Advertisements
Advertisements
Question
What is the area of a triangle with base 4.8 cm and height 3.6 cm?
Advertisements
Solution
We have the base of a triangle = 4.8 cm
Height of a triangle = 3.6 cm
∴ Area of a triangle = `1/2` × base × height
= `1/2xx4.8xx3.6`
= `(4.8xx3.6)/2`
= `17.28/2`
= 8.64 cm2
RELATED QUESTIONS
For what value of k are the points (k, 2 – 2k), (–k + 1, 2k) and (–4 – k, 6 – 2k) are collinear ?
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
The vertices of a ΔABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that `(AD)/(AB) = (AE)/(AC) = 1/4`Calculate the area of the ΔADE and compare it with the area of ΔABC. (Recall Converse of basic proportionality theorem and Theorem 6.6 related to ratio of areas of two similar triangles)
Four points A (6, 3), B (–3, 5), C (4, –2) and D (x, 3x) are given in such a way that `(ΔDBC)/(ΔABC) = 1/2`, find x.
In ∆PQR, PR = 8 cm, QR = 4 cm and PL = 5 cm. 
Find:
(i) the area of the ∆PQR
(ii) QM.
The table given below contains some measures of the right angled triangle. Find the unknown values.
| Base | Height | Area |
| 5 feet | ? | 20 sq.feet |
Show that the ∆ABC is an isosceles triangle if the determinant
Δ = `[(1, 1, 1),(1 + cos"A", 1 + cos"B", 1 + cos"C"),(cos^2"A" + cos"A", cos^2"B" + cos"B", cos^2"C" + cos"C")]` = 0
Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).
Area of triangle PQR is 100 cm2 (see figure). If altitude QT is 10 cm, then its base PR is ______.
Find the missing value:
| Base | Height | Area of Triangle |
| 22 cm | ______ | 170.5 cm2 |
