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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)"`
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उत्तर
Draw AE ⊥ BC, B – E – C.
BC = BD + DC ......[B – D – C]
∴ 20 = 7 + DC
∴ DC = 20 − 7 = 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
In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find `(A(triangleABC))/(A(triangleDCB))`
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 given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:
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)"`
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))`
If ΔXYZ ~ ΔPQR then `"XY"/"PQ" = "YZ"/"QR"` = ?
In fig. BD = 8, BC = 12, B-D-C, then `"A(ΔABC)"/"A(ΔABD)"` = ?
From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?
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.
In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.
Given: PQ ⊥ BC, AD ⊥ BC
Now, A(ΔPQB) = `1/2 xx square xx square`
A(ΔPBC) = `1/2 xx square xx square`
Therefore,
`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`
= `square/square`
