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प्रश्न
In the following figure RP: PK= 3:2, then find the value of A(ΔTRP):A(ΔTPK).
उत्तर
Ratio of the areas of two triangles with common or equal heights is equal to the ratio of their corresponding bases.
`(A(triangleTRP))/(A(triangleTPK))="RP"/"PK"=3/2`
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.
The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.
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 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 the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then `("A(Δ ABC)")/("A(Δ DCB)")` = ?
In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.
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 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., TP = 10 cm, PS = 6 cm. `"A(ΔRTP)"/"A(ΔRPS)"` = ?
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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 [______]
`("A"("PQS"))/("A"("QRS")) = (["______"])/"NR"`,
`100/110 = (["______"])/"NR"`,
NR = [ ______ ] cm
From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?
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)"`
