Advertisements
Advertisements
Question
In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`

Advertisements
Solution
In ∆ABC and ∆ADB,
BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8 ...(Given)
∆ABC and ∆ADB have same base AB. ...(Given)
∴ Areas of triangles with equal bases are proportional to their corresponding heights.
`("A"(∆"ABC"))/("A"(∆"ADB")) = "BC"/"AD"`
∴ `("A"(∆"ABC"))/("A"(∆"ADB")) = 4/8`
∴ `("A"(∆"ABC"))/("A"(∆"ADB")) = 1/2`.
RELATED QUESTIONS
In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ

Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?
The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.
In ∆ABC, B – D – C and BD = 7, BC = 20, then find the following ratio.

`(A(triangleABD))/(A(triangleABC))`
In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio.

\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\]
A roller of diameter 0.9 m and the length 1.8 m is used to press the ground. Find the area of the ground pressed by it in 500 revolutions.
`(pi=3.14)`
Areas of two similar triangles are in the ratio 144 : 49. Find the ratio of their corresponding sides.
In fig., TP = 10 cm, PS = 6 cm. `(A(ΔRTP))/(A(ΔRPS))` = ?
Ratio of corresponding sides of two similar triangles is 4 : 7, then find the ratio of their areas = ?
In fig. BD = 8, BC = 12, B-D-C, then `(A(ΔABC))/(A(ΔABD))` = ?

In fig., PM = 10 cm, A(ΔPQS) = 100 sq. cm, A(ΔQRS) = 110 sq. cm, then NR?

ΔPQS and ΔQRS having seg QS common base.
Areas of two triangles whose base is common are in proportion of their corresponding `square`.
`(A(ΔPQS))/(A(ΔQRS)) = (square)/(NR)`,
`100/110 = (square)/(NR)`,
NR = `square` cm
In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `(A(ΔABC))/(A(ΔBCD))` = ?

From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `(A(ΔABC))/(A(ΔBCD))` = ?

Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:
- Draw two triangles, give the names of all points, and show heights.
- Write 'Given' and 'To prove' from the figure drawn.
If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.
