Advertisements
Advertisements
Question
In ☐ABCD, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area of ☐ABCD.
Advertisements
Solution
Draw perpendicular from C to line AB. Name the point E.
CE = AD = 8 cm
EB = AB − AE = AB − CD = 13 − 9 = 4cm
∴ Area of of a trapezium = `1/2 xx "sum of lengths of parallel sides" xx h`
A (☐ABCD) = `1/2 xx [l(AB) + l(DC)] xx l(AD)`
= `1/2 xx (13 + 9) xx 8`
= `1/2 xx 22 xx 8`
= 11 x 8
= 88 sq. cm
∴ The area of ☐ABCD is 88 sq. cm.
RELATED QUESTIONS
In Fig. 8, the vertices of ΔABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that `(AD)/(AB)=(AE)/(AC)=1/3 `Calculate th area of ADE and compare it with area of ΔABCe.
If P(–5, –3), Q(–4, –6), R(2, –3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.
The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is (`7/2`, y). Find the value of y
Find the area of a triangle whose vertices are
`(at_1^2,2at_1),(at_2^2,2at_2)` and `(at_3^2,2at_3)`
Prove that the points (a, 0), (0, b) and (1, 1) are collinear if `1/a+1/b=1`
Show that the points A(-5,6), B(3,0) and C(9,8) are the vertices of an isosceles right-angled triangle. Calculate its area.
Find the value(s) of p for which the points (3p + 1, p), (p + 2, p – 5) and (p + 1, –p) are collinear ?
Using determinants, find the values of k, if the area of triangle with vertices (–2, 0), (0, 4) and (0, k) is 4 square units.
What is the area of a triangle with base 4.8 cm and height 3.6 cm?
The area of a triangle with vertices A(3, 0), B(7, 0) and C(8, 4) is ______.
