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

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

$\displaystyle \text{ Here } a = 150, \ \ b = 120 \text{and } c = 200$

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{ Here } a = 9, \ \ b = 12 \text{and } c = 15$

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{Perimeter } = 42 \ \text{ cm }$

$\displaystyle \text{Therefore third side } = 42 - 18 - 10 = 14 \ \text{ cm }$

$\displaystyle \text{ Here } a = 18, \ \ b = 10 \text{and } c = 14$

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{ Here } a = 15, \ \ b = 13 \text{and } c = 14$

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

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

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

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

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

$\displaystyle \\$

Question 5: The perimeter of a triangular field is $\displaystyle 540 \ \text{ m}$ and its sides are in the ratio $\displaystyle 25 : 17: 12$. Find the area of the triangle.

$\displaystyle \text{Ratio of sides } = 25:17:12$

$\displaystyle \text{Perimeter } = 540 \ \text{ m}$

$\displaystyle \text{Therefore } 25x + 17x + 12x = 540$

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

Hence the sides are $\displaystyle 250 \text{ cm, } \ 170 \text{ cm } \text{and } 120 \ \text{ cm }$

$\displaystyle \text{ Here } a = 250, \ \ b = 170 \text{and } c = 120$

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{Ratio of sides } = 3:5:7$

$\displaystyle \text{Perimeter } = 300 \ \text{ m}$

$\displaystyle \text{Therefore } 3x + 5x + 7x = 300$

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

Hence the sides are $\displaystyle 60 \text{ cm, } \ 100 \text{ cm } \text{and } 140 \ \text{ cm }$

$\displaystyle \text{ Here } a = 60, \ \ b = 100 \text{and } c = 140$

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{Perimeter } = 240 \ d\text{ m}$

$\displaystyle \text{Therefore third side } = 240 - 78 - 50 = 112 \ d\text{ m}$

$\displaystyle \text{ Here } a = 112, \ \ b = 78 \text{and } c = 50$

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

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

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

$\displaystyle = \sqrt{120 \times 8 \times 42 \times 70} = 1680 \ d\text{ m}^2$

$\displaystyle \text{Altitude } (h) = \frac{2 \times 1680}{50} = 67.2 \ d\text{ m}$

$\displaystyle \\$

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

$\displaystyle \text{ Here } a = 35, \ \ b = 54 \text{and } c = 61$

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{Ratio of sides } = 3:4:5$

$\displaystyle \text{Perimeter } = 144 \ \text{ cm }$

$\displaystyle \text{Therefore } 3x + 4x + 5x = 144$

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

Hence the sides are $\displaystyle 36 \text{ cm, } \ 48 \text{ cm } \text{and } 60 \ \text{ cm }$

$\displaystyle \text{ Here } a = 36, \ \ b = 48 \text{and } c = 60$

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

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

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

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

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

$\displaystyle \\$

Question 10: The perimeter of an isosceles triangle is $\displaystyle 42 \ \text{ cm }$ and its base is $\displaystyle \frac{3}{2}$ times each of the equal sides. Find length of each side of the triangle, area of the triangle and the height of the triangle.

$\displaystyle \text{Perimeter } = 42 \ \text{ cm }$

$\displaystyle \text{Therefore } x + x + \frac{3}{2} x = 42$

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

Hence the sides are $\displaystyle 12 \text{ cm, } \ 12 \text{ cm } \text{and } 18 \ \text{ cm }$

$\displaystyle \text{ Here } a = 12, \ \ b = 12 \text{and } c = 18$

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

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

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

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

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } \triangle ABC$

$\displaystyle \text{ Here } a = 3, \ \ b = 4 \text{and } c = 5$

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

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

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

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

$\displaystyle \text{For } \triangle ACD$

$\displaystyle \text{ Here } a = 5, \ \ b = 4 \text{and } c = 5$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 6 + 9.165 = 15.165 \ \text{ cm}^2$

$\displaystyle \\$

Question 12: The sides of a quadrangular field, taken in order are $\displaystyle 26 \ m, \ \ 27 \ m, \ \ 7 \ \text{ m}$ are $\displaystyle 24 \ \text{ m}$ respectively. The angle contained by the last two sides is a right angle. Find its area.

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

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

$\displaystyle \text{For } \triangle ABC$

$\displaystyle \text{ Here } a = 26, \ \ b = 27 \text{and } c = 25$

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

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

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

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

$\displaystyle \text{For } \triangle ACD$

$\displaystyle \text{ Here } a = 25, \ \ b = 7 \text{and } c = 24$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 291.85 + 84 = 375.85 \ \text{ cm}^2$

$\displaystyle \\$

Question 13: The sides of a quadrilateral, taken in order are $\displaystyle 5, \ 12, \ 14 \text{and } 15$ meters respectively, and the angle contained by the first two sides is a right angle. Find its area.

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

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

$\displaystyle \text{For } \triangle ABC$

$\displaystyle \text{ Here } a = 5, \ \ b = 12 \text{and } c = 13$

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

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

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

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

$\displaystyle \text{For } \triangle ACD$

$\displaystyle \text{ Here } a = 13, \ \ b = 14 \text{and } c = 15$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 30 + 84 = 114 \ \text{ cm}^2$

$\displaystyle \\$

Question 14: A park, in the shape of a quadrilateral $\displaystyle ABCD$, has $\displaystyle \angle C=90^o, AB=9\ m, BC = 12\ m, CD = 5 \ m \text{and } AD = 8 \ \text{ m}$. How much area does it occupy?

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

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

$\displaystyle \text{For } \triangle ABD$

$\displaystyle \text{ Here } a = 9, \ \ b = 13 \text{and } c = 8$

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

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

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

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

$\displaystyle \text{For } \triangle BCD$

$\displaystyle \text{ Here } a = 12, \ \ b = 5 \text{and } c = 13$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 35.49 + 30 = 65.49 \ \text{ cm}^2$

$\displaystyle \\$

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

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

$\displaystyle \text{For } \triangle BCE$

$\displaystyle \text{ Here } a = 17, \ \ b = 26 \text{and } c = 25$

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

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

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

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

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

Area of $\displaystyle Parallelogram \ AECD = 60 \times 24 = 1440 \ \text{ cm}^2$

$\displaystyle \text{Therefore Area of } ABCD = 204 + 1440 = 1644 \ \text{ cm}^2$

$\displaystyle \\$

Question 16: Find the area of a rhombus whose perimeter is $\displaystyle 80 \ \text{ m}$ and one of whose diagonal is $\displaystyle 24 \ \text{ m}$.

$\displaystyle \text{Perimeter } = 80 \ \text{ m}$

Therefore Side $\displaystyle = 20 \ \text{ m}$

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

$\displaystyle \text{For } \triangle ABC$

$\displaystyle \text{ Here } a = 20, \ \ b = 20 \text{and } c = 24$

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

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

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

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

$\displaystyle \text{For } \triangle ACD$

$\displaystyle \text{ Here } a = 24, \ \ b = 20 \text{and } c = 20$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 192 + 192 = 384 \ \text{ m}^2$

$\displaystyle \\$

Question 17: A rhombus sheet, whose perimeter is $\displaystyle 32 \ \text{ m}$ and whose one diagonal is $\displaystyle 10 \ \text{ m}$ long, is painted on both sides at the rate of $\displaystyle Rs. \ 5 \ per \ \text{ m}^2$. Find the cost of painting.

$\displaystyle \text{Perimeter } = 32 \ \text{ m}$

Therefore Side $\displaystyle = 8 \ \text{ m}$

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

$\displaystyle \text{For } \triangle ABC$

$\displaystyle \text{ Here } a = 8, \ \ b = 8 \text{and } c = 10$

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

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

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

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

$\displaystyle \text{For } \triangle ACD$

$\displaystyle \text{ Here } a = 8, \ \ b = 8 \text{and } c = 10$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 31.23 + 31.23 = 62.45 \ \text{ cm}^2$

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{For } \triangle ABD$

$\displaystyle \text{ Here } a = 10, \ \ b = 26 \text{and } c = 24$

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

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

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

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

$\displaystyle \text{For } \triangle BCD$

$\displaystyle \text{ Here } a = 26, \ \ b = 26 \text{and } c = 26$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 120 + 292.72 = 412.72 \ \text{ cm}^2$

$\displaystyle \\$

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

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

$\displaystyle \text{For } \triangle ABD$

$\displaystyle \text{ Here } a = 42, \ \ b = 20 \text{and } c = 34$

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

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

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

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

$\displaystyle \text{For } \triangle BCD$

$\displaystyle \text{ Here } a = 21, \ \ b = 29 \text{and } c = 20$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 336 + 210 = 546 \ \text{ cm}^2$

$\displaystyle \\$

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

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

$\displaystyle \text{Perimeter } = 17 + 8 + 12 + 9 = 46 \ \text{ cm }$

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

$\displaystyle \text{For } \triangle ABC$

$\displaystyle \text{ Here } a = 17, \ \ b = 8 \text{and } c = 15$

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

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

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

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

$\displaystyle \text{For } \triangle ACD$

$\displaystyle \text{ Here } a = 15, \ \ b = 12 \text{and } c = 9$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 60 + 54 = 114 \ \text{ cm}^2$

$\displaystyle \\$

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

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

$\displaystyle \text{For } \triangle ABC$

$\displaystyle \text{ Here } a = 20, \ \ b = 34 \text{and } c = 42$

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

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

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

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

$\displaystyle \text{For } \triangle ACD$

$\displaystyle \text{ Here } a = 42, \ \ b = 20 \text{and } c = 34$

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

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

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

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

$\displaystyle \text{Therefore Area of } ABCD = 336 + 336 = 672 \ \text{ cm}^2$

$\displaystyle \\$