Students can Download Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.2

Question 1.

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

Solution:

ABCD is a parallelogram.

AE ⊥ DC and CF ⊥ AD.

AB = 16 Cm, AE = 8 cm, CF = 10 cm, then AD = ?

(i) Area of ABCD = base × altitude

= DC × AE ( ∵ AB=DC)

= 16 × 8

= 128 sq. cm.

Whether ABCD is quadrilateral, ADCB is a quadrilateral.

∴ Area of ADCB = 128 sq. cm.

Base = AD = ?

Altitude, CF= 100 cm.

Base × Altitude = Area

AD × CF = 128

AD × 10 = 128

∴ AD = \(\frac{128}{10}\)

∴ AD = 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)

Solution:

Given: E, F, G, and H are mid-points of the sides of a parallelogram ABCD,

To Prove: area (EFGH) = \(\frac{1}{2}\) area (ABCD)

Construction: HF is joined.

Proof: Now, AD = BC and AD || BC

EF = \(\frac{1}{2}\) AC

and EF || AC ….. (1)

|||^{ly} in ∆ADC, we have

GH = \(\frac{1}{2}\)AC and

GH || AC …….. (2)

∴ GH = EF and

GH || EF [From (I) and (2)

∴ Quadrilateral, EFGH isa parallelogram.

In parallelogram ABCD

AD = BC and AD || BC [Opposite sides of parallelogram]

∴ \(\frac{1}{2}\) AD = BC

and AD || FC

⇒ HD || FC and

HD || FC

∴ HDFC is a parallelogram Since ∆HGF and parallelogram HDCF are on the same base HF and between the same parallels.

∴ area (∆HGF) = \(\frac{1}{2}\) area (parallelogram HDCF) ………. (3)

|||^{ly}, ar (IHEF) = \(\frac{1}{2}\) ar(parallelogram HABF) …… (4)

Adding (3) and (4), we get

area (∆HGF) + ar (∆HEF)

= \(\frac{1}{2}\) [ar(||^{gm} HDCF) + ar (||^{gm} HABF)]

Hence, ar (||^{gm} EFGH) = \(\frac{1}{2}\) ar(||^{gm} 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 (PAB) = ar (BQC).

Solution:

Data: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

To Prove: ar(∆APB) = ar(∆BQC)

Proof: ABCD is a parallelogram.

∴ AB || DC AB = DC

AD || BC AD = BC

Now ∆APB and ABCD are on same base AB and in between AB || DC

∴ Area(∆APB) = \(\frac{1}{2}\) Area ABCD) ……… (i)

Similarly, ∆BQC andBADC are on the same base BC and in between BC || AD.

∴ Area(∆BQC) = \(\frac{1}{2}\) Area(ABCD) ………. (ii)

From (i) and (ii)

∴ Area(∆APB) = Area (∆BQC).

Question 4.

In Fig., P is a point in the interior of a parallelogram ABCD, Show that

(i) ar(ABP)= 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).

Solution:

Data: P is any point in the interior of a parallelogram ABCD. PA, PB, PC, and PD are joined.

To Prove: (i) ar(∆ABP = ar(∆PCD)= \(\frac{1}{2}\) ar (ABCD)

(ii) ar(∆APD) + ar(∆PBC)= ar(∆APB) + ar(∆PCD)

Construction: AB || XY is drawn through P.

Proof: (i) XY || AB || DC

∴ ABYX and XDCY are parallelograms.

∆APB and ABYX are on base AB and in between AB||XY.

Construction: AD || MN is drawn through P.

Proof: ∆ADP and ADMN lie on the base AD and in between AD || MN

Question 5.

In Fig., 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).

Solution:

Data: PQRS and ABRS are parallelograms and X is any point on side BR.

(i) Parallelogram PQRS and parallelogram ABRS stand on the same base RS and lie between the same parallels SR and PAQB.

ar (PQRS) ar (ABRS)

(ii) AXS and parallelogram ABRS stand on the same base AS and lie between the same parallels AS and RB.

∴ ar (AXS) = \(\frac{1}{2}\) ar (ABRS)

So, ar (AXS) = \(\frac{1}{2}\) ar (PQRS)

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 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:

If AP and AQ are joined PQRS is divided into three triangles, ∆SAP, ∆APQ and ∆ARQ.

∆APQ and PQRS are on base PQ and in between PQ || SR.

Area of ∆PSA + Area of ∆PAQ + Area of ∆QRA

= Area of (PQRS) ……. (i)

W.K.T If a ||^{gram} and a ∆^{le} are on the same base and between the same parallels, then the area of the triangles is half the area of

the parallelogram.

∴ Area (∆PAQ) = \(\frac{1}{2}\) ar (PQRS) ….(ii)

From Equation (1) and (2)

Area of (∆PSA) + Area of (∆QRA) = \(\frac{1}{2}\)

Area (PQRS) …..(iii)

From (i) and (ii).

∴ ar.(∆APQ) = ar.(∆SAP) + ar.(∆ARQ)

∴ The farmer can use the part of the field to sow wheat, i.e. Ar. (∆SAP + ∆ARQ) and in the same area, he can use Area of ∆APQ to sow pulses.