
Answer:
Let us join HF.
In parallelogram ABCD,
AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)
AB = CD (Opposite sides of a parallelogram are equal)
$\Rightarrow \frac{1}{2} \mathrm{AD}=\frac{1}{2} \mathrm{BC}$ and AH || BF
⇒ AH = BF and AH || BF (
H and F are the mid-points of AD and BC)Therefore, ABFH is a parallelogram.
Since ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
∴ Area (ΔHEF) = $\frac{1}{2}$Area (ABFH) … (1)
Similarly, it can be proved that
Area (ΔHGF) = $\frac{1}{2}$Area (HDCF) … (2)
On adding equations (1) and (2), we obtain
Area $(\Delta \mathrm{HEF})+$ Area $(\Delta \mathrm{HGF})=\frac{1}{2}$ Area $(\mathrm{ABFH})+\frac{1}{2}$ Area $(\mathrm{HDCF})$
$=\frac{1}{2}[\text { Area }(\mathrm{ABFH})+\text { Area }(\mathrm{HDCF})]$
$\Rightarrow$ Area $(\mathrm{EFGH})=\frac{1}{2}$ Area $(\mathrm{ABCD})$
Q 3, Ex. 9.2, Page No 159, Areas of Parallelograms & Triangles - NCERT Maths Solutions Class 9th
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:
It can be observed that ΔBQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.
∴Area (ΔBQC) = $\frac{1}{2}$Area (ABCD) … (1)
Similarly, ΔAPB and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.
∴ Area (ΔAPB) = $\frac{1}{2}$Area (ABCD) … (2)
From equation (1) and (2), we obtain
Area (ΔBQC) = Area (ΔAPB)
Q 4, Ex. 9.2, Page No 159 - Areas of Parallelogram & Triangles, NCERT Class 9th Maths Solutions
Question 4:
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that(i) ar (APB) + ar (PCD) = $\frac{1}{2}$ ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
[Hint: Through. P, draw a line parallel to AB]
Answer:
(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.
In parallelogram ABCD,
AB || EF (By construction) … (1)
ABCD is a parallelogram.
∴ AD || BC (Opposite sides of a parallelogram)
⇒ AE || BF … (2)
From equations (1) and (2), we obtain
AB || EF and AE || BF
Therefore, quadrilateral ABFE is a parallelogram.
It can be observed that ΔAPB and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF.
∴ Area (ΔAPB) = $\frac{1}{2}$Area (ABFE) … (3)
Similarly, for ΔPCD and parallelogram EFCD,
Area (ΔPCD) = $\frac{1}{2}$Area (EFCD) … (4)
Adding equations (3) and (4), we obtain
Area $(\Delta \mathrm{APB})+$ Area $(\Delta \mathrm{PCD})=\frac{1}{2}[\text { Area }(\mathrm{ABFE})+\text { Area }(\mathrm{EFCD})]$
Area $(\Delta \mathrm{APB})+$ Area $(\Delta \mathrm{PCD})=\frac{1}{2}$ Area $(\mathrm{ABCD})$..(5)
(ii)
Let us draw a line segment MN, passing through point P and parallel to line segment AD.
In parallelogram ABCD,
MN || AD (By construction) … (6)
ABCD is a parallelogram.
∴ AB || DC (Opposite sides of a parallelogram)
⇒ AM || DN … (7)
From equations (6) and (7), we obtain
MN || AD and AM || DN
Therefore, quadrilateral AMND is a parallelogram.
It can be observed that ΔAPD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.
∴ Area (ΔAPD) = $\frac{1}{2}$Area (AMND) … (8)
Similarly, for ΔPCB and parallelogram MNCB,
Area (ΔPCB) = $\frac{1}{2}$Area (MNCB) … (9)
Adding equations (8) and (9), we obtain
Area $(\Delta \mathrm{APD})+$ Area $(\Delta \mathrm{PCB})=\frac{1}{2}[\text { Area }(\mathrm{AMND})+\text { Area }(\mathrm{MNCB})]$
Area $(\Delta \mathrm{APD})+$ Area $(\Delta \mathrm{PCB})=\frac{1}{2}$ Area $(\mathrm{ABCD})$..(10)
On comparing equations (5) and (10), we obtain
Area (ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area (ΔPCD)
Q 5, Ex. 9.2, Page No 159 - Areas of Parallelograms & Triangles, Class 9th Maths
Question 5:
In the given 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) It can be observed that parallelogram PQRS and ABRS lie on the same base SRand also, these lie in between the same parallel lines SR and PB.
∴ Area (PQRS) = Area (ABRS) … (1)
(ii) Consider ΔAXS and parallelogram ABRS.
As these lie on the same base and are between the same parallel lines AS and BR,
∴ Area (ΔAXS) = $\frac{1}{2}$Area (ABRS) … (2)
From equations (1) and (2), we obtain
Area (ΔAXS) = $\frac{1}{2}$Area (PQRS)
Page No 160:
Question 6:
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field 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?Answer:
From the figure, it can be observed that point A divides the field into three parts. These parts are triangular in shape − ΔPSA, ΔPAQ, and ΔQRA
Area of ΔPSA + Area of ΔPAQ + Area of ΔQRA = Area of parallelogram PQRS … (1)
We know that if a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
∴ Area (ΔPAQ) = $\frac{1}{2}$Area (PQRS) … (2)
From equations (1) and (2), we obtain
Area (ΔPSA) + Area (ΔQRA) = $\frac{1}{2}$Area (PQRS) … (3)
Clearly, it can be observed that the farmer must sow wheat in triangular part PAQ and pulses in other two triangular parts PSA and QRA or wheat in triangular parts PSA and QRA and pulses in triangular parts PAQ.
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