Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Given: ΔABC ∼ ΔDEF
Also, AB = 9 cm, DE = 12 cm.
We now know the area theorem of comparable triangles.
`(ar(ΔABC))/(ar(ΔDEF)) = ((AB)/(DE))^2` ...(i)
Substituting the values in equation (i),
`(ar(ΔABC))/(ar(ΔDEF)) = (9/12)^2`
= `(3/4)^2`
= `9/16`
Hence, the ratio of areas of ΔABC and ΔDEF is 9:16.
APPEARS IN
संबंधित प्रश्न
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"))`
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))`
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)`
If ΔXYZ ~ ΔPQR then `"XY"/"PQ" = "YZ"/"QR"` = ?
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., 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`
