# Telangana SCERT Class 9 Math Solution Chapter 11 Areas Exercise 11.2

## Telangana SCERT Solution Class IX (9) Math Chapter 11 Areas Exercise 11.2

(1) The area of parallelogram ABCD is 36 cm2 . Calculate the height of parallelogram ABEF if AB = 4.2 cm Solution: Given,

Area of parallelogram ABCD = 36 cm2

AB = 4.2 cm

ABEF is another parallelogram

∴ Area of parallelogram ABEF = area of parallelogram ABCD = 36 cm2

Since parallelogram with same base on has equal area

AB is the base both parallelogram

But, h be the height of parallelogram ABEF

∴ Area of parallelogram = b x h

Or, 36 = AB x h

Or, 36 = 4.2 x H

Or, h = 36/4.2

= 8.57 cm

(2) ABCD is a parallelogram. AE is perpendicular on DC and CF is perpendicular on AD. If AB = 10 cm, AE = 8 cm and CF = 12 cm. Find AD. Solution: Given, ABCD is a parallelogram

AE ⊥ DC

AB = 10 cm, AE = 8 cm, CF = 12 cm,

∴ DC = AB = 10 cm [∵ opposite sides of parallelogram are equal]

∴ Area of parallelogram ABCD = b x h

= DC x AE

= 10 x 8

= 80 cm2

∴ If we consider AD as base, CF is the height of parallelogram ABCD

∴ Area of parallelogram ABCD with base AD = b x h

Or, 80 = AD x CF

Or, 80 = AD x 12

Or, Ad = 80/12 = 20/3 = 6.67 cm.

(3) If E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a parallelogram ABCD, show that ar(EFGH) 1/2 ar(ABCD). Given,

ABCD is a ∥ gm

Join AC & HF

B,F,G,H are mid points of sides

AB, BC, CD, DA respectively

∴ In △ABC

By mid point theorem

EF ∥ AC & EF = ½ AC — (i)

Similarly In △ACD,

By mid point theorem

HG ∥ AC & HG = ½ AC — (ii)

Comparing equation (i) & (ii) we get

EF ∥ HG & EP = HG = ½ AC

∴ EFGH is a gm [∵opposite sides are equal and parallel to each other]

Now,

AH = ½ AB; BF = ½ BC

AB = BC [opposite sides of parallelogram ABCD]

∴ AH = BF

AB ∥ BC [opposite of ABCD]

AH ∥ BF [∵ H & F are midpoint of AB & BC parts of parallel lines are parallel to each other]

∴ ABFH is a parallelogram [∵ opposite sided are equal parallel to each other]

Area of △HEF = ½ area of gm ABHF —- (iii)

Since △HEF & gm ABFH has same base

Now, HF is the diagonal of gm EFGH

We know diagonals of gm divides the gm in to two congruent △s

In this case △HEF & △HGF we congruent

∴They have to same area

∴ Area of gm EPGH = area of △HEF + area of △HGF

= 2 x area of △HEF

= 2 x ½ area of gm ABHF – from equation (iii)

= area of gm ABHF – (iv)

Now, gm ABHF is = ½ the area of gm ABCD

Since, the gm ABHF is the gm joined by mid point

HF which divides the gm ABCD in two equal halves.

∴ Area of ∥ gm EFGH = ½ area of ∥ gm ABCD from equation (iv) proved.

(5) P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD show that ar(ΔAPB) = ar Δ(BQC). Solution:

Given, ABCD is a parallelogram

P, Q are two point that lie on sides DC & AD responsibility

∴ Area of △APB = ½ area of gm ABCD

Since of △APD & gm ABCD has the same base AB and △APB lies within the gm ABCD

Also, Area of △BQC = ½ are of gm ABCD

Since △BQC & base same base BC and △BQC lies within the ABCD

Area of △APB = area of △BQC (proved)

(6) P is a point in the interior of a parallelogram ABCD. Show that

(i) ar(ΔAPB) + ar(ΔPCD) 1/2 ar(ABCD)

(ii) ar(ΔAPD) + ar(ΔPBC) = ar(ΔAPB) + ar(ΔPCD)

(Hint : Through P, draw a line parallel to AB) Solution:

Given,

ABCD is a gm

P is an interior point in gm ABCD

Draw a line through P meeting

AB & BC at point Q & R

Respectively parallel to AB

∴ QR ∥ AB ∥ DC [∵ opposite side of gm ABCD are parallel]

QR – AB = DC [∵ parallel lines draw within the are also equal]

∴ △QRB is a ∥ gm [opposite sides are ∥ gm and equal]

DCRQ is a ∥ gm [opposite sides are ∥ gm and equal]

∴ Area of △APB = ½ area of gm AQPB

[∵△APB & ∥ gm as same base AB & △APB lies within ∥ gm △AQRB] – (i)

Again,

Area of △DPC = ½ area of gm DCRQ [∵△DPC to andgm DCRQ has same base & △DPC lies within gm DCRQ] — (ii)

Adding equation (i) & (ii) we get

(i) Area of △APB + area of △DPC = ½ (area of ∥ gm DCQR + area of ∥ gm AQRB)

= ½ (area of gm ABCD) – (ii) (proved)

(ii) ar △APD + ar △PBC + ar △APB + ar △DPC = ar ∥ gm ABCD

Or, ar △APD + ar △PBC + ½ ar ∥ gm ABCD = ar ∥ gm ABCD [From equation (iii)]

Or, ar △APD + ar △PBC = ar ∥ gm ABCD – ½ ar ∥ gm ABCD

Or, ar △APD + ar △PBC = ½ ar ∥ gm ABCD

Or, ar △APD + ar △PBC = ar △APB + ar △DPC [From equation (iii)] Proved

(7) Prove that the area of a trapezium is half the sum of the parallel sides multiplied by the distance between them Solution:

ABCD is a trapezium; AB ∥ DC

AP & BQ are heights of trap. ABCD

∴ AP ⊥ DC, BQ ⊥ DC

ABPQ is clearly a rectangle

Since adjacent angles are 90o and opposite side AP & BQ are parallel to each other and equal in length since AP and BQ are two perpendiculars between two parallel lines.

Area of rectangle = l x b

= PQ x AP

Now,

Area of △APD = ½ x b x l

= ½ x DP x AP

Area of △BQC = ½ x b x h

= ½ x QC x BQ

= ½ x QC x AP [∵AP = BQ opposite sided of rectangle]

∴ Area of trapezium = ar △APD + ar △BQC + ar rectangle ABPQ

= ½ x DP x AP + ½ x QC x AP + PQ x AP

= ½ AP (DP + QC + 2PQ)

= ½ AP (DP + QC + PQ + PQ)

= ½ AP (DC + AB) [∵DP + QC + PQ = DC

PQ = AB, opposite sided of rectangle ABQP]

[AP is the distance between AB & DC]

∴ Ar of hap ABCD = half sum of parallel sides multiplied by the distance between than]

(8.) PQRS and ABRS are parallelograms and X is any point on the side BR. Show that

(i) ar(PQRS) = ar(ABRS)

(ii) ar(ΔAXS) = 1/2 ar(PQRS) Solution:

Given,

PQRS & ABRS are parallelograms

X is any point on side BR

(i) ar (PQRS) = ar (ABRS)

[∵parallelograms under same base & between same parallel lines are equal in area, have SR is the base and PQ ∥ SR since QB is produced from same line PQ]

(ii) ar △AXS = ½ ar gm ABS [∵ △AXS has same base as gm ABRS and lies within the gm ABRS]

ar △AXS = ½ ar gm (PQRS) proved

[∵ar gm (PQRS) = ar gm (ABRS)]

(9) A farmer has a field in the form of a parallelogram PQRS as shown in the figure. He took the mid- point A on RS and joined it to points P and Q. In how many parts of field is divided? What are the shapes of these parts ?

The farmer wants to sow groundnuts which are equal to the sum of pulses and paddy. How should he sow? State reasons?

Solution: Given,

PQRS is a ∥ gm

A is the mid point of RS

∴The field is divided into 3 parts

The three parts are triangles.

Let Consider that the farmer lows ground nets in △PAQ and pulses and paddy in other two triangle – statement 1

∴ ar △PAQ = ½ ar ∥ gm PQRS – (i)

[∵△PAQ has same base as ∥ gm PQRS and △PAQ lies within ∥ gm PQRS]

Or, (△PSA + △AQR + △PAQ) = ar ∥ gm PQRS

Or, ar △PSA + ar △AQR + ½ ar ∥ gm PQRS = ar ∥ gm PQRS [from equation (i)]

Or, ar △PSA + ar △AQR = ar ∥ gm PQRS – ½ ar ∥ gm PQRS

Or, ar △PSA + ar △ABR = ½ ar ∥ gm PQRS

Or, ar △PSA + ar △AQR = ar △PAQ [From equation (i)]

∴ From the above equation we can state that the farmer should saw pulses and paddy in wither of each △PSA and △AQR and ground nuts in △PAQ

Because sum of area of △PSA and △ABQ is aspect to area of △PAQ which statistics statement I

(10) Prove that the area of a rhombus is equal to half of the product of the diagonals. Solution:

ABCD is a rhombus

Diagonals AC & BD intersects at point O

(i)  ∠AOB = ∠AOD = 90o

[∵diagonals of rhombus are perpendicular to each other]

(ii) AB is the common side

(iii) BO = OD [∵diagonals of rhombus bisect each other]

∴ △AOB ≅ △AOD by SAS

Similarly BOC ≅ DOC by SAS

In △AOD & △DOC

(i) ∠AOD = ∠DOC = 90o [∵diagonals of rhombus are perpendicular to each other]

(ii) DO is common side

(iii) AO = DC [∵diagonals of rhombus bisects each other]

△AOD ≅ △DOC by SAS

Normally △AOB ≅ △AOC by SAS

∴ △AOB ≅ △DOC ≅ △BOC ≅ △AOD

Now,

ar △AOB = ½ x b x h = ½ x BO x AO

Now, we know congruent triangles have equal area

ar △AOB = ar △DOC = ar △BOC = ar △AOD –(i)

ar of rhombus = ar △AOB + ar △DOC + ar △BOC + ar △AOD

= 4 x ar △AOB from equation (i)

= 4 x ½ x BO x AO

= 2 x ½ BD x ½ AC

Or, ar of rhombus ABCD = 2 x ½ x BD x ½ AD

[∵ diagonals of rhombus  bisects each other in this base at point O]

= ½ product of diagonal BD and AD

∴ Area of rhombus is half the product of its diagonals. Proved

Updated: October 2, 2021 — 4:31 pm

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