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 $200 \ cm$.

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 $15 \ cm$.

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 $42 \ cm$.

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 $AC$.

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.

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.

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.

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.

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.

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.

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 $AC = 5 \ cm$

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

For $\triangle ABC$

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.

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

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.

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

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?

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

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.

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

For $\triangle BCE$

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 $24 \ m$.

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.

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 $= 62.45 \times 5 \times 2 = 624.45 \ Rs.$

$\\$

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 $26 \ cm$.

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

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 $ABCD = 120 + 292.72 = 412.72 \ cm^2$

$\\$

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 $BD = 20 \ cm$.

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

For $\triangle ABD$

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 $ABCD = 336 + 210 = 546 \ cm^2$

$\\$

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 $AC = 15 \ cm$.

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

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 $ABCD = 60 + 54 = 114 \ cm^2$

$\\$

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.

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

For $\triangle ABC$

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 $ABCD = 336 + 336 = 672 \ cm^2$

$\\$