Note: If a, b, c are the sides of the triangle and s is the semi perimeter, then its area if given by A = \sqrt{s(s-a)(s-b)(s-c)} where s = \frac{a+b+c}{2} . This is Heron’s Formula.

Question 1: Find the area of a triangle whose sides are respectively 150 \ cm, 120 \ cm and 200 \ cm .

Answer:

Here a = 150, \ \ b = 120 and c = 200

and s = \frac{a+b+c}{2} = \frac{150+120+200}{2} = 235

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{235(235-150)(235-120)(235-200)}

= \sqrt{235 \times 85 \times 115 \times 35} = 8966.57 \ cm^2

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Question 2: Find the area of a triangle whose sides are 9 \ cm, 12 \ cm and 15 \ cm .

Answer:

Here a = 9, \ \ b = 12 and c = 15

and s = \frac{a+b+c}{2} = \frac{9+12+15}{2} = 18

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{18(18-9)(18-12)(18-15)}

= \sqrt{18 \times 9 \times 6 \times 3} = 54 \ cm^2

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Question 3: Find the area of a triangle two sides of which are 18 \ cm and 10 \ cm and the perimeter is 42 \ cm .

Answer:

Perimeter = 42 \ cm

Therefore third side = 42 - 18 - 10 = 14 \ cm

Here a = 18, \ \ b = 10 and c = 14

and s = \frac{a+b+c}{2} = \frac{18+10+14}{2} = 21

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{21(21-18)(21-10)(21-14)}

= \sqrt{21 \times 3 \times 11 \times 7} = 54 \ cm^2

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Question 4: In a \triangle ABC, AB=15 \ cm, BC=13 \ cm and AC=14 \ cm . Find the area of \triangle ABC and hence its altitude on AC .

Answer:

Here a = 15, \ \ b = 13 and c = 14

and s = \frac{a+b+c}{2} = \frac{15+13+14}{2} = 21

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{21(21-15)(21-13)(21-14)}

= \sqrt{21 \times 6 \times 8 \times 7} = 84 \ cm^2

Altitude (h) = \frac{2 \times 84}{14} = 12 \ cm

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Question 5: 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.

Answer:

Ratio of sides = 25:17:12

Perimeter = 540 \ m

Therefore 25x + 17x + 12x = 540

\Rightarrow x = \frac{540}{54} = 10

Hence the sides are 250 \ cm, \ 170 \ cm and 120 \ cm

Here a = 250, \ \ b = 170 and c = 120

and s = \frac{a+b+c}{2} = \frac{250+170+120}{2} = 270

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{270(270-250)(270-170)(270-120)}

= \sqrt{270 \times 250 \times 170 \times 120} = 9000 \ m^2

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Question 6: The perimeter of a right triangle is 300 \ m . If its sides are in the ratio 3 : 5 : 7 . Find the area of the triangle.

Answer:

Ratio of sides = 3:5:7

Perimeter = 300 \ m

Therefore 3x + 5x + 7x = 300

\Rightarrow x = \frac{300}{15} = 20

Hence the sides are 60 \ cm, \ 100 \ cm and 140 \ cm

Here a = 60, \ \ b = 100 and c = 140

and s = \frac{a+b+c}{2} = \frac{60+100+140}{2} = 150

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{150(150-60)(150-100)(150-140)}

= \sqrt{150 \times 60 \times 100 \times 140} = 1500 \sqrt{3} \ m^2

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Question 7: The perimeter of a triangular field is 240 \ dm . If two of its sides are 78 \ dm and 50 \ dm , find the length of perpendicular on the side of length 50 \ dm from the opposite vertex.

Answer:

Perimeter = 240 \ dm

Therefore third side = 240 - 78 - 50 = 112 \ dm

Here a = 112, \ \ b = 78 and c = 50

and s = \frac{a+b+c}{2} = \frac{112+78+50}{2} = 120

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{120(120-112)(120-78)(120-50)}

= \sqrt{120 \times 8 \times 42 \times 70} = 1680 \ dm^2

Altitude (h) = \frac{2 \times 1680}{50} = 67.2 \ dm

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Question 8: A triangle has sides 35 \ cm, 54 \ cm and 61 \ cm long. Find its area. Also, find the smallest of its altitudes.

Answer:

Here a = 35, \ \ b = 54 and c = 61

and s = \frac{a+b+c}{2} = \frac{35+54+61}{2} = 75

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{75(75-35)(75-54)(75-61)}

= \sqrt{75 \times 40 \times 21 \times 14} = 939.148 \ cm^2

Smallest Altitude (h) = \frac{2 \times 939.148}{61} = 30.80 \ cm

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Question 9: The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 \ cm . Find the area of the triangle and the height corresponding to the longest side.

Answer:

Ratio of sides = 3:4:5

Perimeter = 144 \ cm

Therefore 3x + 4x + 5x = 144

\Rightarrow x = \frac{144}{12} = 12

Hence the sides are 36 \ cm, \ 48 \ cm and 60 \ cm

Here a = 36, \ \ b = 48 and c = 60

and s = \frac{a+b+c}{2} = \frac{36+48+60}{2} = 72

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{72(72-36)(72-48)(72-60)}

= \sqrt{72 \times 36 \times 24 \times 12} = 864 \ cm^2

Smallest Altitude (h) = \frac{2 \times 864}{60} = 28.80 \ cm

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Question 10: The perimeter of an isosceles triangle is 42 \ cm and its base is \frac{3}{2} times each of the equal sides. Find length of each side of the triangle, area of the triangle and the height of the triangle.2019-01-12_10-09-26

Answer:

Perimeter = 42 \ cm

Therefore x +  x + \frac{3}{2} x = 42

\Rightarrow x = \frac{84}{7} = 12

Hence the sides are 12 \ cm, \ 12 \ cm and 18 \ cm

Here a = 12, \ \ b = 12 and c = 18

and s = \frac{a+b+c}{2} = \frac{12+12+18}{2} = 21

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{21(21-12)(21-12)(21-18)}

= \sqrt{21 \times 9 \times 9 \times 3} = 71.45 \ cm^2

Altitude (h) = \frac{2 \times 71.45}{18} = 7.93 \ cm

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Question 11: Find the area of a quadrilateral ABCD is which AB=3 \ cm BC =4 \ cm  CD = 4 \ cm, DA= 5 \ cm and AC = 5 \ cm

Answer:

Area of ABCD = Area of \triangle ABC + Area of  \triangle ACD

For \triangle ABC 2019-01-12_10-09-49

Here a = 3, \ \ b = 4 and c = 5

and s = \frac{a+b+c}{2} = \frac{3+4+5}{2} = 6

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{6(6-3)(6-4)(6-5)}

= \sqrt{6 \times 3 \times 2 \times 1} = 6 \ cm^2

For \triangle ACD

Here a = 5, \ \ b = 4 and c = 5

and s = \frac{a+b+c}{2} = \frac{5+4+5}{2} = 7

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{7(7-5)(7-4)(7-5)}

= \sqrt{7 \times 2 \times 3 \times 2} = 9.165 \ cm^2

Therefore Area of ABCD = 6 + 9.165 = 15.165 \ cm^2

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Question 12: The sides of a quadrangular field, taken in order are 26 \ m, \ \ 27 \ m, \ \ 7 \ m are 24 \ m respectively. The angle contained by the last two sides is a right angle. Find its area.2019-01-12_10-10-13

Answer:

AC = \sqrt{7^2 + 24^2} = 25 \ m

Area of ABCD = Area of \triangle ABC + Area of  \triangle ACD

For \triangle ABC

Here a = 26, \ \ b = 27 and c = 25

and s = \frac{a+b+c}{2} = \frac{26+27+25}{2} = 39

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{39(39-26)(39-27)(39-25)}

= \sqrt{39 \times 13 \times 12 \times 14} = 291.85 \ cm^2

For \triangle ACD

Here a = 25, \ \ b = 7 and c = 24

and s = \frac{a+b+c}{2} = \frac{25+7+24}{2} = 28

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{28(28-25)(28-7)(28-24)}

= \sqrt{28 \times 3 \times 21 \times 4} = 84 \ cm^2

Therefore Area of ABCD = 291.85 + 84 = 375.85 \ cm^2

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Question 13: The sides of a quadrilateral, taken in order are 5, \ 12, \ 14 and 15 meters respectively, and the angle contained by the first two sides is a right angle. Find its area.2019-01-12_10-10-34

Answer:

AC = \sqrt{5^2 + 12^2} = 13 \ cm

Area of ABCD = Area of \triangle ABC + Area of  \triangle ACD

For \triangle ABC

Here a = 5, \ \ b = 12 and c = 13

and s = \frac{a+b+c}{2} = \frac{5+12+13}{2} = 15

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{15(15-5)(15-12)(15-13)}

= \sqrt{15 \times 10 \times 3 \times 2} = 30 \ cm^2

For \triangle ACD

Here a = 13, \ \ b = 14 and c = 15

and s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = 21

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{21(21-13)(21-14)(21-15)}

= \sqrt{21 \times 8 \times 7 \times 6} = 84 \ cm^2

Therefore Area of ABCD = 30 + 84 = 114 \ cm^2

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Question 14: A park, in the shape of a quadrilateral ABCD , has \angle C=90^o, AB=9\ m, BC = 12\ m, CD = 5 \ m and AD = 8 \ m . How much area does it occupy?2019-01-12_10-11-03

Answer:

BD = \sqrt{12^2 + 5^2} = 13 \ m

Area of ABCD = Area of \triangle ABD + Area of  \triangle BCD

For \triangle ABD

Here a = 9, \ \ b = 13 and c = 8

and s = \frac{a+b+c}{2} = \frac{9+13+8}{2} = 15

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{15(15-9)(15-13)(15-8)}

= \sqrt{15 \times 6 \times 2 \times 7} = 35.49 \ m^2

For \triangle BCD

Here a = 12, \ \ b = 5 and c = 13

and s = \frac{a+b+c}{2} = \frac{12+5+13}{2} = 15

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{15(15-12)(15-5)(15-13)}

= \sqrt{15 \times 3 \times 10 \times 2} = 30 \ cm^2

Therefore Area of ABCD = 35.49 + 30 = 65.49 \ cm^2

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Question 15: Two parallel side of a trapezium are 60 \ cm and 77 \ cm and other sides are 25 \ cm and 26 \ cm . Find the area of the trapezium.

Answer:

Area of ABCD = Area of Parallelogram \ AECD + Area of  \triangle BCE

For \triangle BCE 2019-01-12_10-12-52

Here a = 17, \ \ b = 26 and c = 25

and s = \frac{a+b+c}{2} = \frac{17+26+25}{2} = 34

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{34(34-17)(34-26)(34-25)}

= \sqrt{34 \times 17 \times 8 \times 9} = 204 \ cm^2

Altitude CF (h) = \frac{2 \times 204}{17} = 24 \ cm

Area of Parallelogram \ AECD = 60 \times 24 = 1440 \ cm^2

Therefore Area of ABCD = 204 + 1440 = 1644 \ cm^2

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Question 16: Find the area of a rhombus whose perimeter is 80 \ m and one of whose diagonal is 24 \ m .2019-01-12_10-13-32

Answer:

Perimeter = 80 \ m

Therefore Side = 20 \ m

Area of ABCD = Area of \triangle ABD + Area of  \triangle BCD

For \triangle ABC

Here a = 20, \ \ b = 20 and c = 24

and s = \frac{a+b+c}{2} = \frac{20+20+24}{2} = 32

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{32(32-20)(32-20)(32-24)}

= \sqrt{32 \times 12 \times 12 \times 8} = 192 \ m^2

For \triangle ACD

Here a = 24, \ \ b = 20 and c = 20

and s = \frac{a+b+c}{2} = \frac{24+20+20}{2} = 32

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{32(32-24)(32-20)(32-20)}

= \sqrt{32 \times 8 \times 12 \times 12} = 192 \ cm^2

Therefore Area of ABCD = 192 + 192 = 384 \ m^2

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Question 17: A rhombus sheet, whose perimeter is 32 \ m and whose one diagonal is 10 \ m long, is painted on both sides at the rate of Rs. \ 5 \ per \ m^2 . Find the cost of painting.2019-01-12_10-15-57

Answer:

Perimeter = 32 \ m

Therefore Side = 8 \ m

Area of ABCD = Area of \triangle ABC + Area of  \triangle ACD

For \triangle ABC

Here a = 8, \ \ b = 8 and c = 10

and s = \frac{a+b+c}{2} = \frac{8+8+10}{2} = 13

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{13(13-8)(13-8)(13-10)}

= \sqrt{13 \times 5 \times 5 \times 3} = 31.23 \ cm^2

For \triangle ACD

Here a = 8, \ \ b = 8 and c = 10

and s = \frac{a+b+c}{2} = \frac{8+8+10}{2} = 13

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{13(13-8)(13-8)(13-10)}

= \sqrt{13 \times 5 \times 5 \times 3} = 31.23 \ cm^2

Therefore Area of ABCD = 31.23 + 31.23 = 62.45 \ cm^2

Therefore cost of painting = 62.45 \times 5 \times 2 = 624.45 \ Rs.

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Question 18: Find the area of a quadrilateral ABCD in which AD=24 \ cm, \angle BAD=90^o and BCD forms an equilateral triangle whose each side is equal to 26 \ cm .

Answer:2019-01-12_10-13-58

AB = \sqrt{26^2 - 24^2} = 10 \ cm

Area of ABCD = Area of \triangle ABD + Area of  \triangle BCD

For \triangle ABD

Here a = 10, \ \ b = 26 and c = 24

and s = \frac{a+b+c}{2} = \frac{10+26+24}{2} = 30

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{30(30-10)(30-26)(30-24)}

= \sqrt{30 \times 20 \times 4 \times 6} = 120 \ cm^2

For \triangle BCD

Here a = 26, \ \ b = 26 and c = 26

and s = \frac{a+b+c}{2} = \frac{26+26+26}{2} = 39

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{39(39-26)(39-26)(39-26)}

= \sqrt{39 \times 13 \times 13 \times 13} = 292.72 \ cm^2

Therefore Area of ABCD = 120 + 292.72 = 412.72 \ cm^2

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Question 19: Find the area of a quadrilateral ABCD in which AB = 42 \ cm BC = 21 \ cm CD = 29 \ cm, DA= 34 \ cm and diagonal BD = 20 \ cm .

Answer:

Area of ABCD = Area of \triangle ABD + Area of  \triangle BCD

For \triangle ABD 2019-01-12_10-16-37

Here a = 42, \ \ b = 20 and c = 34

and s = \frac{a+b+c}{2} = \frac{42+20+34}{2} = 48

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{48(48-42)(48-20)(48-34)}

= \sqrt{48 \times 6 \times 28 \times 14} = 336 \ cm^2

For \triangle BCD

Here a = 21, \ \ b = 29 and c = 20

and s = \frac{a+b+c}{2} = \frac{21+29+20}{2} = 35

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{35(35-21)(35-29)(35-20)}

= \sqrt{35 \times 14 \times 6 \times 15} = 210 \ cm^2

Therefore Area of ABCD = 336 + 210 = 546 \ cm^2

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Question 20: Find the perimeter and area of the quadrilateral ABCD in which AB = 17 \ cm. AD =9 \ cm, CD =12 \ cm, \angle ACB = 90^o and AC = 15 \ cm .

Answer:2019-01-12_10-16-49

BC = \sqrt{17^2 - 15^2} = 8 \ cm

Perimeter = 17 + 8 + 12 + 9 = 46 \ cm

Area of ABCD = Area of \triangle ABC + Area of  \triangle ACD

For \triangle ABC

Here a = 17, \ \ b = 8 and c = 15

and s = \frac{a+b+c}{2} = \frac{17+8+15}{2} = 20

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{20(20-17)(20-8)(20-15)}

= \sqrt{20 \times 3 \times 12 \times 5} = 60 \ cm^2

For \triangle ACD

Here a = 15, \ \ b = 12 and c = 9

and s = \frac{a+b+c}{2} = \frac{15+12+9}{2} = 18

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{18(18-15)(18-12)(18-9)}

= \sqrt{18 \times 3 \times 6 \times 9} = 54 \ cm^2

Therefore Area of ABCD = 60 + 54 = 114 \ cm^2

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Question 21: The adjacent sides of a parallelogram ABCD measures 34 \ cm and 20 \ cm , and the diagonal AC measures 42 \ cm . Find the area of the parallelogram.

Answer:

Area of ABCD = Area of \triangle ABC + Area of  \triangle ACD

For \triangle ABC 2019-01-12_10-17-09

Here a = 20, \ \ b = 34 and c = 42

and s = \frac{a+b+c}{2} = \frac{20+34+42}{2} = 48

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{48(48-20)(48-34)(48-42)}

= \sqrt{20 \times 28 \times 14 \times 6} = 336 \ cm^2

For \triangle ACD

Here a = 42, \ \ b = 20 and c = 34

and s = \frac{a+b+c}{2} = \frac{42+20+34}{2} = 48

Therefore area of triangle = \sqrt{s(s-a)(s-b)(s-c)}

= \sqrt{48(48-42)(48-20)(48-34)}

= \sqrt{48 \times 6 \times 28 \times 14} = 336 \ cm^2

Therefore Area of ABCD = 336 + 336 = 672 \ cm^2

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