ML Aggarwal Solution Class 9 Chapter 16 Mensuration EXERCISE 16.1

 EXERCISE 16.1

Question 1

Find the area of a triangle whose base is 6 cm and corresponding height is 4 cm.

Sol :
Base of triangle=6 cm
Height=4 cm
Area of triangle =12×Base ×Height
=12×6×4 cm2
=6×2 cm2
=12 cm2

Question 2

Find the area of a triangle whose sides are
(i) 3 cm, 4 cm and 5 cm
(ii) 29 cm, 20 cm and 21 cm
(iii) 12 cm, 9.6 cm and 7.2 cm
Sol :
(i) Here a=3 cm, b=4 cm and c=5 cm
s=semi perimeter=a+b+c2
=3+4+52 cm=122 cm=6 cm

Area of triangle =s(sa)(sb)(sc)
=6(63)(64)(65)cm2
=6×3×2×1 cm2
=6×6 cm2=6 cm2

(ii) a=29 cm, b=20 cm and c=21 cm
s=semi perimeter =a+b+c2=29+20+212 cm
=702 cm=35 cm

Area of triangle =s(sa)(sb)(sc)
=35(3529)(3520)(3521)cm2
=35×6×15×14 cm2
=7×5×3×2×5×3×7×2 cm2
=7×7×5×5×3×3×2×2 cm2
=7×5×3×2=210 cm2

(iii) a=12 cm , b=9.6 cm and c=7.2 cm
s=semi perimeter=a+b+c2=12+9.6+7.22 cm
=28.82 cm=14.4 cm

Area of triangle =s(sa)(sb)(sc)
=14.4×(14.412)×(14.49.6)×(14.47.2)cm2
=14.4×2.4×4.8×7.2 cm2
=6×2.4×2.4×2×2.4×3×2.4 cm2
=2.4×2.46×6 cm2=2.4×2.4×6 cm2
=34.56 cm2

Question 3

Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. Hence, find the length of the altitude corresponding to the shortest side.

Sol :
Given sides of triangle are 34 cm, 20 cm and 42 cm
i.e. a=34 cm, b=20 cm, and c=42 cm

s=a+b+c2=34+20+422 cm=962 cm
=48 cm

Area of triangle =s( sa)(sb)(sc)
=48(4834)(4820)(4842)cm2
=48×14×28×6 cm2=8×6×14×14×2×6 cm2
=14×68×2 cm2=14×6×16 cm2
=14×6×4 cm2=336 cm2

The shortest side of the triangle is 20 cm
Let h cm be the corresponding altitude , then area of triangle =12× base × height
336=12×20×h
h=336×220 cm
h=33610 cm 
⇒h=33.6 cm

∴The required altitude of the triangle=33.6 cm

Question 4

The sides of a triangular field are 975m, 1050 m and 1125 m. If this field is sold at the rate of Rs. 1000 per hectare, find its selling price. [1 hectare = 10000 m²].
Sol :
Given : Sides of a triangular field are 975 m, 1050 m, and 1125 m
i.e. a=975 m, b=1050 m ,c=1125 m

s=a+b+c2=975+1050+11252 cm =31502 cm
=1575 cm

Area of triangular field =s(sa)(sb)(sc)
=1575(1575975)(15751050)(15751125)m2
=1575×600×525×450 m2
=525×3×150×4×525×150×3 m2
=525×3×1504 m2=525×450×2 m2

=525×90010000 hectare  (∵1 hectare=10000 m2)
=525×90100 hectare =4725100 hectare =47.25 hectare

Selling price of 1 hectare field=Rs 1000
Selling price of 47.25 hectare field= 1000×47.25
=47250

Question 5

The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its area and the perimeter.











Sol :
Here ABC is a right angled triangle BC=12 cm, AB=13 cm

By Pythagoras theorem
⇒AB2=AC2+BC2
⇒(13)2=AC2+(12)2
⇒AC2=(13)2-(12)2
⇒AC2=169-144=25
⇒AC=√25=5 cm

Area of ΔABC=12× base × height
=12×12×5 cm2=30 cm2

Perimeter of ΔABC=AB+BC+CA
=13+12+5=30 cm

Question 6

Find the area of an equilateral triangle whose side is 8 m. Given your answer correct to two decimal places.
Sol :
Side of equilateral triangle=8 cm
Area of equilateral triangle =34( side )2
=34×8×8 m2=3×2×8 m2=1.73×16 m2
=27.71 m2

Question 7

If the area of an equilateral triangle is 81√3 cm² find its. perimeter.
Sol :
Area of equilateral triangle =34( side )2
813=34 (side) 2
( side )2=813×43
⇒(side)2=81×4 cm2
⇒side=√81×4 cm
⇒side=9×2 cm
⇒side=18 cm

Perimeter of equilateral triangle =3×side 
=3×18 cm=54 cm

Question 8

If the perimeter of an equilateral triangle is 36 cm, calculate its area and height.









Sol :
Perimeter of an equilateral triangle=3×side 
⇒36=3×side
⇒side=363 cm
⇒side=12 cm
i.e. AB=BC=CA=12 cm
 
Area of equilateral triangle =34( side )2
=34×(12)2 cm2
=34×12×12 cm2
=3×3×12 cm2
=1.73×36 cm2
=62.4 cm2

In triangle ABD   
⇒AB2=AD2+BD2  (by Pythagoras theorem)
⇒(12)2=AD2+(6)[BD=122=6 cm]
⇒144=AD2+36
⇒AD2=108
⇒AD=√108=10.4

Hence , required height=10.4 cm

Question 9

(i) If the length of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 48 cm, find its area.
(ii) The sides of a triangular plot are in the ratio 3: 5:7 and its perimeter is 300 m. Find its area.
Sol :
Let ABC be the triangle Ratio of the side=3 : 4 : 5
Let the sides be 3x, 4x and 5x
Let a=3x cm, b=4x cm, c=5 cm










∴a+b+c=48
⇒3x+4x+5x=48
⇒12x=48
x=4812
⇒x=4
∴a=3×4 cm=12 cm [∵a=3x cm]
b=4×4 cm=16 cm  [∵b=4x cm]
c=5×4 cm=20 cm [∵c=5x cm]

s=a+b+c2=12+16+202 cm=482 cm=24 cm
∴By Hero's formula

Area of ΔABC=s(sa)(sb)(sc)
=24(2412)(2416)(2420) cm2
=24×12×8×4 cm2 cm2
=12×2×12×4×2×4 cm2
=12×4×2 cm2
=96 cm2


(ii) Side of a triangle are in the ratio=3 : 5 : 7 and perimeter=300 m
∴First side=300×3sum of ratios=300×33+5+7 
=300×315=60 m

Second side =300×515=100 m
and third side =300×715=140 m

Now s= Perimeter 2=3002=150

∴Area=s(sa)(sb)(sc)
=150(15060)(150100)(150140)
=150×90×50×10
=6750000=225×10000×3
=15×1003=15003 cm3
=1500×1.732 m2=2598.0000=2598 m2

Question 10

ABC is a triangle in which AB = AC = 4 cm and ∠ A = 90°. Calculate the area of ∆ABC. Also find the length of perpendicular from A to BC.

Sol :
Given that AB=AC=4 cm
By Pythagoras theorem 
BC2=AB2+AC2
BC2=(4)2+(4)2
BC2=16+16
⇒BC=√32
⇒BC=4√2 cm











Area of ΔABC=12× base × height
=12×AC×AB
=12×4×4=2×4
=8 cm2

Let length of perpendicular from A to BC= h cm
Area of ΔABC=12×BC×h
8=12×42×h
h=8×242
=2×22×22
=4i2=2×2
⇒2×1.41=2.82 cm

Question 11

Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.

Sol :
In ΔABC,
AB=AC=12 cm










But perimeter=30 cm
∴BC=30-(12+12)=30-24=6 cm

Now S=a+b+c2=302=15
∴Area of ΔABC=s(sa)(sb)(sc)
(Using hero's formula)

=15(1512)(1512)(156)
=15×3×3×9=81×15=915 cm2
=9×3.873 cm2=34.857 cm2
=34.86 cm2

Question 12

Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.
Sol :
Base=6 cm, Perimeter=16 cm
Here ABC is isosceles triangle in which 
AB=AC=x (say)
and BC=6 cm










Perimeter of ΔABC=AB+BC+AC
⇒16=x+6+x
⇒16=2x+6
⇒16-6=2x
⇒10=2x
x=102
⇒x=5

i.e. AB=AC=5 cm and BD=12×6=3 cm

In right angled ΔABD
⇒AB2=AD2+BD2
⇒52=AD2+32
⇒25=AD2+9
⇒AD2=25-9
⇒AD2=16
⇒AD=4

Area of ΔABC=12× base × height
=12×6×4 cm2
=3×4 cm2
=12 cm2

Question 13

The sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm².
Sol :
Here ABC be a right angled triangle
AB=5x cm and BC=(3x-1) cm



Area of ΔABC=12×AB×BC
60=12×5x(3x1)
⇒120=5x(3x-1)
⇒120=15x2-5x
⇒0=15x2-5x-120
⇒15x2-5x-120=0
⇒5(3x2-x-24)=0
⇒3x2-x-24=0
⇒3x2-9x+8x-24=0
⇒3x(x-3)+8(x-3)=0
⇒(3x+8)(x-3)=0

Either 3x+8=0 o x-3=0
⇒3x=-8 or x=3
x=83 or x=3

∴x=3  (∵x=83 , is not possible)

AB=5×3 cm=15 cm
BC=(3×3-1) cm
=(9-1) cm=8 cm

In right angled ΔABC,

By Pythagoras theorem
⇒AC2=AB2+BC2
⇒AC2=(15)2+(8)2
⇒AC2=225+64=289
⇒AC2=(17)2
⇒AC=17 cm
Hence , hypotenuse of right angled triangle=17 cm

Question 14

In ∆ ABC, ∠B = 90°, AB = (2A + 1) cm and BC = (A + 1). cm. If the area of the ∆ ABC is 60 cm², find its perimeter.
Sol :
AB=(2x+1) cm , BC=(x+1) cm
Area of ΔABC=12×AB×BC
60×12×(2x+1)×(x+1)
⇒60×2=(2x-1)(x+1)
⇒120=(2x+1)(x+1)
⇒120=2x2+2x+x+1
⇒120=2x2+3x+1
⇒0=2x2+3x+1-120
⇒0=2x2+3x-119
⇒2x2+3x-119=0
⇒2x2+17x-14x-119=0
⇒x(2x+17)-7(2x+17)=0
⇒(x-7)(2x+17)=0










Either x-7=0
or 2x+17=0
or x=7
or x=172 (not possible)
∴AB=(2x+1) cm=(2×7+1)cm
∴(14+1) cm=15 cm

BC=(x+1) cm=7+1=8 cm

In right angled ΔABC
By Pythagoras theorem
⇒AC2=AB2+BC2
⇒AC2=(15)2+(8)2
⇒AC2=225+64
⇒AC2=289
⇒AC=√289=17

Perimeter=AB+BC+AC
=(15+8+17) cm
=40 cm

Question 15

If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find its area.
Sol :
Perimeter of right angled triangle=60 cm
and hypotenuse=25 cm
∴Sum of two sides=60-25=35 cm
Let base=x cm
∴Then altitude=(35-x) cm

But x2+(35-x)2=252 (Pythagoras theorem)
⇒x2+1225+x2-70x=625
⇒2x2-70x+1225-625=0
⇒2x2-70x+600=0
⇒x2-35x+300=0
⇒x2-15x-20x+300=0
⇒x(x-15)-20(x-15)=0
⇒(x-15)(x-20)=0
Either x-15=0, then x=15
or x-20=0, then x=20

∴Sides are 15 cm and 20 cm
and area =12×(base)×(altitude)
=12×15×20
=150 cm2


Question P.Q.

In ∆ ABC, ∠B = 90° and D is mid-point of AC. If AB = 20 cm and BD = 14.5 cm, find the area and the perimeter of ∆ ABC.

Sol :
Here, In ΔABC,
∠B=90° and D is mid point of AC
BD=AD=DC  (By geometry)
∵AC=2BD⇒AC=2×14.5
⇒AC=29..0 cm










By Pythagoras theorem
⇒AC2=AB2+BC2
⇒(29)2=(20)2+BC2
⇒BC2=841-400=441
⇒BC2=√441=21 cm

Area of ΔABC=12×base×height
=12×21×20 cm2
=21×10 cm2
=210 cm2

Perimeter of ΔABC=AB+BC+AC
=(20+21+29) cm=70 cm

Question 16

The perimeter of an isosceles triangle is 40 cm. The base is two third of the sum of equal sides. Find the length of each side.
Sol :
Perimeter of isosceles ΔABC=40 cm
Let each equal side=x
∴base =23(2x)=43x

According to the sum
2x+4x3=40
⇒6x+4x=120
⇒10x=120
x=12010=12
Hence length of each equal side=12cm


Question 17

If the area of an isosceles triangle is 60 cm2 and the length of each of its equal sides is 13 cm, find its base.
Sol :
Area of isosceles triangle=60 cm2
Length of each equal side=13 cm
Let base BC=x cm










Draw AD⟂BC which bisects BC at D
BD=DC=x2 cm

Now in right ΔABD
⇒AB2=BD2+AD2
(13)2=(x2)2+AD2
(Pythagoras theorem)
169=x24+AD2
AD2=169x24...(i)

Area =60cm2
∴AD= Area ×2 base =60×2x=120x...(ii)

Find (i) and (ii)
169x24=(120x)2
676x24=14400x2
⇒676x2-x4=57600
⇒x4-676x2+57600=0
⇒x4-576x2-100x2+57600=0
⇒x2(x2-576)-100(x2-576)=0
⇒(x2-576)(x2-100)=0

Either x2-576=0, then x2=576
⇒x2=(24)2
⇒x=24
or x2-100=0, then x2=100=(10)2
∴x=10
∴Base=10 cm or 24 cm

Question 18

The base of a triangular field is 3 times its height If the cost of cultivating the field at the rate of ₹25 per 100m2 is ₹60000, find its base and height.
Sol :
Cost of cultivating the field at the rate of ₹25 Per 100 m2 is 60000
i.e. Cost of cultivating the field of ₹25 for 100 m2

Cost of cultivating the field of ₹1 for =10025 m2
Cost of cultivating the field of ₹60000 for
10025×60000 m2=4×60000 m2=240000 m2

i.e. Area of field=240000 m2

12× base × height =240000 m2...(1)

Let height of triangular field=h m2
Then, base of triangular field=3h m2

Putting this value in equation (1), we get
12×3h×h=240000
12×3h2=240000
h2=240000×23
h=160000=400
Hence , height of triangular field=400 m and base of triangular field=3×400 m
=1200 m2

Question 19

A triangular park ABC? has sides 120 m, 80 m and 50 m (as shown in the given figure). A gardner Dhania has to put a fence around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹20 per metre leaving a space 3 m wide for a gate on one side.
Sol :









ΔABC is a triangular park whose sides are 120m, 80m and 50m

∴Perimeter of ΔABC=120+80+50=250 m
Portion at which a gate is build=3 m

∴Remaining perimeter=250-3=247 m
Now, length of fence around it=247 m

∴Rate of fencing=20 per m
∴Total cost=20×247=4940 and area of the park

s=a+b+c2=2502
=125

∴Area=s(sa)(sb)(sc)
=125(12550)(12580)(125120)
=125×75×45×5
=5×5×5×5×5×3×3×3×5×5
=5×5×3×515
=37515 m2


Question 20

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (shown in the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?











Sol :
An umberalla is made by stitching 10 triangular piece of cloth of two different colours.
The measurement of each triangular is 20 cm, 50 cm , 50 cm

s2=20+50+502=1202
=60

∴Area of one triangle
=s(sa)(sb)(sc)
=60(6020)(6050)(6050)
=60×40×10×10=240000
=100√24 cm2
=100×2√6 cm2
=200√6 cm2

Now , area of 5 triangular piece of first colour
=5×200√6 m2
=1000√6 cm2

and area of triangular of second colour
=1000√6 cm2

Question 21

(a) In the figure (1) given below, ABC is an equilateral triangle with each side of length 10 cm. In ∆ BCD, ∠D = 90° and CD = 6 cm.
Find the area of the shaded region. Give your answer correct to one decimal place.












(b) In the figure given, ABC is an isosceles right angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.










Sol :
(a) ABC is an equilateral triangle side of equilateral Δ10 cm
Area of equilateral ΔABC
=34×( side )2
=34×(10)2 cm2
=34×100 cm2
=3×25 cm2
=25√3 cm2
=25×1.73 cm2
=43.300 cm2

In right angled ΔBDC
∠D=90° , BC=10 cm, CD=6 cm

By Pythagoras theorem
⇒BD2+DC2=BC2
⇒BD2+(6)2=(10)2
⇒BD2+36=100
⇒BD2=100-36=64
⇒BD=√64=8


Area of ΔBDC=12× base × height 
=12×BD×DC
=12×8×6 cm2
=4×6 cm2=24 cm2

∴Area of shaded portion
=Area of ΔABC-Area of ΔBDC
=43.300 cm2-24 cm2
=19.300 cm2
or 19.3 cm2

(b) AD=AE=3 cm
DB=EC=4 cm

Adding AD+DB=AE+EC=(3+4) cm
⇒AB=AC=7 cm
∵∠A=90°

∴Area of right ΔADE=12AD×AE
=12×3×3=92 cm2

∵ΔBDG is an isosceles right triangle
∴DG2+BG2=BD2
⇒DG2+DG2=(4)2
⇒2DG2=16
DG2=162=8
∴DG=√8 cm

Now area of ΔBDG=12BG×DG
=12DG×DG
=12(8)2
=12×8=4 cm2

Similarly area of isosceles right ΔEFC=4 cm2

Now area of shaded portion =92+4+4
=9+8+82=252 cm2
=12.5 cm2

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