Advertisements
Advertisements
प्रश्न
In ΔABC, B-D-C and BD = 7, BC = 20, then find the following ratio.
(i) `(A(ΔABD))/(A(ΔADC))`
(ii) `(A(ΔABD))/(A(ΔABC))`
(iii) `(A(ΔADC))/(A(ΔABC))`
Advertisements
उत्तर
Draw AE ⊥ BC, B-E-C.
BC = BD + DC ...[B-D-C]
∴ 20 = 7 + DC
∴ DC = 20 – 7
∴ DC = 13
(i) ΔABD and ΔADC have same height AE.
`(A(ΔABD))/(A(ΔADC)) = (BD)/(DC)` ...[Triangles having equal height]
∴ `(A(ΔABD))/(A(ΔADC)) = 7/13`
(ii) ΔABD and ΔABC have same height AE.
`(A(ΔABD))/(A(ΔABC)) = (BD)/(BC)` ...[Triangles having equal height]
∴ `(A(ΔABD))/(A(ΔABC)) = 7/20`
(iii) ΔADC and ΔABC have same height AE.
`(A(ΔADC))/(A(ΔABC)) = (DC)/(BC)` ...[Triangles having equal height]
∴ `(A(ΔADC))/(A(ΔABC)) = 13/20`
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 the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`
In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios.
- `("A"(∆"PQB"))/("A"(∆"PBC"))`
- `("A"(∆"PBC"))/("A"(∆"ABC"))`
- `("A"(∆"ABC"))/("A"(∆"ADC"))`
- `("A"(∆"ADC"))/("A"(∆"PQC"))`
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)"`
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.
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
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.
