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Question 1: The sides of a triangle are in the ratio $12 : 17 : 25$ and its perimeter is $540 \text{ cm}$. Find its area.
Solution: Let sides be $12x, 17x, 25x$. $12x+17x+25x = 540 \Rightarrow 54x = 540 \Rightarrow x = 10$. Sides are $120, 170, 250$. $s = 270$. Area $= \sqrt{270(150)(100)(20)} = 9000 \text{ cm}^2$.
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Question 2: An isosceles triangle has perimeter $30 \text{ cm}$ and each of the equal sides is $12 \text{ cm}$. Find the area of the triangle.
Solution: Third side $= 30 - (12+12) = 6 \text{ cm}$. $s = 15$. Area $= \sqrt{15(3)(3)(9)} = \sqrt{15 \cdot 81} = 9\sqrt{15} \text{ cm}^2$.
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Question 3: Find the area of a triangle two sides of which are $18 \text{ cm}$ and $10 \text{ cm}$ and the perimeter is $42 \text{ cm}$.
Solution: Third side $= 42 - (18+10) = 14 \text{ cm}$. $s = 21$. Area $= \sqrt{21(3)(11)(7)} = \sqrt{21 \cdot 33 \cdot 7} = 21\sqrt{11} \text{ cm}^2$.
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Question 4: A triangular park $ABC$ has sides $120 \text{ m}, 80 \text{ m}$ and $50 \text{ m}$. A gardener has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹ $20$ per metre leaving a space $3 \text{ m}$ wide for a gate on one side.
Solution: $s = 125$. Area $= \sqrt{125(5)(45)(75)} = 375\sqrt{15} \text{ m}^2$. Perimeter $= 250 \text{ m}$. Fencing length $= 250 - 3 = 247 \text{ m}$. Cost $= 247 \times 20 = ₹ 4940$.
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Question 5: The sides of a triangular field are $41 \text{ m}, 40 \text{ m}$ and $9 \text{ m}$. Find the number of rose beds that can be prepared in the field, if each rose bed, on an average, needs $900 \text{ cm}^2$ space.
Solution: $s = 45$. Area $= \sqrt{45(4)(5)(36)} = 180 \text{ m}^2 = 1800000 \text{ cm}^2$. Number of beds $= 1800000 / 900 = 2000$.
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Question 6: Find the area of a quadrilateral $ABCD$ in which $AB = 3 \text{ cm}, BC = 4 \text{ cm}, CD = 4 \text{ cm}, DA = 5 \text{ cm}$ and $AC = 5 \text{ cm}$.
Solution: Area $\triangle ABC$ (Right angled at B) $= 0.5 \times 3 \times 4 = 6$. Area $\triangle ACD$ ($s=7$) $= \sqrt{7(2)(3)(2)} = 2\sqrt{21}$. Total Area $= 6 + 2\sqrt{21} \approx 15.17 \text{ cm}^2$.
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Question 7: A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is $30 \text{ m}$ and its longer diagonal is $48 \text{ m}$, how much area of grass field will each cow be getting?
Solution: Area of one triangle ($30, 30, 48$): $s=54$. Area $= \sqrt{54(24)(24)(6)} = 432$. Total Area $= 2 \times 432 = 864 \text{ m}^2$. Area per cow $= 864/18 = 48 \text{ m}^2$.
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Question 8: An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring $20 \text{ cm}, 50 \text{ cm}$ and $50 \text{ cm}$. How much cloth of each colour is required for the umbrella?
Solution: Area of one piece ($s=60$) $= \sqrt{60(40)(10)(10)} = 200\sqrt{6} \text{ cm}^2$. 5 pieces of each colour. Cloth per colour $= 5 \times 200\sqrt{6} = 1000\sqrt{6} \text{ cm}^2$.
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Question 9: A kite is in the shape of a square with a diagonal $32 \text{ cm}$ and an isosceles triangle of base $8 \text{ cm}$ and sides $6 \text{ cm}$ each is to be made of three different shades. How much paper of each shade has been used in it?
Solution: Square Area $= 0.5 \times 32 \times 32 = 512$. Shade I = Shade II $= 256 \text{ cm}^2$. Triangle Area ($6, 6, 8$): $s=10$. Area $= \sqrt{10(4)(4)(2)} = 8\sqrt{5} \approx 17.92 \text{ cm}^2$ (Shade III).
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Question 10: A field is in the shape of a trapezium whose parallel sides are $25 \text{ m}$ and $10 \text{ m}$. The non-parallel sides are $14 \text{ m}$ and $13 \text{ m}$. Find the area of the field.
Solution: Draw line parallel to side 13m. Forms triangle with sides $13, 14, 15$ ($15 = 25-10$). Area of triangle ($s=21$) $= \sqrt{21(8)(7)(6)} = 84$. Height $= 2 \times 84 / 15 = 11.2 \text{ m}$. Area Trap $= 0.5 \times (25+10) \times 11.2 = 196 \text{ m}^2$.