Advertisements
Advertisements
Question
If ∆ABC ~ ∆DEF such that area of ∆ABC is 9 cm2 and the area of ∆DEF is 16 cm2 and BC = 2.1 cm. Find the length of EF.
Solution
Given ∆ABC ~ ∆DEF
∴ `("Area of" Delta"ABC")/("Area of" Delta"DEF") = "BC"^2/"EF"^2` ...(Square of their corresponding sides)
`9/16 = (2.1)^2/"EF"^2`
`(3/4)^2 = (2.1/"EF")^2`
`3/4 = 2.1/"EF"`
EF = `(4 xx 2.1)/3` = 2.8 cm
Length of EF = 2.8 cm
APPEARS IN
RELATED QUESTIONS
In figure, QA and PB are perpendicular to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm. Find AQ
Given: FB = FD, AE ⊥ FD and FC ⊥ AD.
Prove that: `(FB)/(AD) = (BC)/(ED)`.
In the given figure, DB⊥BC, DE⊥AB and AC⊥BC.
Prove that `(BE)/(DE)=(AC)/(BC)`
In Δ ABC , MN || BC .
If BC = 14 cm and MN = 6 cm , find `("Ar" triangle "AMN")/("Ar" . ("trapezium MBCN"))`
The length of a river in a map is 54cm. if lcm on the map represents 12500m on land, find the length of the river.
The scale of a map is 1 : 200000. A plot of land of area 20km2 is to be represented on the map. Find:
The area in km2 that can be represented by 1 cm2
In the given figure ABC and CEF are two triangles where BA is parallel to CE and AF: AC = 5: 8.
(i) Prove that ΔADF ∼ ΔCEF
(ii) Find AD if CE = 6 cm
(iii) If DF is parallel to BC find area of ΔADF: area of ΔABC.
A girl looks the reflection of the top of the lamp post on the mirror which is 6.6 m away from the foot of the lamppost. The girl whose height is 1.25 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in the same line, find the height of the lamp post.
Construct a triangle similar to a given triangle PQR with its sides equal to `2/3` of the corresponding sides of the triangle PQR (scale factor `2/3 < 1`)
In the given figure the value of x is
