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Page No 238:

Question 1:

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.

Answer:

We have:
Base = 24 cm
Height = 14.5 cm

Now,
Area of triangle=12×Base×Height=12×24×14.5=174 cm2

Page No 238:

Question 2:

The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.

Answer:

Let the height of the triangle be h m.
∴ Base = 3h m
Now,
Area of the triangle = Total CostRate=78358=13.5 ha =135000 m2
We have:
Area of triangle = 135000 m212×Base×Height =13500012×3h×h =135000h2=135000×23h2=90000h =300 m

Thus, we have:
Height = h = 300 m
Base = 3h = 900 m

Page No 238:

Question 3:

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.

Answer:

Let: a=42 cm, b = 34 cm and c=20 cms= a+b+c2=42+34+202=48 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=48(48-42)(48-34)(48-20)=48×6×14×28=4×2×6×6×7×2×7×4=4×2×6×7=336 cm2

We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
Area of triangle = 336 cm212×Base×Height = 336Height = 336×242=16 cm

Page No 238:

Question 4:

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.

Answer:

Let: a=18 cm, b = 24 cm and c=30 cms= a+b+c2=18+24+302=36 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=36(36-18)(36-24)(36-30)=36×18×12×6=12×3×6×3×12×6=12×3×6=216 cm2

We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
Area of triangle = 216 cm212×Base×Height = 216Height = 216×218=24 cm

Page No 238:

Question 5:

Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.

Answer:

Let: a=91 m, b = 98 m and c=105 ms= a+b+c2=91+98+1052=147 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=147(147-91)(147-98)(147-105)=147×56×49×42=7×3×7×2×2×2×7×7×7×7×3×2=7×7×7×2×3×2=4116 m2

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
Area of triangle = 4116 m212×Base×Height = 4116Height = 4116×2105=78.4 m

Page No 238:

Question 6:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle.

Answer:

Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5×5 = 25 m
12×5 = 60 m
13×5 = 65 m

Now,
Let: a=25 m, b = 60 m and c=65 ms= 1502=75 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=75(75-25)(75-60)(75-65)=75×50×15×10=15×5×5×10×15×10=15×5×10=750 m2

Page No 238:

Question 7:

The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of ploughing the field at Rs 18.80 per 10 m2.

Answer:

Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25×10 = 250 m
17×10 = 170 m
12×10 = 120 m

Now,
Let: a=250 m, b =170 m and c=120 ms= 5402=270 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=270(270-250)(270-170)(270-120)=270×20×100×150=30×3×3×20×20×5×30×5=30×3×20×5=9000 m2

Cost of ploughing 10 m2 field = Rs 18.80
Cost of ploughing 1 m2 field = Rs 18.8010

Cost of ploughing 9000 m2 field = 18.8010×9000= Rs 16920

Page No 238:

Question 8:

Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m.

Answer:

(i) Let: a=85 m and b = 154 m Given:Perimeter = 324 mor, a+b+c =324c=324-85-154=85 ms= 3242=162 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=162(162-85)(162-154)(162-85)=162×77×8×77=1296×77×77=36×77×77×36=36×77=2772 m2


(ii) We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:
Area of triangle = 2772 m212×Base×Height = 2772Height = 2772×2154=36 m

Page No 238:

Question 9:

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.

Answer:

We have: a=13 cm and b=20 cmArea of isosceles triangle=b44a2-b2                                         =204×4(13)2-202                                         =5×676-400                                         =5×276                                         =5×16.6                                         =83.06 cm2

Page No 238:

Question 10:

The base of an isosceles triangle measures 80 cm and its area is 360 cm2. Find the perimeter of the triangle.

Answer:


Let PQR be an isosceles triangle and PXQR.
Now,
Area of triangle =360 cm2 12×QR×PX = 360h =72080=9 cmNow, QX = 12×80 = 40 cm and PX = 9 cm
Also,
PQ=QX2+PX2a=402+92=1600+81=1681=41 cm

∴ Perimeter = 80 + 41 + 41  = 162 cm

Page No 238:

Question 11:

The perimeter of an isosceles triangle is 42 cm and its base is 112 times each of the equal sides. Find (i) the length of each side of the triangle, (ii) the area of the triangle, and (iii) the height of the triangle.

Answer:

Let the equal sides of the isosceles triangle be a cm each.
∴ Base of the triangle, b = 32a cm
(i) Perimeter = 42 cm
or, a + a + 32a = 42
or, 2a +32a= 42

2a+32a = 427a2=42a=12 
 
So, equal sides of the triangle are 12 cm each.
Also,
Base = 32a = 32×12=18 cm
(ii)
Area of isosceles triangle = b44a2-b2=184×4(12)2-182        (a= 12 cm and b=18 cm)=4.5×576-324=4.5×252=4.5×15.87=71.42 cm2

(iii)
Area of triangle =71.42 cm212×Base×Height = 71.42Height = 71.42×218=7.94 cm

Page No 238:

Question 12:

If the area of an equilateral triangle is 363 cm2, find its perimeter.

Answer:

Area of equilateral triangle = 34×(Side)234×(Side)2 =363
(Side)2=144Side=12 cm

Thus, we have:
Perimeter = 3 × Side = 3 × 12 = 36 cm

Page No 238:

Question 13:

If the area of an equilateral triangle is 813 cm2, find its height.

Answer:

Area of equilateral triangle = 34×(Side)234×(Side)2 =813
(Side)2=324Side=18 cm

Now, we have:
Height =32×Side=32×18=93 cm

Page No 238:

Question 14:

The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.

Answer:


Let PQR be a right-angled triangle and PQQR.
Now,
PQ=PR2-QR2=502-482=2500-2304=196=14 cm

Area of triangle =12×QR×PQ =12×48×14=336 cm2

Page No 238:

Question 15:

Each side of an equilateral triangle measures 8 cm. Find (i) the area of the triangle, correct to 2 places of decimal and (ii) the height of the triangle, correct to 2 places of decimal. Take 3=1.732.

Answer:

Side of the equilateral triangle = 8 cm

(i)
Area of equilateral triangle = 34×(Side)2=34×(8)2 =1.732×644=27.71 cm2

(ii)
Height =32×Side=32×8=1.732×82=6.93 cm

Page No 238:

Question 16:

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take 3=1.732.

Answer:

Height of the equilateral triangle = 9 cm
Thus, we have:
Height =32×Side9=32×Side Side = 183=183×33=63 cm

Also,
Area of equilateral triangle = 34×(Side)2=34×(63)2 =10843=273=46.76 cm2



Page No 239:

Question 17:

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

Answer:

We know that the triangle is an isosceles triangle.
Thus, we can find out the area of one triangular piece of cloth.
Area of isosceles triangle = b44a2-b2=204×4(50)2-202       (a= 50 cm and b=20 cm)=5×10000-400=5×9600=5×406=2006 =490 cm2

Now,
Area of 1 triangular piece of cloth = 490 cm2
Area of 12 triangular pieces of cloth = 12×490 =5880 cm2

Page No 239:

Question 18:

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 sq cm.

Answer:

Area of one triangular-shaped tile can be found in the following manner:

Let: a=16 cm, b = 12 cm and c=20 cms= a+b+c2=16+12+202=24 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=24(24-16)(24-12)(24-20)=24×8×12×4=6×4×4×4×4×6=6×4×4=96 cm2

Now,
Area of 16 triangular-shaped tiles = 16×96=1536 cm2
Cost of polishing tiles of area 1 cm2 = Rs 1
Cost of polishing tiles of area 1536 cm2 = 1×1536 =Rs 1536

Page No 239:

Question 19:

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

Answer:

We know that ABC is a right-angled triangle.

∴ BC2=AB2-AC2=172-152=289-225=64=8 cm

 Now,Area of triangle ABC=12×Base×Height = 12×BC×AC = 12×8×15 =60 cm2

Let:a=9 cm, b = 15 cm and c=12 cms= a+b+c2=9+15+122=18 cmBy Heron's formula, we have:Area of triangle ACD = s(s-a)(s-b)(s-c)=18(18-9)(18-15)(18-12)=18×9×3×6=6×3×3×3×3×6=6×3×3=54 cm2

Now,
Area of quadrilateral ABCD  = Area of ABC + Area of ACD
                                           = (60 + 54) cm2 =114 cm2
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 17 + 8 + 12 + 9 = 46 cm

Page No 239:

Question 20:

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

Answer:

We know that DBC is a right-angled triangle.
 Area of triangle DBC=12×Base×Height = 12×BC×BD= 12×20×21=210 cm2

Now,
Let: a=42 cm, b = 34 cm and c=20 cms= a+b+c2=42+34+202=48 cmBy Heron's formula, we have:Area of triangle ABD= s(s-a)(s-b)(s-c)=48(48-42)(48-34)(48-20)=48×6×14×28=4×2×6×6×7×2×7×4=4×2×6×7=336 cm2

∴ Area of quadrilateral ABCD = Area of DBC + Area of ABD
                                             =(210 + 336) cm2 = 546 cm2
Also,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 42 + 21 + 29 + 34 = 126 cm

Page No 239:

Question 21:

Find the area of quadrilateral ABCD in which AD = 24 cm, BAD = 90° and ∆BCD is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral. [Given: 3=1.73]]

Answer:

We know that BAD is a right-angled triangle.

∴ AB=BD2-AD2=262-242=676-576=100=10 cm

 Now,Area of triangle BAD=12×Base×Height = 12×AB×AD = 12×10×24 =120 cm2

Also, we know that BDC is an equilateral triangle.
Area of equilateral triangle = 34×(Side)2=34×(26)2=34×676=1693 =292.37 cm2

Now,
Area of quadrilateral ABCD = Area of ABD + Area of BDC
                                         = (120 + 292.37) cm2 = 412.37 cm2

Page No 239:

Question 22:

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

Answer:

Let: a=26 cm, b =30 cm and c=28 cms= a+b+c2=26+30+282=42 cmBy Heron's formula, we have:Area of triangle ABC = s(s-a)(s-b)(s-c)=42(42-26)(42-30)(42-28)=42×16×12×14=14×3×4×4×2×2×3×14=14×4×2×3=336 cm2

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = 2×336=672 cm2



Page No 240:

Question 23:

Find the area of a parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16 cm. [Given: 3=1.73]

Answer:

Let: a=10 cm, b =16 cm and c=14 cms= a+b+c2=10+16+142=20 cmBy Heron's formula, we have:Area of triangle ABC = s(s-a)(s-b)(s-c)=20(20-10)(20-16)(20-14)=20×10×4×6=10×2×10×2×2×3×2=10×2×23=69.2 cm2

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = 2×69.2 cm2=138.4 cm2

Page No 240:

Question 24:

In the given figure ABCD is a quadrilateral in which diagonal BD = 64 cm, AL BD and CMBD such that AL = 16.8 cm and CM = 13.2 cm. Calculate the area of quadrilateral ABCD.

Answer:

Area of ABCD=Area of ABD+Area of BDC=12×BD×AL+12×BD×CM=12×BD(AL+CM)=12×64(16.8+13.2)=32×30=960 cm2

Page No 240:

Question 1:

In a ABC it is given that base = 12 cm and height = 5 cm. Its area is
(a) 60 cm2
(b) 30 cm2
(c) 153 cm2
(d) 45 cm2

Answer:

(b) 30 cm2

Area of triangle = 12×Base×HeightArea of ABC=12×12×5=30 cm2

Page No 240:

Question 2:

The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is
(a) 96 cm2
(b) 120 cm2
(c) 144 cm2
(d) 160 cm2

Answer:

(a) 96 cm2

Let: a=20 cm, b = 16 cm and c=12 cms= a+b+c2=20+16+122=24 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=24(24-20)(24-16)(24-12)=24×4×8×12=6×4×4×4×4×6=6×4×4=96 cm2

Page No 240:

Question 3:

Each side of an equilateral triangle measure 8 cm. The area of the triangle is
(a) 83 cm2
(b) 163 cm2
(c) 323 cm2
(d) 48 cm2

Answer:

(b) 163 cm2
Area of equilateral triangle = 34×(Side)2=34×(8)2=34×64=163 cm2

Page No 240:

Question 4:

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is
(a) 165 cm2
(b) 85 cm2
(c) 163 cm2
(d) 83 cm2

Answer:

(b) 85 cm2
Area of isosceles triangle = b44a2-b2Here, a= 6 cm and b=8 cmThus, we have:84×4(6)2-82=84×144-64=84×80=84×45=85 cm2



Page No 241:

Question 5:

The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is
(a) 8 cm
(b) 30 cm
(c) 4 cm
(d) 11 cm

Answer:

(c) 4 cm
Height of isosceles triangle = 124a2-b2=12452-62       a=5 cm and b=6 cm=12×100-36=12×64=12×8=4 cm

Page No 241:

Question 6:

Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
(a) 510 cm2
(b) 50 cm2
(c) 103 cm2
(d)
75 cm2

Answer:

(b) 50 cm2
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:
Area of triangle = 12×Base×Height=12×10×10=50 cm2

Page No 241:

Question 7:

Each side of an equilateral triangle is 10 cm long. The height of the triangle is
(a) 103 cm
(b) 53 cm
(c) 102 cm
(d)
5 cm

Answer:

(b) 53 cm
Height of equilateral triangle=32×Side=32×10=53 cm

Page No 241:

Question 8:

The height of an equilateral triangle is 6 cm. Its area is
(a) 123 cm2
(b) 63 cm2
(c) 122 cm2
(d)
18 cm2

Answer:

(a) 123 cm2
Height of equilateral triangle = 32×Side6=32×SideSide=123×33=123×3=43  cmNow,Area of equilateral triangle = 34×(Side)2=34×432=34×48=123 cm2

Page No 241:

Question 9:

The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is
(a) 480 m2
(b) 320 m2
(c) 384 m2
(d) 360 m2

Answer:

(c) 384 m2

Let: a=40 m, b = 24 m and c=32 ms= a+b+c2=40+24+322=48 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=48(48-40)(48-24)(48-32)=48×8×24×16=24×2×8×24×8×2=24×8×2=384 m2

Page No 241:

Question 10:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is
(a) 375 cm2
(b) 750 cm2
(c) 250 cm2
(d)
500 cm2

Answer:

(b) 750 cm2

Let the sides of the triangle be 5x cm, 12x cm and 13x cm.
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, the sides of the triangle are 5×5 cm, 12×5 cm and 13×5 cm, i.e., 25 cm, 60 cm and 65 cm.

Now,
Let: a=25 cm, b = 60 cm and c=65 cms= 1502=75 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=75(75-25)(75-60)(75-65)=75×50×15×10=15×5×5×10×15×10=15×5×10=750 cm2

Page No 241:

Question 11:

The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is
(a) 24 cm
(b) 18 cm
(c) 30 cm
(d)
12 cm

Answer:

(a) 24 cm

Let:a=30 cm, b = 24 cm and c=18 cms= a+b+c2=30+24+182=36 cmOn applying Heron's formula, we get:Area of triangle = s(s-a)(s-b)(s-c)=36(36-30)(36-24)(36-18)=36×6×12×18=12×3×12×6×3=12×3×6=216 cm2

The smallest side is 18 cm.
Hence, the altitude of the triangle corresponding to 18 cm is given by:
Area of triangle = 216 cm212×Base×Height = 216Height = 216×218=24 cm

Page No 241:

Question 12:

The base of an isosceles triangle is 16 cm and its area is 48 cm2. The perimeter of the triangle is
(a) 41 cm
(b) 36 cm
(c) 48 cm
(d)
324 cm

Answer:

(b) 36 cm

Let PQR be an isosceles triangle and PXQR.
Now,
Area of triangle =48 cm2 12×QR×PX = 48h =9616=6 cmAlso, QX = 12×24 = 12 cm and PX = 12 cm
PQ=QX2+PX2a=82+62=64+36=100=10 cm

∴ Perimeter = (10 + 10 + 16) cm = 36 cm

Page No 241:

Question 13:

The area of an equilateral triangle is 363 cm2. Its perimeter is
(a) 36 cm
(b) 123 cm
(c) 24 cm
(d) 30
cm

Answer:

(a) 36 cm
Area of equilateral triangle = 34×(Side)234×(Side)2 =363

(Side)2=144Side=12 cm

Now,
Perimeter
 = 3 × Side = 3 × 12 = 36 cm

Page No 241:

Question 14:

Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
(a) 156 cm2
(b) 78 cm2
(c) 60 cm2
(d) 120
cm2

Answer:

(c) 60 cm2
Area of isosceles triangle = b44a2-b2Here, a= 13 cm and b=24 cmThus, we have:244×4(13)2-242=6×676-576=6×100=6×10=60 cm2

Page No 241:

Question 15:

The base of a right triangle. is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is
(a) 168 cm2
(b) 252 cm2
(c) 336 cm2
(d) 504
cm2

Answer:

(c) 336 cm2

Let PQR be a right-angled triangle and PQQR.
Now,
PQ=PR2-QR2=502-482=2500-2304=196=14 cm

Area of triangle =12×QR×PQ =12×48×14=336 cm2

Page No 241:

Question 16:

The area of an equilateral triangle is 813 cm2. Its height is
(a) 93 cm
(b) 63 cm
(c) 183 cm
(d) 9
cm

Answer:

(a) 93 cm
Area of equilateral triangle = 813 cm234×(Side)2=813(Side)2=81×4(Side)2=324Side=18 cmNow,Height = 32×Side=32×18=93 cm



Page No 242:

Question 17:

The difference between the semi-perimeter and the sides of a ABC are 8 cm, 7 cm and 5 cm respectively. The area of the triangle is
(a) 207 cm2
(b) 1014 cm2
(c) 2014 cm2
(d) 140 cm2

Answer:

(c) 2014 cm2
Let the sides of the triangle be a cm, b cm and c cm and the semi-perimeter be s cm.
Given:
s-a=8 cm , s-b=7 cm and s-c=5 cm
By adding all these, we get:
s-a+s-b+s-c=8+7+53s-(a+b+c) =20    2s=a+b+c3s-2s=20s=20 cm

By applying Heron's formula, we get:Area of triangle = s(s-a)(s-b)(s-c)=20×8×7×5=5×4×7×5×4×2=5×4×14=2014 cm2

Page No 242:

Question 18:

For an isosceles right-angled triangle having each of equal sides 'a', we have
I. Area=12a2
II. Perimeter=(2+2)a
III. Hypotenuse = 2a
Which of the following is true?
(a) I only
(b) II only
(c) I and II
(d) I and III

Answer:

(c) I and II
Area of triangle = 12×Base×Height=12×a×a=12a2
Hypotenuse = a2+a2=2a2=2a
Perimeter = a+a+2a=2a+2a=a(2+2)

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Question 19:

For an isosceles triangle having base b and each of the equal sides as a, we have:
I. Area=b·4a2-b24
II. Perimeter = (2a + b)
III. Height=124a2-b2
Which of the following is true?
(a) I only
(b) I and II only
(c) II and III only
(d) I, II and III

Answer:

(d) I, II and III
For an isosceles triangle having base as b and each of the equal sides as a, we have:
I.  Area=b4a2-b24
II.  Perimeter = (2a + b)
III. Height=124a2-b2

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Question 20:

Assertion: Area of an equilateral triangle having each side equal to 4 cm is 43 sq cm.
Reason: Area of an equilateral triangle having each side a is 34a2 sq units.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
Area of equilateral triangle = 34×(Side)2=34×(4)2=34×16=43 cm2
Reason: Area of an equilateral triangle having each side a is 34a2 sq. units.
Hence, it is true.



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Question 21:

Assertion: The area of an isosceles triangle having base = 8 cm and each of the equal sides = 5 cm is 12 cm2.
Reason: The area of an isosceles triangle having each of the equal sides as a and base = b is 14b4a2-b2.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
Area of isosceles triangle = b44a2-b2Here, a= 5 cm and b=8 cmThus, we have:84×4(5)2-82=2×100-64=2×36=2×6=12 cm2

Hence, Assertion is true.

Reason:
Area of an isosceles triangle having each of the equal sides as a and base as b is 14b4a2-b2.
Hence, it is true.

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Question 22:

Assertion: The area of  an equilateral triangle having side 4 cm is 3 cm2.
Reason: The area of an equilateral triangle having each side a is 34a2 sq units.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(d) Assertion is false and Reason is true.

Assertion:
Area of equilateral triangle = 34×(Side)2=34×(4)2=34×16=43 cm2
Hence, Assertion is false.
Reason: Area of an equilateral triangle having each side a is 34a2 square units. Hence, it is true.

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Question 23:

Assertion: The sides of a ABC are in the ratio 2 : 3 : 4 and its perimeter is 36 cm. Then, ar(ABC)=1215 cm2.
Reason: If 2s = (a + b + c), where a, b, c are the sides of a triangle, then its area =(s-a)(s-b)(s-c).
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(c) Assertion is true and Reason is false.
Assertion:
Let the sides of the triangle be 2x cm, 3x cm and 4x cm.
Perimeter = Sum of all sides
or, 36 = 2x + 3x + 4x
or, 9x = 36
or, x = 4
Thus, the sides of the triangle are 2×4 cm, 3×4 cm and 4×4 cm, i.e., 8 cm, 12 cm and 16 cm.

Now,
Let: a=8 cm, b = 12 cm and c=16 cms= 362=18 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=18(18-8)(18-12)(18-16)=18×10×6×2=6×3×5×2×6×2=6×215=1215 cm2
Hence, Assertion is true.
Reason: If 2s = (a + b + c), where a, b and c are the sides of a triangle, then its area =(s-a)(s-b)(s-c).
Hence, it is false.

Area should be s(s-a)(s-b)(s-c).

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Question 24:

Assertion: The area of an isosceles triangle having base = 24 cm and each of the equal sides equal to 13 cm is 60 cm2.
Reason: If 2s = (a + b + c), where a, b, c are the sides of a triangle, then area =s(s-a)(s-b)(s-c).
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
Assertion:
Let: a=24 cm, b = 13 cm and c=13 cms= a+b+c2=24+13+132=25 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=25(25-24)(25-13)(25-13)=25×1×12×12=5×5×12×12=5×12=60 cm2

Reason:
If 2s = (a + b + c), where a, b and c are the sides of a triangle, then area=s(s-a)(s-b)(s-c).

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Question 25:

If the base of an isosceles triangle is 6 cm and its perimeter is 16 cm, then its area is 12 cm2.

Answer:

True
Let the equal sides of the isosceles triangle be a cm each.
We have:
Base of the triangle, b = 6 cm
Perimeter = 16 cm
or, a + a + 6 = 16
or, 2a = 10
or, a = 5 cm

Now,
Area of isosceles triangle = b44a2-b2Here, a= 5 cm and b=6 cmThus, we have:64×4(5)2-62=1.5×100-36=1.5×64=1.5×8=12 cm2

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Question 26:

If each side of an equilateral triangle is 8 cm long, then its area is 203 cm2.

Answer:

False
Area of equilateral triangle=34×(Side)2=34×(8)2=34×64=163 cm2

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Question 27:

If the sides of a triangular field measure 51 m, 37 m and 20 m, then the cost of levelling it at Rs 5 per m2 is Rs 1530.

Answer:

True

Let:a=51 m, b = 37 m and c=20 ms= a+b+c2=51+37+202=54 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=54(54-51)(54-37)(54-20)=54×3×17×34=3×3×3×2×3×17×17×2=17×2×3×3=306 m2
Cost of levelling 1 m2 of area = Rs 5
Cost of levelling 306 m2 of area = 5×306 =Rs 1530



Page No 244:

Question 28:

Match the following columns.

Column I Column II
(a) The lengths of three sides of a triangle are 26 cm, 28 cm and 30 cm. The height corresponding to base 28 cm is ....... cm. (p) 6
(b) The area of an equilateral triangle is 43 cm2. The perimeter of the triangle is ...... cm. (q) 4
(c) If the height of an equilateral triangle is 33 cm, then each side of the triangle measures ...... cm. (r) 24
(d) Let the base of an isosceles triangle be 6 cm and each of the equal sides be 5 cm. Then, its height is ...... cm. (s) 12
(a) ......,
(b) ......,
(c) ......,
(d) ......,

Answer:

(a)
Let:a=26 cm, b =28 cm and c=30 cms= a+b+c2=26+28+302=42 cmBy Heron's formula, we have:Area of triangle ABC = s(s-a)(s-b)(s-c)=42(42-26)(42-28)(42-30)=42×12×16×14=14×3×2×3×2×4×4×14=14×4×2×3=336 cm2

Hence, the height of the triangle corresponding to 28 cm is given by:
Area of triangle =336 cm212×Base×Height = 336Height = 336×228=24 cm


(b)
Area of equilateral triangle = 34×(Side)234×(Side)2 =43
Side2=16Side=4 cm

Perimeter = 3 × Side = 3 × 4 = 12 cm

(c)
Height = 32×SideSide = 2×Height3=2×333=6 cm

(d)
Height of isosceles triangle = 124a2-b2=12452-62       a=5 cm and b=6 cm=12×100-36=12×64=12×8=4 cm

 

Column I Column II
(a) The lengths of three sides of a triangle are 26 cm, 28 cm and 30 cm. The height corresponding to base 28 cm is ....... cm. (r) 24
(b) The area of an equilateral triangle is 43 cm2. The perimeter of the triangle is ...... cm. (s) 12
(c) If the height of an equilateral triangle is 33 cm, then each side of the triangle measures ...... cm. (p) 6
(d) Let the base of an isosceles triangle be 6 cm and each of the equal sides be 5 cm. Then, its height is ...... cm. (q) 4

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Question 29:

A park in the shape of a quadrilateral ABCD has AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m and C = 90°. Find the area of the park.

Answer:

We know that BCD is a right triangle.
∴ BD=BC2+CD2=122+52=144+25=169=13 m

 Now,Area of triangle BCD=12×base×height =12×BC×CD =12×12×5 =30 m2

Let:a=9 m, b = 8 m and c=13 ms= a+b+c2=9+8+132=15 mBy Heron's formula, we have:Area of triangle ABD = s(s-a)(s-b)(s-c)=15(15-9)(15-8)(15-13)=15×6×7×2=5×3×3×2×7×2=3×235=635=6×5.9 =35.4 m2
Now,
Area of quadrilateral ABCD  = Area of BCD + Area of ABD
                                           = (30 + 35.4) m2 = 65.4 m2

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Question 30:

Find the area of a parallelogram ABCD in which AB = 60 cm, BC = 40 cm and AC = 80 cm.

Answer:

Let:a=40 cm, b =80 cm and c=60 cms= a+b+c2=40+80+602=90 cmBy Heron's formula, we have:Area of triangle ABC = s(s-a)(s-b)(s-c)=90(90-40)(90-80)(90-60)=90×50×10×30=30×3×10×5×10×30=30×1015=300×3.87=1161 cm2

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = 2×1161 cm2=2322 cm2



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Question 31:

A piece of land is in the shape of a rhombus ABCD in which each side measures 100 m and diagonal AC is 160 m long. Find the area of the rhombus.

Answer:

Let:a=100 m, b=160 m and c=100 ms=a+b+c2=100+160+1002=180 mBy Heron's formula, we have:Area of triangle ABC=s(s-a)(s-b)(s-c)=180(180-100)(180-160)(180-100)=180×80×20×80=20×3×3×80×20×80=20×3×80=4800 m2

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = 2×4800=9600 m2

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Question 32:

A floral design on a floor is made up of 16 triangular tiles, each having sides 9 cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of Rs 2.50 per cm2.

Answer:

Area of one triangular tile can be found in the following manner:
Let:a=9 cm, b = 28 cm and c=35 cms= a+b+c2=9+28+352=36 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=36(36-9)(36-28)(36-35)=36×27×8×1=9×4×3×9×4×2=9×46=88.2 cm2

∴ Area of 16 triangular tiles = 16×88.2=1411.2 cm2
Cost of polishing 1 cm2 of area = Rs 2.5
Cost of polishing 1411.2 cm2 of area = 2.5×1411.2 =Rs 3528

Page No 245:

Question 33:

A like in the shape of a square with each diagonal 32 cm and having a tail in the shape on an isosceles triangle of base 8 cm and each side 6 cm, is made of three different shades as shown in the figure. How much paper of each shade has been used in it?

Answer:

 Every square is a rhombus. Area of rhombus=12 Ist diagonal×IInd diagonalHere, each diagonal measures 32 cm.Therefore, we can say:Area of square ABCD=12×(Diagonal)2                                =12×32×32                                =512 cm2

We know that a diagonal of a parallelogram divides into two triangles of equal areas. Also, a square is a parallelogram. Therefore, we can say that a diagonal of a square divides into two triangles of equal areas.
∴ Area of ABD = Area of BDC =12×512=256 cm2
Area of isosceles CEF=b44a2-b2Here, a= 6 cm and b=8 cmThus, we have:84×4(6)2-82=84×144-64=84×80=84×45=85 =8×2.24=17.92 cm2



Page No 250:

Question 1:

Each side of an equilateral triangle is 8 cm. Its altitude is
(a) 22 cm
(b) 23 cm
(c) 43 cm
(d) 26 cm

Answer:

(c) 43 cm

Height=32×Side=32×8=43 cm

Page No 250:

Question 2:

The perimeter of an isosceles right-angled triangle having a as each of the equal sides is
(a) (1+2)a
(b) (2+2)a
(c) 3a
(d) (3+2)a

Answer:

(b) (2+2)a
For an isosceles right-angled triangle having each equal side as a, we have:
Hypotenuse = a2+a2=2a2=2a
Perimeter = a+a+2a=2a+2a=(2+2)a

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Question 3:

For an isosceles triangle having base = 12 cm and each of the equal sides equal to 10 cm, the height is
(a) 12 cm
(b) 16 cm
(c) 6 cm
(d) 8 cm

Answer:

(d) 8 cm
Height of isosceles triangle=124a2-b2=124102-122       a=10 cm and b=12 cm=12×400-144=12×256=12×16=8 cm

Page No 250:

Question 4:

Find the area of an equilateral triangle having each side 6 cm.

Answer:

Area of equilateral triangle=34×(Side)2=34×(6)2=34×36=93 cm2

Page No 250:

Question 5:

Using Heron's formula find the area of ABC in which BC = 13 cm, AC = 14 cm and AB = 15 cm.

Answer:

Let:a=13 cm, b =14 cm and c=15 cms= a+b+c2=13+14+152=21 cmBy Heron's formula, we have:Area of triangle ABC=s(s-a)(s-b)(s-c)=21(21-13)(21-14)(21-15)=21×8×7×6=7×3×2×2×2×7×3×2=7×3×2×2=84 cm2

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Question 6:

The sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find the area of the triangle.

Answer:

Let the sides of the triangle be 13x cm, 14x cm and 15x cm.
Perimeter = Sum of all sides
or, 84 = 13x + 14x + 15x
or, 42x = 84
or, x = 2
So, sides of the triangle are 13×2, 14×2 and 15×2, i.e., 26 cm, 28 cm and 30 cm.
Let:a=26 cm, b = 28 cm and c=30 cms= 842=42 cmBy Heron's formula, we have:Area of the triangle=s(s-a)(s-b)(s-c)=42(42-26)(42-28)(42-30)=42×16×14×12=14×3×4×4×14×3×2×2=14×3×4×2=336 cm2

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Question 7:

Find the area of ABC in which BC = 8 cm, AC = 15 cm and AB = 17 cm. Find the length of altitude drawn on AB.

Answer:

Let:a=8 cm, b = 15 cm and c=17 cms= a+b+c2=8+15+172=20 cmBy Heron's formula, we have:Area of triangle ABC=s(s-a)(s-b)(s-c)=20(20-8)(20-15)(20-17)=20×12×5×3=4×5×4×3×5×3=4×5×3=60 cm2

Altitude of the triangle corresponding to AB:
Area of triangle=60 cm212×Base×Height =60Height=60×217=7.06 cm

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Question 8:

An isosceles triangle has perimeter 30 cm and each of its equal sides is 12 cm. Find the area of the triangle.

Answer:

Let the base of the isosceles triangle be b cm.
Now,
Equal side of the triangle = 12 cm

Perimeter = 30 cm
or, 12 + 12 + b = 30
or, 24 + b = 30
or, b = 6 cm
Area of isosceles triangle = b44a2-b2Here, a= 12 cm and b=6 cmThus, we have:=64×4(12)2-62=1.5×576-36=1.5×540=1.5×6×6×15=1.5×615=915 cm2

Page No 250:

Question 9:

The perimeter of an isosceles triangle is 32 cm. The ratio of one of the equal sides to its base is 3 : 2. Find the area of the triangle.

Answer:

Let the equal sides of the isosceles triangle (a) be 3x cm and base of the triangle (b) be 2x cm.

(i) Perimeter = 32 cm
or, 3x + 3x + 2x = 32
or, 8x = 32
or, x = 4

∴ Equal sides of the triangle = 3x = 3(4) = 12 cm
Base of the triangle = 2x = 2(4) = 8 cm
Now,
Area of isosceles triangle=b44a2-b2Here, a=12 cm and b=8 cmThus, we have:=84×4(12)2-82=2×576-64=2×512=2×256×2=2×162=322 cm2



Page No 251:

Question 10:

Given a ∆ABC in which
I. ∠A, ∠B and ∠C are in the ratio 3 : 2 : 1.
II. AB, AC and BC are in the ratio 3:3:23 and AB = 33 cm.
Is ∆ABC a right triangle?
The question give above has two Statements I and II. Answer the question by using instructions given below:
(a) If the question can be answered by one of the given statements only and not by the other.
(b) If the question can be answered by using either statement along.
(c) If the question can be answered by using both the statements but cannot be answered by using either statement.
(d) If the question cannot be answered even by using both the statements together.

Answer:

(b)
Let A, B and C be 3x, 2x and x, respectively.
 3x+2x+x=180°  (Angle sum property)6x=180°x=30°
Now, we have:A=3x=3×30°=90°B=2x=2×30°=60°C=x=30° 

Hence, it is a right triangle at A.
Let the sides of the triangle be 3x cm, 3x cm and 23x cm.
Also,
AB=33 cmor, 3x=33or, x=3
Now,
AC=3x=3×3=3 cmBC=23x=23×3=6 cm

BC2=62=36 cmAB2+AC2=332+32=27+9=36 cmHere,BC2=AB2+AC2Thus, it is a right-angled triangle.

So, it can be answered using either statement.

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Question 11:

In the given figure ABC and ∆DBC have the same base BC such that AB = 120 m, AC = 122 m, BC = 22 m, BD = 24 m and CD = 26 m. Find the area of the shaded region.

Answer:

Let:a=22 m, b=122 m and c=120 ms= a+b+c2=22+122+1202=132 mBy Heron's formula, we have:Area of triangle ABC = s(s-a)(s-b)(s-c)=132(132-22)(132-122)(132-120)=132×110×10×12=11×12×11×10×10×12=11×12×10=1320 m2

Now,
Let:a=22 m, b =24 m and c=26 ms= a+b+c2=22+24+262=36 mBy Heron's formula, we have:Area of triangle OBC = s(s-a)(s-b)(s-c)=36(36-22)(36-24)(36-26)=36×14×12×10=12×3×7×2×12×5×2=12×2105=24105=246 m2

Area of the shaded portion=Area of ABC-Area ofOBC                                           =(1320-246) m2                                           =1074 m2

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Question 12:

A point O is taken inside an equilateral ABC. If OLBC, OMAC and ONAB such that OL = 14 cm, OM = 10 cm and ON = 6 cm, find the area of ∆ABC.

Answer:

Let each side of ABC be x cm. ar(ABC)=ar(BOC)+ar(AOC)+ar(AOB)                             =12×BC×OL+12×AC×OM+12×AB×ON                             =12×x×14+12×x×10+12×x×6                             =12x14+10+6                              =12x×30                              =15x cm2ar(ABC)=15xcm2or,34x2=15xor,3x2=60xor, 3x2-60x=0or,3x(x-203)=0or, 3x=0 and (x-203)=0or, x=0  (Side cannnot be 0.)and, x=203ar(ABC)=15x=15×203=3003 cm2



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