In Fig. 6.11, if DE||BC, find the ratio of ar (△ADE) and ar (△DECB)
Solution:
Given, DE||BC
We have to find the ratio of ar (△ADE) and ar (△DECB).
From the figure,
DE = 6 cm
BC = 12 cm
In △ADE and △ABC,
∠A = ∠A = common angle
By the property of similar triangles,
The corresponding angles are equal.
∠ABC = ∠ADE
∠BCA = ∠AED
AAA criterion states that if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angle will also be equal.
By AAA criterion, the triangles ABC and ADE are similar.
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Area of △ADE/Area of △ABC = DE²/BC²
= (6)²/(12)²
= 1/4
Let the area of △ADE = a
So, the area of △ABC = 4a
Area of quadrilateral DECB = area of △ABC - area of △ADE
So, area of quadrilateral DECB = 4a - a = 3a
Now, Area of △ADE/Area of quadrilateral DECB = a/3a
= 1/3
Therefore, the ratio of the area of △ADEto the area of quadrilateral DECB is 1:3
✦ Try This: If DE||BC, find the ratio of ar (△ADE) and ar (△DECB) when DE = 6 cm and BC = 8 cm
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6
NCERT Exemplar Class 10 Maths Exercise 6.3 Problem 8
In Fig. 6.11, if DE||BC, find the ratio of ar (△ADE) and ar (△DECB)
Summary:
In Fig. 6.11, if DE||BC, the ratio of ar (△ADE) and ar (△DECB) is 1:3
☛ Related Questions:
- ABCD is a trapezium in which AB||DC and P and Q are points on AD and BC, respectively such that PQ|| . . . .
- Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller t . . . .
- In a triangle PQR, N is a point on PR such that QN ⊥ PR . If PN. NR = QN², prove that ∠PQR = 90°
visual curriculum