Advertisements
Advertisements
प्रश्न
Find the missing value:
| Base | Height | Area of parallelogram |
| ______ | 15 cm | 154.5 cm2 |
Advertisements
उत्तर
| Base | Height | Area of parallelogram |
| 10.3 | 15 cm | 154.5 cm2 |
Explanation:
b = ?
h = 15 cm
Area = 154.5 cm2
b × 15 = 154.5
b = `154.5/15`
b = 10.3 cm
Therefore, the base of such parallelogram is 10.3 cm.
APPEARS IN
संबंधित प्रश्न
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.

Find the area of the following triangle:

For what value of a the point (a, 1), (1, –1) and (11, 4) are collinear?
In a ΔABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ΔABC and hence its altitude on AC ?
Show that the points A(3, 0), B(6, 4) and C(–1, 3) are the vertices of an isosceles right triangle.
Find the value of y for which the points A(–3, 9), B(2, y) and C(4, –5) are collinear.
Find a relation between x and y, if the points A(x, y), B(–5, 7) and C(–4, 5) are collinear.
The table given below contains some measures of the right angled triangle. Find the unknown values.
| Base | Height | Area |
| ? | 12 m | 24 sq.m |
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
The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC.
