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
CF ⊥ AD
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,
In quad ABFH
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 and∥ gm 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
In △ADQ & △AOD
(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 BD x AD
= ½ product of diagonal BD and AD
∴ Area of rhombus is half the product of its diagonals. Proved
Nice help full
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