9 questions · timed · auto-graded
Solution:
$\text{Area of }\triangle=\frac{1}{2}\times\text{base}\times\text{height}$ $=\frac{1}{2}\times4\times4=8\text{cm}^2$
Hence, the given statement is true.
Solution:
$\text{Area of equilateral }\triangle=\frac{\sqrt{3}}{4}(\text{side})^2$ $=\frac{\sqrt{3}}{4}(8)^2=\frac{\sqrt{3}}{4}\times64=16\sqrt{3}\text{cm}^2$ Hence, the given statement is false.Solution:
We know that, area of a triangle $=\frac{1}{2}(\text{Base}\times\text{Height})$ Here Base = 4cm and Height = 6cm $\therefore$ Area of a triangle $=\frac{1}{2}\times4\times6=12\text{cm}^2$Solution:
Let equal of an isosceles triangle be b. $\therefore$ Perimeter of a triangle, $2\text{s}=\text{b}+\text{b}+5$ $\big[\because\ 2\text{s}=\text{a}+\text{b}+\text{c}\big]$ $\therefore\ 11=2\text{b}+5$ $\Rightarrow2\text{b}=11-5$ $\Rightarrow2\text{b}=6$ $\Rightarrow\text{b}=\frac{6}{2}=3\text{cm}$ We know that, area of an isosceles triangle, $=\frac{\text{a}}{4}\sqrt{4\text{b}^2-\text{a}^2}$ Here, sides of triangle are a = 5cm and b = 3cm $\therefore$ Area of an isosceles triangle $=\frac{5\sqrt{4(3)^2-(5)^2}}{4}$ $=\frac{5\sqrt{4\times9-25}}{4}$ $=5\frac{\sqrt{36-25}}{4}$ $=\frac{5\sqrt{11}}{4}\text{cm}^2$Solution:
The base of the parallelogram is 10cm and the corresponding altitude is 3.5cm. Area of || gm = base × corresponding altitude = 10cm × 3.5cm = 35cm2 Hence, the given statement is false.In a triangle, the sides are given as 11cm, 12cm and 13cm. The length of the altitude is 10.25cm corresponding to the side having length 12cm.
The area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a.
Solution:
We see a regular hexagon is divided into six equilateral triangles. So, the area of the regular hexagon is divided side ‘a’ is the sum of area of the six equilateral triangles with side a.
Hence, the given statement that the area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a, is false.The cost of levelling the ground in the form of a triangle having the sides 51m, 37m and 20m at the rate of Rs. 3per m2 is Rs. 918.
Solution:
Let sides of a triangle be a = 51m, b = 37m and c = 20m Now, semi-perimeter of triangle, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{51+37+20}{2}$ $=\frac{108}{2}=54\text{m}$ $\therefore$ Area of a triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula] $=\sqrt{54(54-51)(54-37)(54-20)}$ $=\sqrt{54\times3\times17\times34}$ $=\sqrt{9\times3\times2\times3\times17\times17\times2}$ $=3\times3\times2\times17=306\text{m}^2$ $\because$ Cost of levelling per m2 = Rs. 3 $\therefore$ Cost of leavelling per 306 m2 = 3 × 306 = Rs. 918