Advertisements
Advertisements
प्रश्न
Ratio of areas of ∆MNO, ∆MOP and ∆MPQ in the following figure is ______.
विकल्प
2 : 1 : 3
1 : 3 : 2
2 : 3 : 1
1 : 2 : 3
Advertisements
उत्तर
Ratio of areas of ∆MNO, ∆MOP and ∆MPQ in the following figure is 2 : 1 : 3.
Explanation:
We know that, area of triangle = `1/2` × base × height
So, area of triangle MNO = `1/2` × NO × MO
= `1/2` × 5 × 4
= `1/2` × 20
= 10 cm2
Area of triangle MOP = `1/2` × MO × OP
= `1/2` × 5 × 2
= `1/2` × 10
= 5 cm2
Area of triangle MPQ = `1/2` × MO × PQ ...[MP = MO]
= `1/2` × 5 × 6
= `1/2` × 30
= 15 cm2
So, the ratio of area = 10 : 5 : 15 ...[Divide each by 5]
Then we get, 2 : 1 : 3
APPEARS IN
संबंधित प्रश्न
In Fig. 6, ABC is a triangle coordinates of whose vertex A are (0, −1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the mid-point of BC, find the areas of ∆ABC and ∆DEF.
Find the area of a triangle whose vertices are A(3, 2), B (11, 8) and C(8, 12).
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)
Prove that the points A(7, 10), B(–2, 5) and C(3, –4) are the vertices of an isosceles right triangle.
Find the centroid of ΔABC whose vertices are A(–1, 0), B(5, –2) and C(8, 2).
A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ΔADE.
Show that the following points are collinear:
A(2, –2), B(–3, 8) and C(–1, 4)
Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear ?
Find the area of the following triangle:
If `D((-1)/2, 5/2), E(7, 3)` and `F(7/2, 7/2)` are the midpoints of sides of ∆ABC, find the area of the ∆ABC.
