Advertisements
Advertisements
प्रश्न
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)
उत्तर
We have,
ΔACB ~ ΔAPQ
Then,
`"AC"/"AP"="CB"/"PQ"="AB"/"AQ"` [Corresponding parts of similar Δ are proportional]
`rArr"AC"/2.8=10/5=6.5/"AQ"`
`rArr"AC"/2.8=10/5` and `10/5=6.5/"AQ"`
`rArr"AC"=10/5xx2.8` and `"AQ"=6.5xx5/10`
⇒ AC = 5.6 cm and AQ = 3.25 cm
By area of similar triangle theorem
`("Area"(triangleACB))/("Area"(triangleAPQ))/"BC"^2/"PQ"^2`
`=10^2/5^2`
`=100/25`
= 4
APPEARS IN
संबंधित प्रश्न
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.
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.
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.
In Figure, DE || BC If DE = 4 cm, BC = 6 cm and Area (ΔADE) = 16 cm2, find the area of ΔABC.
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.
If D is a point on the side AB of ΔABC such that AD : DB = 3.2 and E is a Point on BC such that DE || AC. Find the ratio of areas of ΔABC and ΔBDE.
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.
The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.
In the given figure, D is the mid-point of BC, then the value of `(coty^circ)/(cotx^circ)` is ______.
