Advertisements
Advertisements
प्रश्न
Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:
- Draw two triangles, give the names of all points, and show heights.
- Write 'Given' and 'To prove' from the figure drawn.
Advertisements
उत्तर
i. The triangles are as follows:
ii. Given: AP = DQ
To prove: `(A(ΔABC))/(A(ΔDEF)) = (BC)/(EF)`
APPEARS IN
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))`
In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).
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 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 ∆ABC, B – D – C and BD = 7, BC = 20, then find the following ratio.
`(A(triangleABD))/(A(triangleABC))`
In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio.
\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\]
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))`.
A roller of diameter 0.9 m and the length 1.8 m is used to press the ground. Find the area of the ground pressed by it in 500 revolutions.
`(pi=3.14)`
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 [______]
`("A"("PQS"))/("A"("QRS")) = (["______"])/"NR"`,
`100/110 = (["______"])/"NR"`,
NR = [ ______ ] cm
In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?
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)"`
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`
