| (i) 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 \[384\text{ }{{m}^{2}}.\] |
| (ii) The area of a quadrilateral ABCD in which B = 3 cm, BC= 4 cm, CD =4 cm, DA = 5 cm and AC = 5 cm is \[18\text{ }c{{m}^{2}}.\] |
| (iii) An advertisement board is in the form of an isosceles triangle with its sides equal to 12 m, 10 m and 10 m. The cost of painting it at ` 2.25 per \[{{m}^{2}}\]is \[18\text{ }c{{m}^{2}}.\]112. |
| (iv) Heron's formula cannot be used to calculate area of quadrilaterals. |
A)
(i) (ii) (iii) (iv) T F F T
B)
(i) (ii) (iii) (iv) F T F F
C)
(i) (ii) (iii) (iv) T F T F
D)
(i) (ii) (iii) (iv) T F F F
Correct Answer: D
Solution :
(i) True: a = 40 m, b = 24 m and c = 32 m \[s=\frac{1}{2}(40+24+32)=48\,m\] Area \[=\sqrt{s(s-a)(s-b)(s-c)}\] \[=\sqrt{48\times 8\times 24\times 16}=384\,{{m}^{2}}\] (II) False: In \[\Delta \Alpha CD,\]we have \[s=\frac{a+b+c}{2}=\frac{5+5+4}{2}=7\,cm\] Area \[(\Delta \Alpha CD)\] \[=\sqrt{7\times 2\times 2\times 3}\] \[=\sqrt{84}=2\sqrt{21}\,c{{m}^{2}}\] In \[\Delta \Alpha CB,\] \[s=\frac{a+b+c}{2}=\frac{5+4+3}{2}=6\,cm\] Area \[(\Delta \Alpha CB)=\sqrt{6\times 1\times 1\times 3\times 2}=\sqrt{36}=6\,c{{m}^{2}}\] Area of quadrilateral ABCD = Area \[(\Delta \Alpha CD)+\]Area \[(\Delta \Alpha CB)\] \[=2\sqrt{21}+6=2(3+\sqrt{21})c{{m}^{2}}\] (III) False: \[s=\frac{a+b+c}{2}\] \[=\frac{10+10+12}{2}=\frac{32}{2}=16\,m\] Area of advertisement board \[=\sqrt{16+6\times 6\times 4}=6\times 8=48\,{{m}^{2}}\] Cost of painting= `\[(48\times 2.25)\]= ` 108. (IV) False: Heron?s formula can be used to calculate area of quadrilaterals dividing it into tow triangles.
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