Advertisements
Advertisements
प्रश्न
The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.
उत्तर
We have,
ΔABC ~ ΔPQR
AD = 6 cm
And, PS = 9 cm
By area of similar triangle theorem
`("Area"(triangleABC))/("Area"(trianglePQR))="AB"^2/"PQ"^2` ........(i)
In ΔABD and ΔPQS
∠B = ∠Q [ΔABC ~ ΔPQR]
∠ADB = ∠PSQ [Each 90°]
Then, ΔABD ~ ΔPQS [By AA similarity]
`therefore"AB"/"PQ"="AD"/"PS"` [Corresponding parts of similar Δ are proportional]
`rArr"AB"/"PQ"=6/9`
`rArr"AB"/"PQ"=2/3` ......(ii)
Compare equations (i) and (ii)
`("Area"(triangleABC))/("Area"(trianglePQR))=(2/3)^2=4/9`
APPEARS IN
संबंधित प्रश्न
In the given figure, DE || BC and DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that `(ar(ABC))/(ar(DBC)) = (AO)/(DO)`
Triangles ABC and DEF are similar If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of ΔABC and ΔDEF.
The areas of two similar triangles are 169 cm2 and 121 cm2 respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.
In ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ΔAPQ and trapezium BPQC.
The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5 cm, find the length of QR.
Prove that in a right-angle triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
In the given figure, D is the mid-point of BC, then the value of `(coty^circ)/(cotx^circ)` is ______.
In the adjoining figure, ΔADB ∼ ΔBDC. Prove that BD2 = AD × DC.
