Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर १
Let ABC and PQR be two right triangles with AD ⊥ BC and PS ⊥ QR.
BC = 9, AD = 5, PS = 6 and QR = 10.
The ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.
`("A"(Δ"ABC"))/("A"(Δ"PQR")) = ("AD" × "BC")/("PS" × "QR")`
∴ `("A"(Δ"ABC"))/("A"(Δ"PQR")) = (5 × 9)/(6 × 10)`
∴ `("A"(Δ"ABC"))/("A"(Δ"PQR")) = 3/4`
Hence, Ratio of Area of △ABC : Area of △PQR = 3 : 4.
उत्तर २
Let the base, height, and area of the first triangle be b1, h1, and A1 respectively. Let the base, height and area of the second triangle be b2, h2, and A2 respectively.
b1 = 9, h1 = 5, b2 = 10 and h2 = 6.
The ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.
`("A"_1)/("A"_2) = ("b"_1 × "h"_1)/("b"_2 × "h"_2)`
∴ `("A"_1)/("A"_2) = (9 × 5)/(10 × 6)`
∴ `("A"_1)/("A"_2) = 3/4`
The ratio of the areas of the triangles is 3 : 4.
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).
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 the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:
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"))`
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 ?
In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then `(A(Δ ABC))/(A(Δ DCB))` = ?
In ∆ABC, B – D – C and BD = 7, BC = 20, then find the following ratio.
`(A(triangleABD))/(A(triangleABC))`
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))` = ?
From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `(A(ΔABC))/(A(ΔBCD))` = ?
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`
