Advertisements
Advertisements
प्रश्न
Vertices of given triangles are taken in order and their areas are provided aside. Find the value of ‘p’.
| Vertices | Area (sq.units) |
| (0, 0), (p, 8), (6, 2) | 20 |
Advertisements
उत्तर
Let the vertices be A(0, 0) B(p, 8), c(6, 2)
Area of a triangle = 20 sq.units
`1/2[(x_1y_2 + x_2y_3 + x_3y_1) - (x_2y_1 + x_3y_2 + x_1y_3)]` = 20
`1/2[(0 + 2"p" + 0) - (0 + 48 + 0)]` = 20
`1/2[2"p" - 48]` = 20
2p – 48 = 40
⇒ 2p = 40 + 48
p = `88/2` = 44
The value of p = 44
APPEARS IN
संबंधित प्रश्न
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
A rectangular plot of land measures 45 m x 30 m. A boundary wall of height 2.4 m is built all around the plot at a distance of 1 m from the plot. Find the area of the inner surface of the boundary wall.
The length of a rectangle is twice the side of a square and its width is 6 cm greater than the side of the square. If the area of the rectangle is three times the area of the square; find the dimensions of each.
Calculate the area of quadrilateral ABCD, in which ∠ABD = 90°, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.
Trapezium given below; find its area.
Trapezium given below; find its area.
The figure given below shows the cross-section of a concrete structure. Calculate the area of cross-section if AB = 1.8 cm, CD = 0.6 m, DE = 0.8 m, EF = 0.3 m and AF = 1.2 m.
Two adjacent sides of a parallelogram are 24 cm and 18 cm. If the distance between the longer sides is 12 cm; find the distance between the shorter sides.
A quadrilateral field of unequal has a longer diagonal with 140m. The perpendiculars from opposite vertives upon this diagonal are 20m and 14m. Find the area of the field.
When proving that a quadrilateral is a trapezium, it is necessary to show
