Advertisements
Advertisements
प्रश्न
In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio.

\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\]
Advertisements
उत्तर
Construction: Draw a perpendicular from vertex A to line BC.

\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)} = \frac{\frac{1}{2} \times AX \times DC}{\frac{1}{2} \times AX \times BC}\]
\[ = \frac{DC}{BC}\]
\[ = \frac{13}{20} \left( \because DC = BC - BD \right)\]
APPEARS IN
संबंधित प्रश्न
The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.
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

In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.

In ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio.

`"A(∆ ABD)"/"A(∆ ADC)"`
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))`
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))` = ?

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.
