Advertisements
Advertisements
Question
Two parallel side of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.
Advertisements
Solution
Given that two parallel sides of trapezium are AB = 77 and CD = 60 cm
Other sides are BC = 26 m and AD = 25 cm.
Join AE and CF
Now, DE ⊥ AB and CF ⊥ AB
∴ DC = EF = 60 cm
Let AE = x
⇒ BF = 77 – 60 – x = 17 – x
`In ΔADE, DE^2 = AD^2 – AE^2 = 25^2 – x^2` [โต Pythagoras theorem]

And in ΔBCF, `CF^2= BC^2 – BF^2` [โต By Pythagoras theorem]
`⇒25=sqrt(26^2-(17-x)^2)`
`⇒25^2-x^2=25^2-(289-x^2-34-x)` [ โต`(a-b)^2=a^2-2ab+b^2` ]
`⇒265-x^2=676-289-x^2+34x`
`34x=238`
`x=7`
`∴ DE =sqrt(25^2-x^2)=sqrt(625-7^2)=sqrt(516)=24cm`
∴ Area of trapezium = `1/2`(๐ ๐ข๐ ๐๐ ๐๐๐๐๐๐๐๐ ๐ ๐๐๐๐ )×โ๐๐๐โ๐ก=`1/2`(60×77)×24=`1644cm^2`
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?
Area of PQRS = Area of PQR + Area of ΔPQS = (6+9.166)๐๐2=15.166๐๐2
Let Δ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.
If each side of a triangle is doubled, the find percentage increase in its area.
Mark the correct alternative in each of the following:
The sides of a triangle are 16 cm, 30 cm, 34 cm. Its area is
The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is
The base and hypotenuse of a right triangle are respectively 5 cm and 13 cm long. Its area is ______.
The lengths of the sides of Δ ABC are consecutive integers. It Δ ABC has the same perimeter as an equilateral triangle with a side of length 9 cm, what is the length of the shortest side of ΔABC?
The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is ______.
The length of each side of an equilateral triangle having an area of `9sqrt(3)`cm2 is ______.
