Advertisements
Advertisements
Question
A park, in the shape of a quadrilateral ABCD, has ∠C = 900, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m How much area does it occupy?
Advertisements
Solution
Given sides of a quadrilaterals are AB = 9, BC = 12, CD = 05, DA = 08
Let us joint BD
In ΔBCD applying Pythagoras theorem.
`BD^2=BC^2+CD^2`
=`(12)^2+(5)^2`
=144+25
=169
𝐵𝐷=13𝑚
Area of ΔBCD = `1/2`×𝐵𝐶×𝐶𝐷=[`1/2`×12×5]`m^2=30m^2`
or ΔABD
`S=sqrt(perimeter)/2=sqrt(9+8+13)/2=15cm`
By heron’s formula`sqrt(s(s-a)(s-b)(s-c))`
Area of the triangle =`sqrt(15(15-9)(15-8)(15-13))m^2`
=`35.496 + 30 m^2`
= `65.5 m^2` (approximately)
APPEARS IN
RELATED QUESTIONS
A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Radha made a picture of an aeroplane with coloured papers as shown in the given figure. Find the total area of the paper used.
If each side of a triangle is doubled, the find percentage increase in its area.
The sides of a triangle are 7 cm, 9 cm and 14 cm. Its area is
If the area of an isosceles right triangle is 8 cm2, what is the perimeter of the triangle?
The area of an equilateral triangle with side `2sqrt(3)` cm is ______.
If the area of an equilateral triangle is `16sqrt(3)` cm2, then the perimeter of the triangle is ______.
The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm2 is ______.
The area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a.
In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude is 10.25 cm corresponding to the side having length 12 cm.
