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प्रश्न
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.
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उत्तर
Let the height of the larger triangle be h1 and that of the smaller triangle be h2.
The ratio of the areas of two triangles with a common base is equal to the ratio of their corresponding heights.
`"Area(larger Triangle)"/"Area(smaller Triangle)"="h"_1/"h"_2`
`4/3=2/"h"_2`
`4xx "h"_2=3xx2`
`therefore"h"_2=(3xx2)/4=6/4`
`therefore"h"_2=1.5" cm"`
The corresponding height of the smaller triangle is 1.5 cm.
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संबंधित प्रश्न
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In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.
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`(A(triangleABD))/(A(triangleABC))`
In the given, seg BE ⊥ seg AB and seg BA ⊥ seg AD.
if BE = 6 and AD = 9 find `(A(Δ ABE))/(A(Δ BAD))`.
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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.
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In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?
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In ΔABC, B − D − C and BD = 7, BC = 20, then find the following ratio.
(i) `"A(ΔABD)"/"A(ΔADC)"`
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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`
