RS Aggarwal Solutions Class 9 Ex 7A

Q.1: The height of an equilateral triangle measures 9 cm. Find its area correct to 2 decimal places. (Take root 3 is 3=1.732)

Sol:

Let, a be the side of an equilateral triangle.

Therefore, the height of an equilateral triangle = 34×a units

Height of an equilateral triangle = 9cm (given)

32a=9
a=9×23

a=9×2×33×3 [rationalizing the denominator ]

9×2×33
a=63
base=63

Area of the equilateral triangle = 12×base×height

= 12×63×9 = 273

Area of the equilateral triangle = 27×1.732=46.764 = 46.76 cm2

 

Q.2: An umbrella is made by stitching 12 triangular pieces of cloth, each measuring 50cm×20cm×50cm. Find the area of the cloth used in it.

 

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Sol:

Let a = 50 cm, b = 20 cm and c = 50 cm

Now,

S = 12(a+b+c)

S = 12(50+20+50)=1202=60cm

Therefore, area of triangular piece of cloth =S(Sa)(Sb)(Sc)

= 60(6050)(6020)(6050)

= 60×10×40×10

= 6×10×10×4×10×10

= 10×10×10×10×2×2×2×3

= 10×10×26

= 2006=200×2.45=490cm2

Therefore, the area of one piece of cloth = 490 cm2

Now, the area of 12 pieces = 12×490cm2=5880cm2

 

Q.3: A floral design on a floor is made up of 16 tiles, each triangular in shape having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles at Re 1 per cm2

 

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Sol:

Let, a = 16 cm, b = 12 cm and c = 20 cm

Now,

S = 12(a+b+c)

S = 12(16+12+20)=482=24cm

Area of one triangular tile =S(Sa)(Sb)(Sc)

= 24(2416)(2412)(2420)

= 2×2×2×2××2×2×2×2×2×2×3×3

= 2×2×2×2×2×3

Area of one tile = 96 cm2

area of all tiles = 96×16=1536cm2

Cost of polishing the tiles per sq. cm = Re. 1

Thus, the total cost of polishing all the tiles = Rs. (1×1536)=Rs.1536

 

Q.4: Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12cm, ACB=90 and AC = 15 cm.

 

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Sol:

Consider the right triangle ABC.

Now, by Pythagoras theorem, we have

BC = AB2AC2 = 172152 = 289225

Therefore, BC = 64 = 8 cm

Now, the Perimeter of quad. ABCD = 17 + 9 + 12 + 8 = 46 cm

Area of triangle ABC=12×base×height

ABC=12×8×15

= 60 cm2

For area of triangle, ACD:

Let, a = 15 cm, b = 12 cm and c = 9 cm

Now,

S = 12(a+b+c)

S = 12(15+12+9)=362=18cm

Area of one triangular tile  =S(Sa)(Sb)(Sc)

=18(1815)(1812)(189)
=18×3×6×9
=18×18×3×3

= 18×3=54cm2

Thus the area of quadrilateral ABCD= area of ABC + area of ACD

= 60 + 54 = 114 cm2

 

Q.5: Find the perimeter and area of the quadrilateral ABCD in which AB = 42cm, BC = 21cm, CD = 29cm, DA=34 cm and CBD=90.              

Sol:

Perimeter of quad. ABCD = 34 + 29 + 21 + 42 = 126 cm

Area of triangle BCD = 12×base×height

Area of triangle BCD = 12×20×21=210cm2

For area of triangle ABC,

Let, a = 42 cm, b = 20 cm and c = 34 cm

Now,

S = 12(a+b+c)

S = 12(42+20+34)=962=48cm

Area of triangle =S(Sa)(Sb)(Sc)

= 48(4842)(4820)(4834)

= 48×6×28×14

= 16×3×3×2×2×14×14

=4×3×2×14=336cm2

Area of quadrilateral ABCD = area of ABD + area of BCD

Thus, the area of quadrilateral ABCD = 336 + 210 = 546 cm2

 

Q.6: Find the area of the quadrilateral ABCD in which AB = 24 cm, BAC=90 and BCD is an equilateral triangle having each side equals to 26cm. Also, find the perimeter of the quadrilateral. (Given 3=1.73 )

 

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Sol:

Consider the right triangle ABD,

By pythagoras theorem, we have

AB=BD2AD2 = 262242 = 676576

Therefore, AB = 100 =10 cm

base=10cm

Area of the triangle ABD = 12×base×height

area of ABD=12×10×24

AreaofABD=120cm2

Area of equilateral triangle BCD = 34a2

1.734(26)2 = 292.37 cm2

Area of quad. ABCD = area of ABD + area of BCD

= 120 + 292.37 = 412.37 cm2

Q.7: Find the area of parallelogram ABCD in which AB = 28 cm, BC = 26cm and diagonal AC = 30 cm.

 

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Sol:

Consider the triangle ABC,

Let, a = 26 cm, b = 30 cm and c = 28 cm

Now,

S = 12(a+b+c)

S = 12(26+30+28)=842=42cm

Area of triangle ABC = S(Sa)(Sb)(Sc)

= 42(4226)(4330)(4228)

= 42×16×12×14

= 14×3×16×4×3×14

= 14×14×3×3×16×4

= 4×3×2×14=336cm2

In a parallelogram, diagonal divides the parallelogram in two equal area

Therefore area of quadrilateral ABCD= area of ABC + area of ACD

= Area of area of ABC×2

= 336 ×2 = 672 cm2

 

Q.8: Find area of parallelogram ABCD in which AB = 14 cm, BC = 10cm and AC = 16 cm. (Given 3=1.73)

 

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Sol:

Consider the triangle ABC,

Let, a = 10 cm, b = 16 cm and c = 14 cm

S = 12(a+b+c)

S = 12(10+16+14)=402=20cm

Area of triangle ABC = S(Sa)(Sb)(Sc)

= 20(2010)(2016)(2014)

= 20×10×4×6

= 10×2×10×4×3×2

= 14×10×4×2×2×2×3

= 10×2×2×3=403cm2

In a parallelogram, diagonal divides the parallelogram in two equal areas

Therefore, the area of quadrilateral ABCD= area of ABC + area of ACD

= Area of area of ABC×2

= 403×2cm2

= 803cm2 = 138.4 cm2

 

Q.9: In the given figure ABCD is a quadrilateral in which diagonal BD =64 cm ALBD and CMBD such that AL= 16cm and CM =16.8 cm. Calculate the area of quadrilateral ABCD.

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Sol:

Area of triangle ABD = 12×base×height

= 12×BL×AL

= 12×64×16.8 = 537.6 cm2

Area of triangle BCD = 12×base×height

= 12×BD×CM

= 12×64×13.2 = 422.4 cm2

Therefore, the area of quadrilateral ABCD= Area of ABD + BCD = 537.6 + 422.4 = 960 cm2


Practise This Question

The following figure shows the elliptical path of a planet about the sun. The two shaded parts have equal areas. If t1, t2 be the time taken by the planet to go from A to B and from C to D respectively–