Triangles
Category : 7th Class
We know that the triangle is a closed region which is bounded by three line segments.
In the following figure perimetre of \[\Delta ABC=AB+BC+CA.\]
\[=\frac{1}{2}\times b\times h\]
In the given figure area of \[\Delta PQR\text{ }=\frac{1}{2}\times QR\text{ }\times PS\]
Where s represents half of the perimetre and is equal to \[\frac{a+b+c}{2}\] a, b, c are the length of different sides of a triangle.
The above formula is also known as Heron's formula.
Where a is the length of side of an equilateral triangle
Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.
(a) 51cm3
(b) 52cm3
(c) 53cm4
(d) 54cm2
(e) None of these
Answer: (d)
Explanation
Here, a = 9 cm, b = 12 cm and c = 15 cm
\[\therefore S=\frac{a+b+c}{2}=\frac{9+12+15}{2}=\frac{36}{2}=18\]
Area\[=\sqrt{s(s-a)(s-b)(s-c)}\]\[=\sqrt{s(s-a)(s-b)(s-c)}\]
\[=\sqrt{18\text{ }\times 9\times 6\text{ }\times 3}=54\text{ }c{{m}^{2}}\]
A field in the form of a right triangle with hypotenuse 10 m and one side 8 m. Find the are a of the field.
(a)\[22{{m}^{2}}\]
(b)\[23{{m}^{2}}\]
(c)\[24{{m}^{2}}\]
(d) \[25{{m}^{2}}\]
(e) None of these
Answer: (c)
Given\[AC=10\text{ }m,\text{ }AB=8\text{ }m\]
By Pythagoras theorem, \[B{{C}^{2}}=A{{C}^{2}}-A{{B}^{2}}=100-64=36\Rightarrow BC=6\text{ }cm\]
Area of \[=\frac{1}{2}\times BC\times AB=\frac{1}{2}\times 6\times 8=24{{m}^{2}}\]
The base of a triangular field is three times of its altitude. If the cost of .catering the field at Rs. 96 per hectare is Rs. 3600 then find the measure of the base and height.
(a) (1500m, 500m)
(b) (900 m, 300 m)
(c) (500m, 1500m)
(d) (400 m, 1200 m)
(e) None of these
Answer: (c)
The cost of cutting grass from a triangular field at Rs. 45 per 100 m2 is Rs.900. Find its height if twice the base of the triangle is 5 times the height.
(a) 30m
(b) 40m
(c) 50m
(d) 10m
(e) None of these
Answer: (b)
The area of an equilateral triangle is \[173.2\text{ }c{{m}^{2}}.\] Find the perimeter of the triangle.
(a) 27.5cm
(b) 4.5cm
(c) 60cm
(d) 10cm
(e) None of these
Answer: (c)
Quadrilaterals
It is the closed geometrical figure which is bounded by four line segments.
The perimeter of a quadrilateral = sum of all sides
In the given figure, Perimeter of quadntateral
\[ABCD\text{ }=AB+BC+CD+DA\]
The area of quadrilateral is equal to the sum of area of two triangles formed by joining the diagonal. In the given figure area of quadrilateral PQRS = sum of the area of triangle PQR and PRS
Now we emphasis on areas of some special quadrilateral
Rectangle
It is a quadrilateral whose opposite sides are equal and all angles are right angle.
Perimetre of rectangle = 2(Length + breadth)
Area\[~=\text{ }Length\text{ }\times breadth\]
\[Length=\frac{breadth}{area},Breadth=\frac{lenadth}{area}\]
Diagonal \[=\sqrt{{{(length)}^{2}}+{{(breadth)}^{2}}}\]
Square
It is a quadrilateral whose sides are equal and all angles are right angle.
Perimeter of a square \[=4x\]side
Area of a square \[=sid{{e}^{2}}=\frac{1}{2}\times {{(diagonal)}^{2}}\]
\[Diagonal=\text{ }side\sqrt{2}\]
Parallelogram
It is a quadrilateral whose opposite sides are parallel and equal.
Perimeter \[=2x\]sum of length of adjacent side.
Area = base x corresponding height.
Rhombs
It is parallelogram whose all sides are equal.
Perimeter \[=4x\]side
Area = base \[x\] vertical height or \[\left( \frac{1}{2} \right)\] product of diagonals
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