Advertisements
Advertisements
प्रश्न
ABC is a triangle in which ∠A =90°, AN⊥ BC, BC = 12 cm and AC = 5cm. Find the ratio of the areas of ΔANC and ΔABC.
उत्तर
In ΔANC and ΔABC
∠C = ∠C [Common]
∠ANC = ∠BAC [Each 90°]
Then, ΔANC ~ ΔBAC [By AA similarity]
By area of similarity triangle theorem
`("Area"(triangleANC))/("Area"(triangleBAC))="AC"^2/"BC"^2`
`=5^2/12^2`
`=25/144`
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
D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.
In figure below ΔACB ~ ΔAPQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm,
find CA and AQ. Also, find the area (ΔACB): area (ΔAPQ)
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.
The areas of two similar triangles are 25 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.
AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (ΔADE): Area (ΔABC) = 3: 4
∆ABC and ∆DEF are equilateral triangles. If A(∆ABC) : A(∆DEF) = 1 : 2 and AB = 4, find DE.
In the given figure 1.66, seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2 DP, then Find A(◻DPQE) by completing the following activity.
If ΔABC is similar to ΔDEF such that 2 AB = DE and BC = 8 cm then EF is equal to ______.
In ΔABC, seg DE || side BC. If 2A(ΔADE) = A(⬜ DBCE), find AB : AD and show that BC = `sqrt(3)` DE.